Decomposition of the logarithm. Natural logarithm, ln x function

Logarithm of b (b\u003e 0) to base a (a\u003e 0, a ≠ 1) Is the exponent to which the number a must be raised to get b.

The logarithm of b to base 10 can be written as lg (b), and the logarithm base e (natural logarithm) is ln (b).

Often used when solving problems with logarithms:

Properties of logarithms

There are four main properties of logarithms.

Let a\u003e 0, a ≠ 1, x\u003e 0, and y\u003e 0.

Property 1. Logarithm of the product

Logarithm of the product is equal to the sum of the logarithms:

log a (x ⋅ y) \u003d log a x + log a y

Property 2. Logarithm of the quotient

Logarithm of the quotient is equal to the difference of logarithms:

log a (x / y) \u003d log a x - log a y

Property 3. Logarithm of the degree

Logarithm of degree is equal to the product of the power by the logarithm:

If the base of the logarithm is in power, then another formula works:

Property 4. Logarithm of the root

This property can be obtained from the property of the logarithm of the degree, since the root of the nth degree is equal to the degree 1 / n:

The formula for the transition from a logarithm in one base to a logarithm in another base

This formula is also often used when solving various problems for logarithms:

A special case:

Comparison of logarithms (inequalities)

Suppose we have 2 functions f (x) and g (x) under logarithms with the same bases and there is an inequality sign between them:

To compare them, you first need to look at the base of the logarithms a:

If a\u003e 0, then f (x)\u003e g (x)\u003e 0
If 0< a < 1, то 0 < f(x) < g(x)

How to solve problems with logarithms: examples

Logarithm tasks included in the USE in mathematics for grade 11 in task 5 and task 7, you can find tasks with solutions on our website in the relevant sections. Also, tasks with logarithms are found in the bank of tasks in mathematics. All examples can be found through the site search.

What is a logarithm

Logarithms have always been considered a difficult topic in high school maths. There are many different definitions of the logarithm, but most textbooks somehow use the most difficult and unfortunate ones.

We will define the logarithm simply and clearly. To do this, create a table:

So, we have before us powers of two.

Logarithms - properties, formulas, how to solve

If you take the number from the bottom line, then you can easily find the degree to which you have to raise the two to get this number. For example, to get 16, you need to raise two to the fourth power. And to get 64, you need to raise two to the sixth power. This can be seen from the table.

And now - actually, the definition of the logarithm:

base a from argument x is the power to which the number a must be raised to get the number x.

Notation: log a x \u003d b, where a is the base, x is the argument, b is actually what the logarithm is.

For example, 2 3 \u003d 8 ⇒log 2 8 \u003d 3 (log base 2 of 8 is three, since 2 3 \u003d 8). With the same success log 2 64 \u003d 6, since 2 6 \u003d 64.

The operation of finding the logarithm of a number in a given base is called. So, let's add a new line to our table:

2 1	2 2	2 3	2 4	2 5	2 6
2	4	8	16	32	64
log 2 2 \u003d 1	log 2 4 \u003d 2	log 2 8 \u003d 3	log 2 16 \u003d 4	log 2 32 \u003d 5	log 2 64 \u003d 6

Unfortunately, not all logarithms are calculated so easily. For example, try to find log 2 5. The number 5 is not in the table, but logic dictates that the logarithm will lie somewhere on the segment. Because 2 2< 5 < 2 3 , а чем больше степень двойки, тем больше получится число.

Such numbers are called irrational: the numbers after the decimal point can be written indefinitely, and they are never repeated. If the logarithm turns out to be irrational, it is better to leave it that way: log 2 5, log 3 8, log 5 100.

It is important to understand that the logarithm is an expression with two variables (base and argument). At first, many are confused about where the basis is and where is the argument. To avoid annoying misunderstandings, just take a look at the picture:

Before us is nothing more than the definition of the logarithm. Remember: logarithm is the degreeto which the base must be raised to get the argument. It is the base that is raised to the power - in the picture it is highlighted in red. It turns out that the base is always at the bottom! I tell this wonderful rule to my students at the very first lesson - and no confusion arises.

How to count logarithms

We figured out the definition - it remains to learn how to count logarithms, i.e. get rid of the log sign. To begin with, we note that two important facts follow from the definition:

Argument and radix must always be greater than zero. This follows from the definition of the degree by a rational indicator, to which the definition of the logarithm is reduced.
The base must be different from one, since one is still one to any degree. Because of this, the question “to what degree one should raise one to get a two” is meaningless. There is no such degree!

Such restrictions are called range of valid values (ODZ). It turns out that the ODZ of the logarithm looks like this: log a x \u003d b ⇒x\u003e 0, a\u003e 0, a ≠ 1.

Note that there is no restriction on the number b (logarithm value). For example, the logarithm may well be negative: log 2 0.5 \u003d −1, because 0.5 \u003d 2 −1.

However, now we are considering only numerical expressions, where you do not need to know the ODV of the logarithm. All restrictions have already been taken into account by the task compilers. But when the logarithmic equations and inequalities come in, the DHS requirements will become mandatory. Indeed, at the base and in the argument there can be very strong constructions that do not necessarily correspond to the above restrictions.

Now let's look at the general scheme for calculating logarithms. It consists of three steps:

Represent base a and argument x as a power with the smallest possible base greater than one. Along the way, it's better to get rid of decimals;
Solve the equation for variable b: x \u003d a b;
The resulting number b will be the answer.

That's all! If the logarithm turns out to be irrational, this will be seen already at the first step. The requirement for the base to be greater than one is very relevant: this reduces the likelihood of error and greatly simplifies calculations. Similarly, with decimal fractions: if you immediately convert them to ordinary ones, there will be many times less errors.

Let's see how this scheme works with specific examples:

A task. Calculate the logarithm: log 5 25

Let's represent the base and the argument as a power of five: 5 \u003d 5 1; 25 \u003d 5 2;

Let's compose and solve the equation:
log 5 25 \u003d b ⇒ (5 1) b \u003d 5 2 ⇒5 b \u003d 5 2 ⇒ b \u003d 2;

Received the answer: 2.

A task. Calculate the logarithm:

A task. Calculate the logarithm: log 4 64

Let's represent the base and the argument as a power of two: 4 \u003d 2 2; 64 \u003d 2 6;
Let's compose and solve the equation:
log 4 64 \u003d b ⇒ (2 2) b \u003d 2 6 ⇒2 2b \u003d 2 6 ⇒2b \u003d 6 ⇒ b \u003d 3;
Received the answer: 3.

A task. Calculate the logarithm: log 16 1

Let's represent the base and the argument as a power of two: 16 \u003d 2 4; 1 \u003d 2 0;
Let's compose and solve the equation:
log 16 1 \u003d b ⇒ (2 4) b \u003d 2 0 ⇒2 4b \u003d 2 0 ⇒4b \u003d 0 ⇒ b \u003d 0;
Received the answer: 0.

A task. Calculate the log of: log 7 14

We represent the base and the argument as a power of seven: 7 \u003d 7 1; 14 is not represented as a power of seven, since 7 1< 14 < 7 2 ;
It follows from the previous paragraph that the logarithm is not considered;
The answer is no change: log 7 14.

A small note on the last example. How do you ensure that a number is not an exact power of another number? It's very simple - just factor it into prime factors. If the expansion has at least two different factors, the number is not an exact power.

A task. Find out if the exact powers of the number are: 8; 48; 81; 35; 14.

8 \u003d 2 2 2 \u003d 2 3 is the exact degree, since there is only one factor;
48 \u003d 6 · 8 \u003d 3 · 2 · 2 · 2 · 2 \u003d 3 · 2 4 - is not an exact degree, since there are two factors: 3 and 2;
81 \u003d 9 9 \u003d 3 3 3 3 3 \u003d 3 4 - the exact degree;
35 \u003d 7 · 5 - again not an exact degree;
14 \u003d 7 2 - again not an exact degree;

Note also that the prime numbers themselves are always exact powers of themselves.

Decimal logarithm

Some logarithms are so common that they have a special name and designation.

of x is the base 10 logarithm, i.e. the power to which the number 10 must be raised to get the number x. Designation: lg x.

For example, lg 10 \u003d 1; lg 100 \u003d 2; lg 1000 \u003d 3 - etc.

From now on, when a phrase like "Find lg 0.01" appears in a textbook, you should know: this is not a typo. This is the decimal logarithm. However, if you are not used to such a designation, you can always rewrite it:
log x \u003d log 10 x

Everything that is true for ordinary logarithms is also true for decimal.

Natural logarithm

There is another logarithm, which has its own notation. In a way, it is even more important than decimal. This is the natural logarithm.

of argument x is the logarithm base e, i.e. the power to which the number e must be raised to get the number x. Designation: ln x.

Many will ask: what else is the number e? This is an irrational number, its exact meaning cannot be found and written down. I will give only the first figures:
e \u003d 2.718281828459 ...

We will not delve into what this number is and why it is needed. Just remember that e is the base of the natural logarithm:
ln x \u003d log e x

Thus, ln e \u003d 1; ln e 2 \u003d 2; ln e 16 \u003d 16 - etc. On the other hand, ln 2 is an irrational number. In general, the natural logarithm of any rational number is irrational. Except, of course, units: ln 1 \u003d 0.

For natural logarithms, all the rules are true that are true for ordinary logarithms.

Logarithm. Properties of the logarithm (power of the logarithm).

How do I represent a number as a logarithm?

We use the definition of a logarithm.

The logarithm is an indicator of the degree to which the base must be raised to get the number under the sign of the logarithm.

Thus, in order to represent some number c in the form of a logarithm to the base a, it is necessary to put the power with the same base as the base of the logarithm under the sign of the logarithm, and write this number c in the exponent:

Absolutely any number can be represented in the form of a logarithm - positive, negative, whole, fractional, rational, irrational:

In order not to confuse a and c under stressful conditions of a control or exam, you can use this rule to memorize:

what is below goes down, what is above goes up.

For example, you want to represent the number 2 as a logarithm to base 3.

We have two numbers - 2 and 3. These numbers are the base and the exponent, which we will write under the sign of the logarithm. It remains to determine which of these numbers needs to be written down, to the base of the degree, and which - up, to the exponent.

The base 3 in the logarithm is at the bottom, which means that when we represent two in the form of a logarithm to the base 3, 3 will also be written down to the base.

2 stands above the three. And in writing the power of two, we write it above the three, that is, in the exponent:

Logarithms. First level.

Logarithms

Logarithm positive number b by reason awhere a\u003e 0, a ≠ 1, is called the exponent to which the number must be raised a, To obtain b.

Definition of the logarithm can be briefly written like this:

This equality is valid for b\u003e 0, a\u003e 0, a ≠ 1. It is usually called logarithmic identity.
The action of finding the logarithm of a number is called by taking the logarithm.

Logarithm properties:

Logarithm of the product:

Logarithm of the quotient of division:

Replacing the base of the logarithm:

Logarithm of the degree:

Logarithm of the root:

Power logarithm:

Decimal and natural logarithms.

Decimal logarithm numbers call the base 10 logarithm of this number and write & nbsp lg b
Natural logarithm numbers call the base logarithm of that number ewhere e - an irrational number, approximately equal to 2.7. In this case, they write ln b.

Other notes on algebra and geometry

Basic properties of logarithms

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 bases: log a x and log a y. Then they can be added and subtracted, and:

log a x + log a y \u003d log a (x y);
log a x - log a y \u003d 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 \u003d log 6 (4 9) \u003d log 6 36 \u003d 2.

A 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 \u003d log 2 (48: 3) \u003d log 2 16 \u003d 4.

A 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 \u003d log 3 (135: 5) \u003d log 3 27 \u003d 3.

As you can see, the original expressions are composed of "bad" logarithms, which are not separately counted. But after transformations, completely normal numbers are obtained. Many tests are based on this fact. 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 if the ODL of the logarithm is observed: a\u003e 0, a ≠ 1, x\u003e 0. And one more thing: learn to apply all formulas not only from left to right, but also vice versa, i.e. it is possible to enter the numbers in front of the sign of the logarithm into the logarithm itself.

How to solve logarithms

This is what is most often required.

A 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 \u003d 6 log 7 49 \u003d 6 2 \u003d 12

A task. Find the meaning of the expression:

Note that the denominator contains the logarithm, the base and the argument of which are exact powers: 16 \u003d 2 4; 49 \u003d 7 2. We have:

I think the last example needs 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 will remain 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 for 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 be given. Then for any number c such that c\u003e 0 and c ≠ 1, the following equality holds:

In particular, if we put c \u003d x, we get:

From the second formula it follows that it is possible to swap the base and the argument of the logarithm, but the whole expression is "reversed", that is, 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 solving 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:

A 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 \u003d log 5 2 4 \u003d 4log 5 2; log 2 25 \u003d log 2 5 2 \u003d 2 log 2 5;

Now let's "flip" the second logarithm:

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.

A 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:

Now let's get rid of the decimal logarithm by moving to the new base:

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 the exponent in the argument. The 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:.

In fact, what happens if the number b is raised to such a power that the number b to this power 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 the transition to a new base, the basic logarithmic identity is sometimes the only possible solution.

A task. Find the meaning of the expression:

Note that log 25 64 \u003d log 5 8 - just moved the square from the base and the logarithm argument. Taking into account the rules for multiplying degrees with the same base, we get:

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 \u003d 1 is. Remember once and for all: the logarithm to any base a from this base is equal to one.
log a 1 \u003d 0 is. The base a can be anything, but if the argument is one, the logarithm is zero! Because a 0 \u003d 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.

Today we will talk about logarithm formulas and give indicative solution examples.

By themselves, they imply decision templates according to the basic properties of logarithms. Before applying the formulas of the logarithms for the solution, we recall for you, first all the properties:

Now, based on these formulas (properties), we show examples of solving logarithms.

Examples of solving logarithms based on formulas.

Logarithm a positive number b in base a (denoted by log a b) is the exponent to which a must be raised to get b, while b\u003e 0, a\u003e 0, and 1.

According to the definition, log a b \u003d x, which is equivalent to a x \u003d b, therefore log a a x \u003d x.

Logarithms, examples:

log 2 8 \u003d 3, because 2 3 \u003d 8

log 7 49 \u003d 2, because 7 2 \u003d 49

log 5 1/5 \u003d -1, because 5 -1 \u003d 1/5

Decimal logarithm is the usual logarithm, at the base of which is 10. It is denoted as lg.

log 10 100 \u003d 2, because 10 2 \u003d 100

Natural logarithm - also the usual logarithm is the logarithm, but already with the base e (e \u003d 2.71828 ... is an irrational number). It is designated as ln.

It is advisable to remember the formulas or properties of logarithms, because we will need them in the future when solving logarithms, logarithmic equations and inequalities. Let's try each formula once again with examples.

Basic logarithmic identity
a log a b \u003d b
8 2log 8 3 \u003d (8 2log 8 3) 2 \u003d 3 2 \u003d 9
The logarithm of the product is equal to the sum of the logarithms
log a (bc) \u003d log a b + log a c
log 3 8.1 + log 3 10 \u003d log 3 (8.1 * 10) \u003d log 3 81 \u003d 4
The logarithm of the quotient is equal to the difference of the logarithms
log a (b / c) \u003d log a b - log a c
9 log 5 50/9 log 5 2 \u003d 9 log 5 50-log 5 2 \u003d 9 log 5 25 \u003d 9 2 \u003d 81
Properties of the power of a logarithm and the base of a logarithm
The exponent of the logarithm of the number log a b m \u003d mlog a b

The exponent of the base of the logarithm log a n b \u003d 1 / n * log a b

log a n b m \u003d m / n * log a b,

if m \u003d n, we get log a n b n \u003d log a b

log 4 9 \u003d log 2 2 3 2 \u003d log 2 3
Moving to a new foundation
log a b \u003d log c b / log c a,
if c \u003d b, we get log b b \u003d 1

then log a b \u003d 1 / log b a

log 0.8 3 * log 3 1.25 \u003d log 0.8 3 * log 0.8 1.25 / log 0.8 3 \u003d log 0.8 1.25 \u003d log 4/5 5/4 \u003d -1

As you can see, the formulas for logarithms are not as complicated as they seem. Now, having considered examples of solving logarithms, we can move on to logarithmic equations. We will consider examples of solving logarithmic equations in more detail in the article: "". Do not miss!

If you still have questions about the solution, write them in the comments to the article.

Note: we decided to get education in another class, study abroad as an option for the development of events.

As you know, when multiplying expressions with powers, their exponents always add up (a b * a c \u003d a b + c). This mathematical law was derived 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 want 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. In simple and accessible language.

Definition in mathematics

The logarithm is an expression of the following form: log ab \u003d 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 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 is very simple, you need to find such a degree so that from 2 to the desired degree you get 8. Having done some calculations in your mind, we get the number 3! And right, 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 distinct types of logarithmic expressions:

Natural logarithm ln a, where the base is Euler's number (e \u003d 2.7).
Decimal a, base 10.
Logarithm of any number b to base a\u003e 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 \u200b\u200bof 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, you cannot divide numbers by zero, and it is also impossible to extract an even root of 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 be 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\u003e 0, then a b\u003e 0, it turns out that "c" must also be greater than zero.

How to solve logarithms?

For example, given the task of finding the answer to the equation 10 x \u003d 100. It is very easy, you need to choose such a degree, raising the number ten to which we get 100. This, of course, 10 2 \u003d 100.

Now let's represent this expression as a logarithmic one. We get log 10 100 \u003d 2. When solving logarithms, all actions almost converge to find the power to which it is necessary to introduce the base of the logarithm 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, larger values \u200b\u200bwill require a power table. It can be used even by those who do not understand anything 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 \u200b\u200bof the numbers are defined, which are the answer (a c \u003d b). Let's 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 \u003d 81 can be written as the logarithm of 81 to base 3, equal to four (log 3 81 \u003d 4). For negative powers, the rules are the same: 2 -5 \u003d 1/32, we write it as a logarithm, we get log 2 (1/32) \u003d -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 take a 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)\u003e 3 - it is a logarithmic inequality, since the unknown value "x" is under the sign of the logarithm. And also in the expression, two values \u200b\u200bare 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 \u003d √9) imply one or more specific numerical values \u200b\u200bin the answer, while solving the inequality determines both the range of admissible values \u200b\u200band 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 \u200b\u200bof 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 basic identity looks like this: a logaB \u003d B. It applies only 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) \u003d log d s 1 + log d s 2. In this case, a prerequisite is: d, s 1 and s 2\u003e 0; a ≠ 1. You can give a proof for this formula of logarithms, with examples and a solution. Let log as 1 \u003d f 1 and log as 2 \u003d f 2, then a f1 \u003d s 1, a f2 \u003d s 2. We get that s 1 * s 2 \u003d a f1 * a f2 \u003d a f1 + f2 (properties of powers ), and then by definition: log a (s 1 * s 2) \u003d f 1 + f 2 \u003d 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) \u003d 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 \u003d 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 rests on natural postulates. Let's take a look at the proof.

Let log a b \u003d t, it turns out a t \u003d b. If we raise both parts to the power m: a tn \u003d b n;

but since a tn \u003d (a q) nt / q \u003d b n, therefore log a q b n \u003d (n * t) / t, then log a q b n \u003d 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 mandatory part of exams in mathematics. For admission to the university or passing 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, you should find out whether it is possible to simplify the expression or bring it to a general form. 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 of natural logarithms, you need 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 \u003d log 2 (4 * 128) \u003d log 2 512. The answer is 9.
log 4 8 \u003d log 2 2 2 3 \u003d 3/2 log 2 2 \u003d 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 \u200b\u200bout 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 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 exam. Let's see how such tasks are solved.

Given log 2 (2x-1) \u003d 4. Solution:
rewrite the expression, simplifying it a little log 2 (2x-1) \u003d 2 2, by the definition of the logarithm, we get that 2x-1 \u003d 2 4, therefore 2x \u003d 17; x \u003d 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 exponent is taken out by the factor, which is under the sign of the logarithm and as its base, the expression remaining under the logarithm must be positive.

basic properties.

logax + logay \u003d loga (x y);
logax - logay \u003d loga (x: y).

identical grounds

Log6 4 + log6 9.

Now let's complicate the task a little.

Examples of solving logarithms

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:

Of course, all these rules make sense if the ODL of the logarithm is observed: a\u003e 0, a ≠ 1, x\u003e

A task. Find the meaning of the expression:

Moving to a new foundation

Let the logarithm be given. Then for any number c such that c\u003e 0 and c ≠ 1, the following equality holds:

A task. Find the meaning of the expression:

Basic properties of the logarithm

1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.

The exponent is 2.718281828…. To remember the exponent, you can study the rule: the exponent is 2.7 and twice the year of birth of Leo Nikolaevich Tolstoy.

Basic properties of logarithms

Knowing this rule, you will know both the exact value of the exponent and the date of birth of Leo Tolstoy.

Examples for logarithms

Logarithm expressions

Example 1.
and). x \u003d 10ac ^ 2 (a\u003e 0, c\u003e 0).

By properties 3.5 we calculate

4. Where .

Example 2. Find x if

Example 3. Let the value of the logarithms be given

Evaluate log (x) if

Basic properties of logarithms

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 bases: logax and logay. Then they can be added and subtracted, and:

logax + logay \u003d loga (x y);
logax - logay \u003d loga (x: y).

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:

Since the bases of the logarithms are the same, we use the sum formula:
log6 4 + log6 9 \u003d log6 (4 9) \u003d log6 36 \u003d 2.

A task. Find the value of the expression: log2 48 - log2 3.

The bases are the same, we use the difference formula:
log2 48 - log2 3 \u003d log2 (48: 3) \u003d log2 16 \u003d 4.

A task. Find the value of the expression: log3 135 - log3 5.

Again the bases are the same, so we have:
log3 135 - log3 5 \u003d log3 (135: 5) \u003d log3 27 \u003d 3.

Removing the exponent from the logarithm

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.

A task. Find the value of the expression: log7 496.

Let's get rid of the degree in the argument using the first formula:
log7 496 \u003d 6 log7 49 \u003d 6 2 \u003d 12

A task. Find the meaning of the expression:

Note that the denominator contains the logarithm, the base and the argument of which are exact powers: 16 \u003d 24; 49 \u003d 72. We have:

I think the last example needs clarification. Where did the logarithms disappear? Until the very last moment, we work only with the denominator.

Formulas for logarithms. Logarithms are examples of solutions.

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: log2 7. Since log2 7 ≠ 0, we can cancel the fraction - the denominator will remain 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

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 be given. Then for any number c such that c\u003e 0 and c ≠ 1, the following equality holds:

In particular, if we put c \u003d x, we get:

From the second formula it follows that it is possible to swap the base and the argument of the logarithm, but the whole expression is "reversed", that is, 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 solving 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:

A task. Find the value of the expression: log5 16 log2 25.

Note that the arguments of both logarithms contain exact degrees. Let's take out the indicators: log5 16 \u003d log5 24 \u003d 4log5 2; log2 25 \u003d log2 52 \u003d 2log2 5;

Now let's "flip" the second logarithm:

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.

A task. Find the value of the expression: log9 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:

Now let's get rid of the decimal logarithm by moving to the new base:

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 the exponent in the argument. The 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:.

Like the formulas for the transition to a new base, the basic logarithmic identity is sometimes the only possible solution.

A task. Find the meaning of the expression:

Note that log25 64 \u003d log5 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:

If someone is not in the know, it was a real problem from the exam 🙂

Logarithmic unit and logarithmic zero

logaa \u003d 1 is. Remember once and for all: the logarithm to any base a from this base is equal to one.
loga 1 \u003d 0 is. The base a can be anything, but if the argument is one, the logarithm is zero! Because a0 \u003d 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.

Basic properties of the logarithm

The given properties need to be known, since, on their basis, almost all problems are solved and examples are associated with logarithms. The rest of the exotic properties can be derived by mathematical manipulation of these formulas

1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.

When calculating the formulas for the sum and difference of logarithms (3.4) are encountered quite often. The rest are somewhat complex, but in a number of tasks they are indispensable for simplifying complex expressions and calculating their values.

Common cases of logarithms

Some of the common logarithms are those in which the base is even ten, exponential or two.
The base ten logarithm is usually called the decimal logarithm and is simply denoted lg (x).

From the recording it is clear that the basics are not written in the recording. For example

The natural logarithm is the logarithm based on the exponent (denoted by ln (x)).

The exponent is 2.718281828…. To remember the exponent, you can study the rule: the exponent is 2.7 and twice the year of birth of Leo Nikolaevich Tolstoy. Knowing this rule, you will know both the exact value of the exponent and the date of birth of Leo Tolstoy.

And another important base two logarithm is

The derivative of the logarithm of the function is equal to one divided by the variable

The integral or antiderivative of the logarithm is determined by the dependence

The given material is enough for you to solve a wide class of problems related to logarithms and logarithms. To assimilate the material, I will give only a few common examples from the school curriculum and universities.

Examples for logarithms

Logarithm expressions

Example 1.
and). x \u003d 10ac ^ 2 (a\u003e 0, c\u003e 0).

By properties 3.5 we calculate

2.
By the property of the difference of logarithms, we have

3.
Using properties 3,5 we find

4. Where .

A seemingly complex expression using a number of rules is simplified to the form

Finding the values \u200b\u200bof logarithms

Example 2. Find x if

Decision. For the calculation, we apply to the last term 5 and 13 of the properties

Substitute and grieve

Since the bases are equal, we equate the expressions

Logarithms. First level.

Let the value of the logarithms be given

Evaluate log (x) if

Solution: Let us logarithm the variable to write the logarithm through the sum of terms

This is where the acquaintance with logarithms and their properties just begins. Practice calculations, enrich your practical skills - you will soon need this knowledge to solve logarithmic equations. Having studied the basic methods for solving such equations, we will expand your knowledge for another equally important topic - logarithmic inequalities ...

Basic properties of logarithms

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 bases: logax and logay. Then they can be added and subtracted, and:

logax + logay \u003d loga (x y);
logax - logay \u003d loga (x: y).

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:

A task. Find the value of the expression: log6 4 + log6 9.

Since the bases of the logarithms are the same, we use the sum formula:
log6 4 + log6 9 \u003d log6 (4 9) \u003d log6 36 \u003d 2.

A task. Find the value of the expression: log2 48 - log2 3.

The bases are the same, we use the difference formula:
log2 48 - log2 3 \u003d log2 (48: 3) \u003d log2 16 \u003d 4.

A task. Find the value of the expression: log3 135 - log3 5.

Again the bases are the same, so we have:
log3 135 - log3 5 \u003d log3 (135: 5) \u003d log3 27 \u003d 3.

Removing the exponent from the logarithm

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.

How to solve logarithms

This is what is most often required.

A task. Find the value of the expression: log7 496.

Let's get rid of the degree in the argument using the first formula:
log7 496 \u003d 6 log7 49 \u003d 6 2 \u003d 12

A task. Find the meaning of the expression:

Note that the denominator contains the logarithm, the base and the argument of which are exact powers: 16 \u003d 24; 49 \u003d 72. We have:

Now let's look at the basic fraction. The numerator and denominator contain the same number: log2 7. Since log2 7 ≠ 0, we can cancel the fraction - the denominator will remain 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

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 be given. Then for any number c such that c\u003e 0 and c ≠ 1, the following equality holds:

In particular, if we put c \u003d x, we get:

From the second formula it follows that it is possible to swap the base and the argument of the logarithm, but the whole expression is "reversed", that is, 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 solving 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:

A task. Find the value of the expression: log5 16 log2 25.

Note that the arguments of both logarithms contain exact degrees. Let's take out the indicators: log5 16 \u003d log5 24 \u003d 4log5 2; log2 25 \u003d log2 52 \u003d 2log2 5;

Now let's "flip" the second logarithm:

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.

A task. Find the value of the expression: log9 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:

Now let's get rid of the decimal logarithm by moving to the new base:

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 the exponent in the argument. The 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:.

Like the formulas for the transition to a new base, the basic logarithmic identity is sometimes the only possible solution.

A task. Find the meaning of the expression:

Note that log25 64 \u003d log5 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:

If someone is not in the know, it was a real problem from the exam 🙂

Logarithmic unit and logarithmic zero

logaa \u003d 1 is. Remember once and for all: the logarithm to any base a from this base is equal to one.
loga 1 \u003d 0 is. The base a can be anything, but if the argument is one, the logarithm is zero! Because a0 \u003d 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.