The ratio of natural logarithms. How to solve logarithms - step by step instructions for solving
Logarithms, like any numbers, can be added, subtracted and transformed in every way. But since logarithms are not exactly ordinary numbers, there are rules here, which are called basic properties.
It is imperative to know these rules - no serious logarithmic problem can be solved without them. In addition, there are very few of them - everything can be learned in one day. So let's get started.
Addition and subtraction of logarithms
Consider two logarithms with the same base: log a x and log a y... Then they can be added and subtracted, and:
- log a x+ log a y= log a (x · y);
- log a x- log a y= log a (x : y).
So, the sum of the logarithms is equal to the logarithm of the product, and the difference is the logarithm of the quotient. Please note, the key point here is - identical grounds... If the reasons are different, these rules do not work!
These formulas will help you calculate a logarithmic expression even when its individual parts are not counted (see the lesson "What is a logarithm"). Take a look at the examples - and see:
Log 6 4 + log 6 9.
Since the bases of the logarithms are the same, we use the sum formula:
log 6 4 + log 6 9 = log 6 (4 9) = log 6 36 = 2.
Task. Find the value of the expression: log 2 48 - log 2 3.
The bases are the same, we use the difference formula:
log 2 48 - log 2 3 = log 2 (48: 3) = log 2 16 = 4.
Task. Find the value of the expression: log 3 135 - log 3 5.
Again the bases are the same, so we have:
log 3 135 - log 3 5 = log 3 (135: 5) = log 3 27 = 3.
As you can see, the original expressions are composed of "bad" logarithms, which are not separately counted. But after transformations, quite normal numbers are obtained. Many are built on this fact. test papers... But what control - such expressions in all seriousness (sometimes - practically unchanged) are offered on the exam.
Removing the exponent from the logarithm
Now let's complicate the task a little. What if the base or argument of the logarithm is based on a degree? Then the exponent of this degree can be taken out of the sign of the logarithm according to the following rules:
It's easy to see that the last rule follows the first two. But it is better to remember it all the same - in some cases it will significantly reduce the amount of computation.
Of course, all these rules make sense when observing the ODV of the logarithm: a > 0, a ≠ 1, x> 0. And one more thing: learn to apply all formulas not only from left to right, but also vice versa, ie. you can enter the numbers in front of the sign of the logarithm into the logarithm itself. This is what is most often required.
Task. Find the value of the expression: log 7 49 6.
Let's get rid of the degree in the argument using the first formula:
log 7 49 6 = 6 log 7 49 = 6 2 = 12
Task. Find the meaning of the expression:
[Figure caption]
Note that the denominator contains the logarithm, the base and argument of which are exact powers: 16 = 2 4; 49 = 7 2. We have:
[Figure caption] I think the last example needs some clarification. Where did the logarithms disappear? Until the very last moment, we work only with the denominator. We presented the base and the argument of the logarithm standing there in the form of degrees and brought out the indicators - we got a "three-story" fraction.
Now let's look at the basic fraction. The numerator and denominator contain the same number: log 2 7. Since log 2 7 ≠ 0, we can cancel the fraction - the denominator remains 2/4. According to the rules of arithmetic, the four can be transferred to the numerator, which was done. The result was the answer: 2.
Moving to a new foundation
Speaking about the rules of addition and subtraction of logarithms, I specifically emphasized that they only work for the same bases. What if the reasons are different? What if they are not exact powers of the same number?
Formulas for the transition to a new foundation come to the rescue. Let us formulate them in the form of a theorem:
Let the logarithm log a x... Then for any number c such that c> 0 and c≠ 1, the equality is true:
[Figure caption]
In particular, if we put c = x, we get:
[Figure caption]
From the second formula it follows that it is possible to swap the base and the argument of the logarithm, but in this case the whole expression is "reversed", i.e. the logarithm appears in the denominator.
These formulas are rarely found in common numerical expressions. It is possible to assess how convenient they are only when deciding logarithmic equations and inequalities.
However, there are tasks that are generally not solved except by the transition to a new foundation. Consider a couple of these:
Task. Find the value of the expression: log 5 16 log 2 25.
Note that the arguments of both logarithms contain exact degrees. Let's take out the indicators: log 5 16 = log 5 2 4 = 4log 5 2; log 2 25 = log 2 5 2 = 2 log 2 5;
Now let's "flip" the second logarithm:
[Figure caption]Since the product does not change from the permutation of the factors, we calmly multiplied the four and the two, and then dealt with the logarithms.
Task. Find the value of the expression: log 9 100 · lg 3.
The base and argument of the first logarithm are exact degrees. Let's write this down and get rid of the metrics:
[Figure caption]Now let's get rid of decimal logarithm by going to a new base:
[Figure caption]Basic logarithmic identity
Often in the process of solving it is required to represent a number as a logarithm to a given base. In this case, the formulas will help us:
In the first case, the number n becomes an indicator of the degree standing in the argument. Number n can be absolutely anything, because it is just the value of the logarithm.
The second formula is actually a paraphrased definition. It is called that: basic logarithmic identity.
Indeed, what happens if the number b to such a power that the number b to this degree gives the number a? That's right: you get this very number a... Read this paragraph carefully again - many people "hang" on it.
Like the formulas for transition to a new base, the basic logarithmic identity is sometimes the only possible solution.
Task. Find the meaning of the expression:
[Figure caption]
Note that log 25 64 = log 5 8 - just moved the square out of the base and the logarithm argument. Taking into account the rules for multiplying degrees with the same base, we get:
[Figure caption]If someone is not in the know, it was a real problem from the exam :)
Logarithmic unit and logarithmic zero
In conclusion, I will give two identities that can hardly be called properties - rather, they are consequences of the definition of the logarithm. They are constantly encountered in problems and, surprisingly, create problems even for "advanced" students.
- log a a= 1 is the logarithmic unit. Remember once and for all: logarithm to any base a from this very base is equal to one.
- log a 1 = 0 is logarithmic zero. Base a can be anything, but if the argument contains one - the logarithm is zero! because a 0 = 1 is a direct consequence of the definition.
That's all the properties. Be sure to practice putting them into practice! Download the cheat sheet at the beginning of the lesson, print it out, and solve the problems.
As you know, when multiplying expressions with powers, their exponents always add up (a b * a c = a b + c). This mathematical law was deduced by Archimedes, and later, in the 8th century, the mathematician Virasen created a table of whole indicators. It was they who served for the further discovery of logarithms. Examples of using this function can be found almost everywhere where you need to simplify a cumbersome multiplication by simple addition. If you spend 10 minutes reading this article, we will explain to you what logarithms are and how to work with them. Simple and accessible language.
Definition in mathematics
The logarithm is an expression of the following form: log ab = c, that is, the logarithm of any non-negative number (that is, any positive) "b" based on its base "a" is the power "c", to which the base "a" must be raised, in order to end up get the value "b". Let's analyze the logarithm using examples, for example, there is an expression log 2 8. How to find the answer? It's very simple, you need to find such a degree so that from 2 to the desired degree you get 8. After doing some calculations in your mind, we get the number 3! And rightly so, because 2 to the power of 3 gives the number 8 in the answer.
Varieties of logarithms
For many pupils and students, this topic seems complicated and incomprehensible, but in fact, logarithms are not so scary, the main thing is to understand their general meaning and remember their properties and some rules. There are three separate species logarithmic expressions:
- Natural logarithm ln a, where the base is Euler's number (e = 2.7).
- Decimal a, base 10.
- Logarithm of any number b to base a> 1.
Each of them is solved in a standard way, including simplification, reduction and subsequent reduction to one logarithm using logarithmic theorems. To obtain the correct values of the logarithms, you should remember their properties and the sequence of actions when solving them.
Rules and some restrictions
In mathematics, there are several rules-restrictions that are accepted as an axiom, that is, they are not negotiable and are true. For example, numbers cannot be divided by zero, and it is also impossible to extract an even root from negative numbers... Logarithms also have their own rules, following which you can easily learn to work even with long and capacious logarithmic expressions:
- the base "a" must always be greater than zero, and at the same time not equal to 1, otherwise the expression will lose its meaning, because "1" and "0" in any degree are always equal to their values;
- if a> 0, then a b> 0, it turns out that "c" must also be greater than zero.
How do you solve logarithms?
For example, given the task to find the answer to the equation 10 x = 100. It is very easy, you need to choose such a power, raising the number ten to which we get 100. This, of course, 10 2 = 100.
Now let's imagine given expression in the form of a logarithmic. We get log 10 100 = 2. When solving logarithms, all actions practically converge to find the power to which it is necessary to introduce the base of the logarithm in order to get the given number.
To accurately determine the value of an unknown degree, it is necessary to learn how to work with the table of degrees. It looks like this:
As you can see, some exponents can be guessed intuitively if you have a technical mindset and knowledge of the multiplication table. However, for large values a table of degrees is required. It can be used even by those who know nothing at all about complex mathematical topics. The left column contains numbers (base a), the top row of numbers is the power c to which the number a is raised. At the intersection in the cells, the values of the numbers are defined, which are the answer (a c = b). Take, for example, the very first cell with the number 10 and square it, we get the value 100, which is indicated at the intersection of our two cells. Everything is so simple and easy that even the most real humanist will understand!
Equations and inequalities
It turns out that under certain conditions the exponent is the logarithm. Therefore, any mathematical numerical expression can be written as a logarithmic equality. For example, 3 4 = 81 can be written as the logarithm of 81 to base 3, equal to four (log 3 81 = 4). For negative powers, the rules are the same: 2 -5 = 1/32, we write it as a logarithm, we get log 2 (1/32) = -5. One of the most fascinating areas of mathematics is the topic of "logarithms". We will consider examples and solutions of equations a little below, immediately after studying their properties. Now let's look at what inequalities look like and how to distinguish them from equations.
An expression of the following form is given: log 2 (x-1)> 3 - it is logarithmic inequality, since the unknown value "x" is under the sign of the logarithm. And also in the expression, two values are compared: the logarithm of the required number to base two is greater than the number three.
The most important difference between logarithmic equations and inequalities is that equations with logarithms (for example, logarithm 2 x = √9) imply one or more specific numerical values in the answer, while solving the inequality determines both the range of admissible values and the points breaking this function. As a consequence, the answer is not a simple set of separate numbers, as in the answer to the equation, but a continuous series or set of numbers.
Basic theorems on logarithms
When solving primitive tasks to find the values of the logarithm, its properties may not be known. However, when it comes to logarithmic equations or inequalities, first of all, it is necessary to clearly understand and apply in practice all the basic properties of logarithms. We will get acquainted with examples of equations later, let's first analyze each property in more detail.
- The main identity looks like this: a logaB = B. It only applies if a is greater than 0, not equal to one, and B is greater than zero.
- The logarithm of the product can be represented in the following formula: log d (s 1 * s 2) = log d s 1 + log d s 2. In this case, a prerequisite is: d, s 1 and s 2> 0; a ≠ 1. You can give a proof for this formula of logarithms, with examples and a solution. Let log as 1 = f 1 and log as 2 = f 2, then a f1 = s 1, a f2 = s 2. We obtain that s 1 * s 2 = a f1 * a f2 = a f1 + f2 (properties of powers ), and further by definition: log a (s 1 * s 2) = f 1 + f 2 = log a s1 + log as 2, which is what was required to prove.
- The logarithm of the quotient looks like this: log a (s 1 / s 2) = log a s 1 - log a s 2.
- The theorem in the form of a formula takes the following form: log a q b n = n / q log a b.
This formula is called the "property of the degree of the logarithm". It resembles the properties of ordinary degrees, and it is not surprising, because all mathematics is based on natural postulates. Let's take a look at the proof.
Let log a b = t, it turns out a t = b. If we raise both parts to the power of m: a tn = b n;
but since a tn = (a q) nt / q = b n, hence log a q b n = (n * t) / t, then log a q b n = n / q log a b. The theorem is proved.
Examples of problems and inequalities
The most common types of logarithm problems are examples of equations and inequalities. They are found in almost all problem books, and are also included in the compulsory part of exams in mathematics. To enter the university or pass the entrance examinations in mathematics, you need to know how to correctly solve such tasks.
Unfortunately, there is no single plan or scheme for solving and determining the unknown value of the logarithm, however, certain rules can be applied to each mathematical inequality or logarithmic equation. First of all, it is necessary to find out whether it is possible to simplify the expression or reduce to general view... You can simplify long logarithmic expressions if you use their properties correctly. Let's get to know them soon.
When solving logarithmic equations, it is necessary to determine what kind of logarithm is in front of us: an example of an expression can contain a natural logarithm or decimal.
Here are examples ln100, ln1026. Their solution boils down to the fact that you need to determine the degree to which the base 10 will be equal to 100 and 1026, respectively. For solutions natural logarithms it is necessary to apply logarithmic identities or their properties. Let's look at the examples of solving logarithmic problems of different types.
How to use logarithm formulas: with examples and solutions
So, let's look at examples of using the main theorems on logarithms.
- The property of the logarithm of the product can be used in tasks where it is necessary to decompose a large value of the number b into simpler factors. For example, log 2 4 + log 2 128 = log 2 (4 * 128) = log 2 512. The answer is 9.
- log 4 8 = log 2 2 2 3 = 3/2 log 2 2 = 1.5 - as you can see, applying the fourth property of the power of the logarithm, it was possible to solve a seemingly complex and unsolvable expression. You just need to factor the base into factors and then take the power values out of the sign of the logarithm.
Tasks from the exam
Logarithms are often found in entrance exams, especially a lot of logarithmic problems in the Unified State Exam (state exam for all school graduates). Usually, these tasks are present not only in part A (the easiest test part of the exam), but also in part C (the most difficult and voluminous tasks). The exam assumes exact and perfect knowledge of the topic "Natural logarithms".
Examples and solutions to problems are taken from the official versions of the Unified State Exam. Let's see how such tasks are solved.
Given log 2 (2x-1) = 4. Solution:
rewrite the expression, simplifying it a little log 2 (2x-1) = 2 2, by the definition of the logarithm we get that 2x-1 = 2 4, therefore 2x = 17; x = 8.5.
- It is best to convert all logarithms to one base so that the solution is not cumbersome and confusing.
- All expressions under the sign of the logarithm are indicated as positive, therefore, when the exponent of the expression, which is under the sign of the logarithm and as its base, is taken out by a factor, the expression remaining under the logarithm must be positive.
As society developed and production became more complex, mathematics also developed. Moving from simple to complex. From the usual accounting by the method of addition and subtraction, with their repeated repetition, we came to the concept of multiplication and division. Reducing the repetitive operation of multiplication has become the concept of exponentiation. The first tables of the dependence of numbers on the base and the number of raising to a power were compiled back in the 8th century by the Indian mathematician Varasen. From them, you can count the time of occurrence of logarithms.
Historical sketch
The revival of Europe in the 16th century also stimulated the development of mechanics. T a large amount of computation was required related to multiplication and division of multidigit numbers. Ancient tables did a great service. They made it possible to replace complex operations with simpler ones - addition and subtraction. A big step forward was the work of the mathematician Michael Stiefel, published in 1544, in which he realized the idea of many mathematicians. This made it possible to use tables not only for degrees in the form of primes, but also for arbitrary rational ones.
In 1614, the Scotsman John Napier, developing these ideas, first introduced the new term "logarithm of a number". New complex tables were compiled to calculate the logarithms of sines and cosines, as well as tangents. This greatly reduced the work of astronomers.
New tables began to appear, which were successfully used by scientists for three centuries. It took a long time before the new operation in algebra acquired its finished form. The definition of the logarithm was given and its properties were studied.
Only in the 20th century, with the advent of the calculator and the computer, mankind abandoned the ancient tables that had been successfully working for the 13th century.
Today we call the logarithm of b base a the number x, which is a power of a, to make the number b. This is written in the form of a formula: x = log a (b).
For example, log 3 (9) will be 2. This is obvious if you follow the definition. If 3 is raised to the power of 2, then we get 9.
So, the formulated definition sets only one restriction, the numbers a and b must be real.
Varieties of logarithms
The classical definition is called the real logarithm and is actually a solution to the equation a x = b. Option a = 1 is borderline and is of no interest. Note: 1 is equal to 1 in any degree.
Real value of the logarithm defined only when radix and argument are greater than 0, and the radix must not be equal to 1.
A special place in the field of mathematics play logarithms, which will be named depending on the magnitude of their base:
Rules and restrictions
The fundamental property of logarithms is the rule: the logarithm of the product is equal to the logarithmic sum. log abp = log a (b) + log a (p).
As a variant of this statement will be: log с (b / p) = log с (b) - log с (p), the quotient function is equal to the difference of functions.
From the previous two rules it is easy to see that: log a (b p) = p * log a (b).
Other properties include:
Comment. Don't make a common mistake - the logarithm of the sum is not equal to the sum of the logarithms.
For many centuries, the operation of finding the logarithm has been a rather laborious task. Mathematicians used known formula logarithmic polynomial decomposition theory:
ln (1 + x) = x - (x ^ 2) / 2 + (x ^ 3) / 3 - (x ^ 4) / 4 +… + ((-1) ^ (n + 1)) * (( x ^ n) / n), where n - natural number greater than 1, which determines the accuracy of the calculation.
Logarithms with other bases were calculated using the theorem on the transition from one base to another and the property of the logarithm of the product.
Since this method is very time consuming and when solving practical problems difficult to implement, then we used pre-compiled tables of logarithms, which greatly accelerated the whole work.
In some cases, specially compiled graphs of logarithms were used, which gave less accuracy, but significantly accelerated the search. desired value... The curve of the function y = log a (x), built over several points, allows using a regular ruler to find the values of the function at any other point. For a long time, engineers used the so-called graph paper for these purposes.
In the 17th century, the first auxiliary analog computing conditions appeared, which by XIX century acquired a finished look. The most successful device was named slide rule... With all the simplicity of the device, its appearance has significantly accelerated the process of all engineering calculations, and it is difficult to overestimate this. Currently, few people are already familiar with this device.
The advent of calculators and computers made it meaningless to use any other device.
Equations and inequalities
For solutions different equations and inequalities using logarithms, the following formulas apply:
- Transition from one base to another: log a (b) = log c (b) / log c (a);
- Consequently previous version: log a (b) = 1 / log b (a).
To solve inequalities, it is useful to know:
- The value of the logarithm will be positive only if the base and the argument are both greater or less than one; if at least one condition is violated, the value of the logarithm will be negative.
- If the logarithm function is applied to the right and left sides of the inequality, and the base of the logarithm is greater than one, then the inequality sign is preserved; otherwise, it changes.
Examples of tasks
Let's consider several options for using logarithms and their properties. Examples with solving equations:
Consider the option of placing the logarithm in power:
- Problem 3. Calculate 25 ^ log 5 (3). Solution: under the conditions of the problem, the record is similar to the following (5 ^ 2) ^ log5 (3) or 5 ^ (2 * log 5 (3)). Let's write it differently: 5 ^ log 5 (3 * 2), or the square of a number as an argument to a function can be written as the square of the function itself (5 ^ log 5 (3)) ^ 2. Using the properties of logarithms, this expression is 3 ^ 2. Answer: as a result of the calculation, we get 9.
Practical use
Being a purely mathematical tool, it seems far from real life that the logarithm suddenly became very important for describing objects the real world... It is difficult to find a science where it is not applied. This fully applies not only to natural, but also humanitarian fields of knowledge.
Logarithmic dependencies
Here are some examples of numerical dependencies:
Mechanics and Physics
Historically, mechanics and physics have always evolved using mathematical methods research and at the same time served as a stimulus for the development of mathematics, including logarithms. The theory of most laws of physics is written in the language of mathematics. We will give only two examples of the description of physical laws using the logarithm.
It is possible to solve the problem of calculating such a complex quantity as the speed of a rocket using the Tsiolkovsky formula, which laid the foundation for the theory of space exploration:
V = I * ln (M1 / M2), where
- V is the final speed of the aircraft.
- I is the specific impulse of the engine.
- M 1 is the initial mass of the rocket.
- M 2 is the final mass.
Another important example- this is the use in the formula of another great scientist Max Planck, which serves to assess the equilibrium state in thermodynamics.
S = k * ln (Ω), where
- S - thermodynamic property.
- k is Boltzmann's constant.
- Ω is the statistical weight of different states.
Chemistry
Less obvious would be the use of formulas in chemistry containing the ratio of logarithms. We will also give only two examples:
- Nernst equation, the condition of the redox potential of the medium in relation to the activity of substances and the equilibrium constant.
- The calculation of such constants as the autoprolysis index and the acidity of the solution are also not complete without our function.
Psychology and biology
And it’s completely incomprehensible what psychology has to do with it. It turns out that the strength of sensation is well described by this function as the inverse ratio of the value of the intensity of the stimulus to the lower value of the intensity.
After the above examples, it is no longer surprising that the topic of logarithms is widely used in biology. Volumes can be written about biological forms corresponding to logarithmic spirals.
Other areas
It seems that the existence of the world is impossible without connection with this function, and it rules all the laws. Especially when the laws of nature are associated with a geometric progression. It is worth referring to the MatProfi website, and there are many such examples in the following areas of activity:
The list can be endless. Having mastered the basic laws of this function, you can plunge into the world of infinite wisdom.
With this video, I begin a long series of tutorials on logarithmic equations. Now before you are three examples at once, on the basis of which we will learn to solve the most simple tasks, which are called so - protozoa.
log 0.5 (3x - 1) = −3
lg (x + 3) = 3 + 2 lg 5
Let me remind you that the simplest logarithmic equation is the following:
log a f (x) = b
In this case, it is important that the variable x is present only inside the argument, that is, only in the function f (x). And the numbers a and b are exactly numbers, and in no case are functions containing the variable x.
Basic solution methods
There are many ways to solve such designs. For example, most of the teachers in the school suggest this way: Immediately express the function f (x) by the formula f ( x) = a b. That is, when you meet the simplest construction, you can go straight to the solution without additional actions and constructions.
Yes, of course, the decision will turn out to be correct. However, the problem with this formula is that most students do not understand, where it comes from and why we raise the letter a to the letter b.
As a result, I often see very offensive mistakes when, for example, these letters are swapped. This formula you need to either understand or cram, and the second method leads to mistakes at the most inappropriate and most crucial moments: at exams, tests, etc.
That is why I propose to all my students to abandon the standard school formula and use the second approach to solve logarithmic equations, which, as you probably already guessed from the name, is called canonical form.
The idea behind the canonical form is simple. Let's take another look at our problem: on the left we have log a, while the letter a means exactly a number, and in no case a function containing a variable x. Therefore, this letter is subject to all restrictions that are imposed on the base of the logarithm. namely:
1 ≠ a> 0
On the other hand, from the same equation, we see that the logarithm should be equal to the number b, and no restrictions are imposed on this letter, because it can take any values - both positive and negative. It all depends on what values the function f (x) takes.
And here we remember our wonderful rule that any number b can be represented as a logarithm to the base a from a to the power of b:
b = log a a b
How do you remember this formula? It's very simple. Let's write the following construction:
b = b 1 = b log a a
Of course, all the restrictions that we wrote down at the beginning arise. Now let's use the basic property of the logarithm, and introduce the factor b as the power of a. We get:
b = b 1 = b log a a = log a a b
As a result, the original equation will be rewritten as follows:
log a f (x) = log a a b → f (x) = a b
That's all. The new function no longer contains the logarithm and is solved using standard algebraic techniques.
Of course, someone will now object: why bother to come up with some canonical formula, why perform two additional unnecessary steps, if you could immediately go from the initial construction to the final formula? Yes, even then, that the majority of students do not understand where this formula comes from and, as a result, regularly make mistakes when applying it.
But this sequence of actions, consisting of three steps, allows you to solve the original logarithmic equation, even if you do not understand where the final formula comes from. By the way, this record is called the canonical formula:
log a f (x) = log a a b
The convenience of the canonical form also lies in the fact that it can be used to solve a very wide class of logarithmic equations, and not just the simplest ones that we are considering today.
Solution examples
Now let's look at real-life examples. So, we decide:
log 0.5 (3x - 1) = −3
Let's rewrite it like this:
log 0.5 (3x - 1) = log 0.5 0.5 −3
Many students are in a hurry and try to immediately raise the number 0.5 to the power that came to us from the original problem. Indeed, when you are already well trained in solving such problems, you can immediately follow this step.
However, if you are just starting to study this topic now, it is better not to rush anywhere in order not to make offensive mistakes. So, we have before us the canonical form. We have:
3x - 1 = 0.5 −3
This is no longer a logarithmic equation, but a linear one with respect to the variable x. To solve it, let's first deal with the number 0.5 to the −3 power. Note that 0.5 is 1/2.
(1/2) −3 = (2/1) 3 = 8
Everything decimals convert to normal when you solve a logarithmic equation.
We rewrite and get:
3x - 1 = 8
3x = 9
x = 3
That's it, we got an answer. The first task has been solved.
Second task
Let's move on to the second task:
As you can see, this equation is no longer the simplest one. If only because the difference is on the left, and not one single logarithm in one base.
Therefore, you need to somehow get rid of this difference. In this case, everything is very simple. Let's take a close look at the bases: on the left is the number under the root:
General recommendation: in all logarithmic equations, try to get rid of radicals, that is, from entries with roots and go to power functions, simply because the exponents of these degrees are easily taken out of the sign of the logarithm, and ultimately such a notation greatly simplifies and speeds up the calculations. So let's write it this way:
Now we recall the remarkable property of the logarithm: from the argument, as well as from the base, you can derive degrees. In the case of grounds, the following occurs:
log a k b = 1 / k loga b
In other words, the number that stood in the degree of the base is carried forward and at the same time turns over, that is, it becomes the reciprocal. In our case, there was a degree of foundation with an exponent of 1/2. Therefore, we can render it as 2/1. We get:
5 2 log 5 x - log 5 x = 18
10 log 5 x - log 5 x = 18
Please note: in no case should you get rid of the logarithms at this step. Remember the mathematics of grades 4-5 and the procedure: first, multiplication is performed, and only then addition and subtraction. In this case, we subtract one of the same from 10 elements:
9 log 5 x = 18
log 5 x = 2
Now our equation looks like it should. it simplest design and we solve it with the canonical form:
log 5 x = log 5 5 2
x = 5 2
x = 25
That's all. The second task has been solved.
Third example
Let's move on to the third task:
lg (x + 3) = 3 + 2 lg 5
Let me remind you the following formula:
lg b = log 10 b
If for some reason you are confused by the log b, then when performing all the calculations, you can simply log 10 b. You can work with decimal logarithms in the same way as with others: take out degrees, add and represent any numbers in the form lg 10.
It is these properties that we will now use to solve the problem, since it is not the simplest one that we wrote down at the very beginning of our lesson.
To begin with, note that the factor 2 before lg 5 can be introduced and becomes a power of the base 5. In addition, the free term 3 is also representable as a logarithm - this is very easy to observe from our notation.
Judge for yourself: any number can be represented as log base 10:
3 = log 10 10 3 = log 10 3
Let's rewrite the original problem taking into account the received changes:
lg (x - 3) = lg 1000 + lg 25
log (x - 3) = log 1000 25
lg (x - 3) = lg 25,000
Before us is the canonical form again, and we got it, bypassing the stage of transformations, that is, the simplest logarithmic equation did not appear anywhere in our country.
This is exactly what I talked about at the very beginning of the lesson. The canonical form allows you to solve a wider class of problems than the standard school formula given by most school teachers.
Well, that's all, we get rid of the sign of the decimal logarithm, and we get a simple linear construction:
x + 3 = 25,000
x = 24,997
Everything! The problem has been solved.
A note on scope
Here I would like to make an important remark about the scope of definition. Surely now there are students and teachers who will say: "When we solve expressions with logarithms, it is imperative to remember that the argument f (x) must be greater than zero!" In this regard, a logical question arises: why in none of the considered problems did we require this inequality to be fulfilled?
Do not worry. No extra roots will arise in these cases. And this is another great trick that allows you to speed up the solution. Just know that if in a problem the variable x occurs only in one place (or rather, in a single argument of a single logarithm), and nowhere else in our case is there a variable x, then write the domain not necessary because it will run automatically.
Judge for yourself: in the first equation we got that 3x - 1, that is, the argument should be equal to 8. This automatically means that 3x - 1 will be greater than zero.
With the same success, we can write that in the second case x must be equal to 5 2, that is, it is certainly greater than zero. And in the third case, where x + 3 = 25,000, that is, again obviously greater than zero. In other words, the domain is automatically satisfied, but only if x occurs only in the argument of only one logarithm.
That's all you need to know to solve the simplest problems. This rule alone, together with transformation rules, will allow you to solve a very wide class of problems.
But let's be honest: in order to finally understand this technique, in order to learn how to apply the canonical form of the logarithmic equation, it is not enough just to watch one video tutorial. Therefore, download the options for independent decision that are attached to this video tutorial and start solving at least one of these two independent works.
It will take you just a few minutes. But the effect of such training will be much higher compared to if you just watched this video tutorial.
I hope this tutorial will help you understand logarithmic equations. Use the canonical form, simplify expressions using rules for working with logarithms - and no problem will be scary for you. And I have everything for today.
Consideration of the scope
Now let's talk about the domain of the logarithmic function, as well as how this affects the solution of logarithmic equations. Consider a construction of the form
log a f (x) = b
Such an expression is called the simplest - there is only one function in it, and the numbers a and b are exactly numbers, and in no case is it a function that depends on the variable x. It is solved very simply. You just need to use the formula:
b = log a a b
This formula is one of the key properties of the logarithm, and when substituted into our original expression, we get the following:
log a f (x) = log a a b
f (x) = a b
This is a familiar formula from school textbooks. Many students will probably have a question: since in the original expression the function f (x) is under the log sign, the following restrictions are imposed on it:
f (x)> 0
This limitation is in effect because the logarithm of negative numbers does not exist. So, maybe because of this limitation, you should introduce a check for answers? Perhaps they need to be substituted in the source?
No, in the simplest logarithmic equations an additional check is unnecessary. And that's why. Take a look at our final formula:
f (x) = a b
The fact is that the number a is in any case greater than 0 - this requirement is also imposed by the logarithm. The number a is the base. In this case, no restrictions are imposed on the number b. But it doesn't matter, because no matter what degree we raise positive number, we will still get a positive number at the output. Thus, the requirement f (x)> 0 is fulfilled automatically.
What is really worth checking is the scope of the function under the log sign. There may be rather complicated structures, and in the process of solving them, you must definitely follow them. Let's see.
First task:
First step: transform the fraction on the right. We get:
We get rid of the sign of the logarithm and get the usual irrational equation:
Of the resulting roots, only the first one suits us, since the second root is less than zero. The only answer will be the number 9. That's it, the problem is solved. No additional checks that the expression under the sign of the logarithm is greater than 0 is not required, because it is not just greater than 0, but by the condition of the equation it is equal to 2. Therefore, the requirement “greater than zero” is fulfilled automatically.
Let's move on to the second task:
Everything is the same here. We rewrite the construction, replacing the three:
We get rid of the signs of the logarithm and get an irrational equation:
We square both sides, taking into account the restrictions, and we get:
4 - 6x - x 2 = (x - 4) 2
4 - 6x - x 2 = x 2 + 8x + 16
x 2 + 8x + 16 −4 + 6x + x 2 = 0
2x 2 + 14x + 12 = 0 |: 2
x 2 + 7x + 6 = 0
We solve the resulting equation through the discriminant:
D = 49 - 24 = 25
x 1 = −1
x 2 = −6
But x = −6 does not suit us, because if we substitute this number into our inequality, we get:
−6 + 4 = −2 < 0
In our case, it is required that it be greater than 0 or, in extreme cases, equal. But x = −1 suits us:
−1 + 4 = 3 > 0
The only answer in our case is x = −1. That's the whole solution. Let's go back to the very beginning of our calculations.
The main takeaway from this lesson is that you do not need to check the constraints for a function in the simplest logarithmic equations. Because in the process of solving all the constraints are met automatically.
However, this in no way means that you can forget about checking altogether. In the process of working on a logarithmic equation, it may well turn into an irrational one, which will have its own restrictions and requirements for the right-hand side, as we have seen today on two different examples.
Feel free to solve such problems and be especially careful if there is a root in the argument.
Logarithmic equations with different bases
We continue to study logarithmic equations and analyze two more rather interesting tricks with which it is fashionable to solve more complex structures... But first, let's remember how the simplest tasks are solved:
log a f (x) = b
In this notation, a and b are exactly numbers, and in the function f (x) the variable x must be present, and only there, that is, x must be only in the argument. We will transform such logarithmic equations using the canonical form. To do this, note that
b = log a a b
Moreover, a b is exactly the argument. Let's rewrite this expression as follows:
log a f (x) = log a a b
This is exactly what we are trying to achieve, so that both the left and the right are the logarithm to the base a. In this case, we can, figuratively speaking, strike out the signs of log, and from the point of view of mathematics, we can say that we are simply equating the arguments:
f (x) = a b
As a result, we will get a new expression that will be much easier to solve. Let's apply this rule to our tasks today.
So the first construct:
First of all, note that there is a fraction on the right with log in the denominator. When you see such an expression, it will not be superfluous to remember the wonderful property of logarithms:
Translated into Russian, this means that any logarithm can be represented as a quotient of two logarithms with any base s. Of course, 0< с ≠ 1.
So: this formula has one wonderful special case when the variable c is equal to the variable b. In this case, we get a construction of the form:
It is this construction that we observe from the sign to the right in our equation. Let's replace this construction with log a b, we get:
In other words, compared to the original problem, we have swapped the argument and the base of the logarithm. Instead, we had to flip the fraction.
We recall that any degree can be derived from the base according to the following rule:
In other words, the coefficient k, which is the degree of the base, is taken out as an inverted fraction. Let's take it out as an inverted fraction:
The fractional factor cannot be left in front, because in this case we will not be able to represent this record as a canonical form (after all, in the canonical form, there is no additional factor in front of the second logarithm). Therefore, let's add the fraction 1/4 to the exponent argument:
Now we equate arguments whose bases are the same (and we really have the same bases), and write:
x + 5 = 1
x = −4
That's all. We got the answer to the first logarithmic equation. Please note: in the original problem, the variable x occurs only in one log, and it is in its argument. Therefore, there is no need to check the domain, and our number x = −4 is indeed the answer.
Now let's move on to the second expression:
lg 56 = lg 2 log 2 7 - 3lg (x + 4)
Here, in addition to the usual logarithms, we will have to work with lg f (x). How to solve such an equation? It may seem to an untrained student that this is some kind of toughness, but in fact, everything is solved in an elementary way.
Take a close look at the term lg 2 log 2 7. What can we say about it? The reasons and arguments for log and lg are the same, and that should be suggestive. Let's remember again how the degrees are taken out from under the sign of the logarithm:
log a b n = nlog a b
In other words, what was the power of the number b in the argument becomes a factor in front of log itself. Let's use this formula to express lg 2 log 2 7. Don't be intimidated by lg 2 - this is the most common expression. You can rewrite it like this:
All the rules that apply to any other logarithm are true for it. In particular, the factor in front can be added to the degree of the argument. Let's write:
Very often students do not see this action point blank, because it is not good to enter one log under the sign of another. In fact, there is nothing criminal about this. Moreover, we get a formula that can be easily calculated if you remember an important rule:
This formula can be considered both as a definition and as one of its properties. In any case, if you transform a logarithmic equation, you should know this formula in the same way as representing any number in the form of log.
We return to our task. We rewrite it taking into account the fact that the first term to the right of the equal sign will simply be equal to lg 7. We have:
lg 56 = lg 7 - 3lg (x + 4)
Let's move lg 7 to the left, we get:
lg 56 - lg 7 = −3lg (x + 4)
Subtract the expressions on the left because they have the same base:
lg (56/7) = −3lg (x + 4)
Now let's take a close look at the equation we got. It is practically the canonical form, but there is a factor of −3 on the right. Let's put it in the right lg argument:
log 8 = log (x + 4) −3
Before us is the canonical form of the logarithmic equation, so we cross out the signs of lg and equate the arguments:
(x + 4) −3 = 8
x + 4 = 0.5
That's all! We have solved the second logarithmic equation. In this case, no additional checks are required, because in the original problem x was present in only one argument.
Let me reiterate the key points of this tutorial.
The main formula that is studied in all the lessons on this page dedicated to solving logarithmic equations is the canonical form. And don't be intimidated by the fact that most school textbooks teach you to solve such problems in a different way. This tool works very effectively and allows you to solve a much wider class of problems than the simplest ones that we studied at the very beginning of our lesson.
In addition, it will be useful to know the basic properties for solving logarithmic equations. Namely:
- The formula for the transition to one base and the special case when we flip log (this was very useful to us in the first problem);
- The formula for adding and removing degrees from the sign of the logarithm. Here, many students freeze and do not see at close range that the exponential and inserted degree itself can contain log f (x). Nothing wrong with that. We can introduce one log by the sign of the other and at the same time significantly simplify the solution of the problem, which we observe in the second case.
In conclusion, I would like to add that it is not necessary to check the scope in each of these cases, because everywhere the variable x is present only in one sign of log, and at the same time it is in its argument. As a result, all requirements of the scope are met automatically.
Variable radix problems
Today we will look at logarithmic equations, which for many students seem to be non-standard, if not completely unsolvable. It is about expressions based not on numbers, but on variables and even functions. We will solve such constructions using our standard technique, namely, through the canonical form.
To begin with, let's remember how the simplest problems are solved, which are based on ordinary numbers. So, the simplest is a construction of the form
log a f (x) = b
To solve such problems, we can use the following formula:
b = log a a b
We rewrite our original expression and get:
log a f (x) = log a a b
Then we equate the arguments, that is, we write:
f (x) = a b
Thus, we get rid of the log sign and solve the already common problem. In this case, the roots obtained in the solution will be the roots of the original logarithmic equation. In addition, the record, when both the left and the right are on the same logarithm with the same base, is called the canonical form. It is to such a record that we will try to reduce today's constructions. So let's go.
First task:
log x - 2 (2x 2 - 13x + 18) = 1
Replace 1 with log x - 2 (x - 2) 1. The degree that we observe in the argument is, in fact, the number b that stood to the right of the equal sign. Thus, we will rewrite our expression. We get:
log x - 2 (2x 2 - 13x + 18) = log x - 2 (x - 2)
What do we see? Before us is the canonical form of the logarithmic equation, so we can safely equate the arguments. We get:
2x 2 - 13x + 18 = x - 2
But the decision doesn't end there, because given equation is not equivalent to the original. After all, the resulting construction consists of functions that are defined on the entire number line, and our initial logarithms are not defined everywhere and not always.
Therefore, we must write down the scope separately. Let's not be smart and first write down all the requirements:
First, the argument of each of the logarithms must be greater than 0:
2x 2 - 13x + 18> 0
x - 2> 0
Secondly, the base must not only be greater than 0, but also different from 1:
x - 2 ≠ 1
As a result, we get the system:
But do not be alarmed: when processing logarithmic equations, such a system can be significantly simplified.
Judge for yourself: on the one hand, we are required that the quadratic function be greater than zero, and on the other hand, this quadratic function is equated to a certain linear expression, which is also required to be greater than zero.
In this case, if we require that x - 2> 0, then the requirement 2x 2 - 13x + 18> 0 will automatically be satisfied. Therefore, we can safely cross out the inequality containing quadratic function... Thus, the number of expressions contained in our system will be reduced to three.
Of course, we could just as well cross out the linear inequality, that is, cross out x - 2> 0 and require that 2x 2 - 13x + 18> 0. But you must admit that solving the simplest linear inequality is much faster and easier, than quadratic, even under the condition that as a result of solving this entire system, we get the same roots.
In general, try to optimize your computations whenever possible. And in the case of logarithmic equations, cross out the most difficult inequalities.
Let's rewrite our system:
Here is such a system of three expressions, with two of which we, in fact, have already figured out. Let's write it out separately quadratic equation and solve it:
2x 2 - 14x + 20 = 0
x 2 - 7x + 10 = 0
Before us is the given square trinomial and, therefore, we can use Vieta's formulas. We get:
(x - 5) (x - 2) = 0
x 1 = 5
x 2 = 2
And now we return to our system and find that x = 2 does not suit us, because we are required that x be strictly greater than 2.
But x = 5 suits us perfectly: the number 5 is greater than 2, and at the same time 5 is not equal to 3. Therefore, the only solution this system will be x = 5.
That's it, the problem has been solved, including taking into account the ODZ. Let's move on to the second equation. Here we will find more interesting and informative calculations:
The first step: just like last time, we bring the whole thing to the canonical form. For this, we can write the number 9 as follows:
You don't have to touch the root with the root, but it's better to transform the argument. Let's go from root to rational exponent. Let's write down:
Let me not rewrite our entire large logarithmic equation, but just immediately equate the arguments:
x 3 + 10x 2 + 31x + 30 = x 3 + 9x 2 + 27x + 27
x 2 + 4x + 3 = 0
Before us is the newly given square trinomial, we use Vieta's formulas and write:
(x + 3) (x + 1) = 0
x 1 = −3
x 2 = −1
So, we got the roots, but no one guaranteed us that they would fit the original logarithmic equation. After all, the log signs impose additional restrictions (here we would have to write the system, but due to the cumbersomeness of the whole structure, I decided to calculate the domain separately).
First of all, remember that the arguments must be greater than 0, namely:
These are the requirements imposed by the domain of definition.
Immediately, we note that since we equate the first two expressions of the system to each other, then we can delete any of them. Let's delete the first one because it looks more threatening than the second one.
In addition, note that the solution to the second and third inequalities will be the same sets (the cube of some number is greater than zero, if this number itself is greater than zero; similarly with a root of the third degree - these inequalities are completely analogous, so one of them we can cross it out).
But with the third inequality, this will not work. Let's get rid of the radical sign on the left, for which we will build both parts into a cube. We get:
So we get the following requirements:
- 2 ≠ x> −3
Which of our roots: x 1 = −3 or x 2 = −1 meets these requirements? Obviously, only x = −1, because x = −3 does not satisfy the first inequality (since our inequality is strict). So, returning to our problem, we get one root: x = −1. That's all, the problem is solved.
Once again, the key points of this task:
- Feel free to apply and solve logarithmic equations using the canonical form. Students who make such a notation, and do not go directly from the original problem to a construction like log a f (x) = b, allow much less mistakes than those who are in a hurry somewhere, skipping intermediate steps of calculations;
- As soon as a variable base appears in the logarithm, the problem ceases to be the simplest one. Therefore, when solving it, it is necessary to take into account the domain of definition: the arguments must be greater than zero, and the bases must not only be greater than 0, but also they must not be equal to 1.
There are different ways to impose the final requirements on the final answers. For example, you can solve the whole system containing all the requirements for the domain of definition. On the other hand, you can first solve the problem itself, and then remember about the domain of definition, work it out separately in the form of a system and superimpose it on the resulting roots.
Which way to choose when solving a specific logarithmic equation is up to you. In any case, the answer will be the same.
The logarithm of a positive number b to base a (a> 0, a is not equal to 1) is a number c such that a c = b: log a b = c ⇔ a c = b (a> 0, a ≠ 1, b> 0) & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp
Please note: the logarithm of a non-positive number is undefined. In addition, the base of the logarithm must be a positive number, not equal to 1. For example, if we square -2, we get the number 4, but this does not mean that the logarithm to the base -2 of 4 is 2.
Basic logarithmic identity
a log a b = b (a> 0, a ≠ 1) (2)It is important that the areas of definition of the right and left sides of this formula are different. The left-hand side is defined only for b> 0, a> 0, and a ≠ 1. Right part is defined for any b, but does not depend on a at all. Thus, the use of the basic logarithmic "identity" in solving equations and inequalities can lead to a change in the GDV.
Two obvious consequences of the definition of a logarithm
log a a = 1 (a> 0, a ≠ 1) (3)log a 1 = 0 (a> 0, a ≠ 1) (4)
Indeed, when raising the number a to the first power, we get the same number, and when raising it to the zero power, we get one.
Logarithm of the product and the logarithm of the quotient
log a (b c) = log a b + log a c (a> 0, a ≠ 1, b> 0, c> 0) (5)Log a b c = log a b - log a c (a> 0, a ≠ 1, b> 0, c> 0) (6)
I would like to warn schoolchildren against thoughtless use of these formulas when solving logarithmic equations and inequalities. When they are used "from left to right", the ODZ narrows, and when you move from the sum or difference of logarithms to the logarithm of the product or quotient, the ODV expands.
Indeed, the expression log a (f (x) g (x)) is defined in two cases: when both functions are strictly positive, or when f (x) and g (x) are both less than zero.
Transforming this expression into the sum log a f (x) + log a g (x), we are forced to limit ourselves only to the case when f (x)> 0 and g (x)> 0. There is a narrowing of the range of permissible values, and this is categorically unacceptable, since it can lead to the loss of solutions. A similar problem exists for formula (6).
The degree can be expressed outside the sign of the logarithm
log a b p = p log a b (a> 0, a ≠ 1, b> 0) (7)And again I would like to call for accuracy. Consider the following example:
Log a (f (x) 2 = 2 log a f (x)
The left-hand side of the equality is defined, obviously, for all values of f (x), except zero. The right side is only for f (x)> 0! Taking the degree out of the logarithm, we again narrow the ODV. The reverse procedure expands the range of valid values. All these remarks apply not only to degree 2, but also to any even degree.
The formula for the transition to a new base
log a b = log c b log c a (a> 0, a ≠ 1, b> 0, c> 0, c ≠ 1) (8)That rare case when the ODV does not change during the transformation. If you have reasonably chosen a base c (positive and not equal to 1), the formula for transition to a new base is completely safe.
If we choose the number b as the new base c, we get an important special case of formula (8):
Log a b = 1 log b a (a> 0, a ≠ 1, b> 0, b ≠ 1) (9)
Some simple examples with logarithms
Example 1. Calculate: lg2 + lg50.
Solution. lg2 + lg50 = lg100 = 2. We used the formula for the sum of logarithms (5) and the definition of the decimal logarithm.
Example 2. Calculate: lg125 / lg5.
Solution. lg125 / lg5 = log 5 125 = 3. We used the formula for transition to a new base (8).
Table of formulas related to logarithms
| a log a b = b (a> 0, a ≠ 1) |
| log a a = 1 (a> 0, a ≠ 1) |
| log a 1 = 0 (a> 0, a ≠ 1) |
| log a (b c) = log a b + log a c (a> 0, a ≠ 1, b> 0, c> 0) |
| log a b c = log a b - log a c (a> 0, a ≠ 1, b> 0, c> 0) |
| log a b p = p log a b (a> 0, a ≠ 1, b> 0) |
| log a b = log c b log c a (a> 0, a ≠ 1, b> 0, c> 0, c ≠ 1) |
| log a b = 1 log b a (a> 0, a ≠ 1, b> 0, b ≠ 1) |
