Useful Tips

Division of logarithms with the same base

Our experienced team of editors and researchers contributed to this article and tested it for accuracy and completeness.

A team of content managers carefully monitors the work of editors to ensure that each article meets our high quality standards.

Actions with logarithms may seem rather complicated, but, as with power functions or polynomials, you just need to know the basic rules. There are very few of them: to divide the logarithms with the same base or decompose the logarithm of the quotient, it is enough to use a couple of basic properties of the logarithms.

Logarithm Addition and Subtraction

Consider two logarithms with the same base: log a x and log a y. Then they can be added and subtracted, moreover:

  1. log a x + log a y = log a (x · y),
  2. 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 equal grounds . If the grounds are different, these rules do not work!

These formulas will help to calculate the logarithmic expression even when its individual parts are not counted (see the lesson "What is the logarithm"). Take a look at the examples and see:

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 made up of “bad” logarithms that are not counted separately. But after the transformations, quite normal numbers are obtained. On this fact many tests are built. Yes, control - such expressions in all seriousness (sometimes - almost unchanged) are offered at the exam.

Removing exponent from the logarithm

Now let's complicate the task a bit. What if there is a degree in the base or argument of the logarithm? Then an indicator of this degree can be taken out of the logarithm according to the following rules:

It is easy to see that the last rule follows their first two. But it’s better to remember it all the same - in some cases this will significantly reduce the amount of computation.

Of course, all these rules make sense when observing the ODZ logarithm: a> 0, a ≠ 1, x> 0. And also: learn to apply all formulas not only from left to right, but also vice versa, i.e. you can enter the numbers in front 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 by the first formula:
log 7 49 6 = 6 log 7 49 = 6 2 = 12

Task. Find the value of the expression:

Note that the denominator is the logarithm, the base and argument of which are exact degrees: 16 = 2 4, 49 = 7 2. We have:

[Caption]

I think the last example needs clarification. Where did the logarithms disappear? Until the very last moment, we work only with the denominator. They presented the basis and argument of the logarithm there in the form of degrees and carried out indicators - they received a “three-story” fraction.

Now let's look at the main fraction. The numerator and denominator have the same number: log 2 7. Since log 2 7 ≠ 0, we can reduce the fraction - 2/4 will remain in the denominator. According to the rules of arithmetic, the four can be transferred to the numerator, which was done. The result was the answer: 2.

Transition to a new foundation

Speaking about the rules of addition and subtraction of logarithms, I specifically emphasized that they work only on the same grounds. But what if the grounds 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. We formulate them in the form of a theorem:

Let the logarithm of log a x be given. Then for any number c such that c> 0 and c ≠ 1, the equality

[Caption]

In particular, if we put c = x, we get:

[Caption]

From the second formula it follows that you can swap the base and the argument of the logarithm, but at the same time the whole expression is “flipped”, i.e. the logarithm is in the denominator.

These formulas are rarely found in ordinary numerical terms. It is possible to evaluate how convenient they are only when solving logarithmic equations and inequalities.

However, there are tasks that cannot be solved at all 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. We take out the indicators: log 5 16 = log 5 2 4 = 4log 5 2, log 2 25 = log 2 5 2 = 2log 2 5,

And now, “flip” the second logarithm:

[Caption]

Since the product does not change from the permutation of the factors, we calmly multiplied the four and the two, and then figured out the logarithms.

Task. Find the value of the expression: log 9 100 · log 3.

The base and argument of the first logarithm are exact degrees. We write this and get rid of the indicators:

[Caption]

Now we will get rid of the decimal logarithm, moving to a new base:

[Caption]

Basic logarithmic identity

Often in the process of solving it is required to represent the number as a logarithm for a given basis. In this case, the formulas will help us:

In the first case, the number n becomes an indicator of the degree 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 rephrased definition. This is what it is called: the basic logarithmic identity.

In fact, what happens if the number b is raised to such an extent that the number b in this degree gives the number a? That's right: this is the very number a. Carefully read this paragraph again - many on it "hang."

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

Task. Find the value of the expression:

Note that log 25 64 = log 5 8 - just took out the square from the base and the argument of the logarithm. Given the rules of multiplication of degrees with the same base, we get:

[Caption]

If someone is not in the know, this was a real challenge from the exam :)

Logarithmic unit and logarithmic zero

In conclusion, I will give two identities that can hardly be called properties - rather, these are consequences of the definition of the logarithm. They are constantly found in tasks and, surprisingly, create problems even for “advanced” students.

  1. log a a = 1 is a logarithmic unit. Remember once and for all: the logarithm for any base a from this base itself is equal to one.
  2. log a 1 = 0 is a logarithmic zero. The base a can be anything, but if the argument is 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 applying them in practice! Download the cheat sheet at the beginning of the lesson, print it - and solve problems.

As you know, when multiplying expressions with degrees, their indicators always add up (a b * a c = a b + c). This mathematical law was deduced by Archimedes, and later, in the VIII century, the mathematician Virasen created a table of integer indices. It was they who served for the further discovery of the logarithms. Examples of using this function can be found almost everywhere where it is necessary to simplify the 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 affordable language.

Definition in math

A 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 degree of "c", into which it is necessary to raise the base "a", so that in the end get the value of "b". Let's analyze the logarithm with examples, let's say there is an expression log 2 8. How to find the answer? Very simple, you need to find such a degree that from 2 to the desired degree get 8. Having done some calculations in the mind, we get the number 3! And it is true, because 2 in degree 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 the 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 types of logarithmic expressions:

  1. The natural logarithm of ln a, where the base is the Euler number (e = 2.7).
  2. Decimal a, where the base is the number 10.
  3. The logarithm of any number b at the base a> 1.

Each of them is solved in a standard way, including simplification, reduction and subsequent reduction to a single logarithm using logarithmic theorems. To obtain the correct values ​​of the logarithms, one should remember their properties and the sequence of actions when solving them.

Rules and some restrictions

In mathematics, there are several rules, restrictions, which are accepted as an axiom, that is, are not subject to discussion and are true. For example, it is impossible to divide the numbers by zero, and it is still impossible to extract the root of an even degree 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" should 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" are always equal in any degree to their values,
  • if a> 0, then a b> 0, it turns out that "c" must be greater than zero.

How to solve logarithms?

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

Now let's imagine this expression as a logarithmic. We get log 10 100 = 2. When solving the logarithms, all actions practically converge to find the degree to which you need to enter the base of the logarithm to get the given number.

To accurately determine the value of an unknown degree, you must learn to work with a table of degrees. It looks like this:

As you can see, some degree indicators can be guessed intuitively if there is a technical mentality and knowledge of the multiplication table. However, for large values, a degree table is required. Even those who do not understand anything at all in complex mathematical topics can use it. The left column shows the numbers (base a), the top row of numbers is the value of degree c to which the number a is raised. At the intersection, the values ​​of numbers that are the answer (a c = b) are defined in the cells. 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 real humanities will understand!

Equations and Inequalities

It turns out that under certain conditions the exponent is the logarithm. Therefore, any mathematical numerical expressions can be written in the form of a logarithmic equality. For example, 3 4 = 81 can be written as the logarithm of 81 on base 3, which is four (log 3 81 = 4). For negative degrees, the rules are the same: 2 -5 = 1/32 we write in the form of a logarithm, we get log 2 (1/32) = -5. One of the most fascinating sections of mathematics is the theme of "logarithms." We will consider examples and solutions of equations just below, immediately after studying their properties. Now let's look at how inequalities look and how to distinguish them from equations.

An expression of the following form is given: log 2 (x-1)> 3 - it is a logarithmic inequality, since the unknown value of "x" is under the sign of the logarithm. And also in the expression two quantities are compared: the logarithm of the desired number on the basis of two is greater than the number three.

The most important difference between logarithmic equations and inequalities is that equations with logarithms (for example, the logarithm of 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 result, the answer is not a simple set of individual numbers as in the answer of the equation, but a continuous series or set of numbers.

Basic logarithm theorems

When solving primitive tasks on finding 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 put into 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.

  1. The basic identity looks like this: logaB = B. It applies only when a is greater than 0, not equal to one, and B is greater than zero.
  2. 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. Moreover, a prerequisite is: d, s 1 and s 2> 0, and ≠ 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 get that s 1 * s 2 = a f1 * a f2 = a f1 + f2 (properties of degrees ), and then by definition: log a (s 1 * s 2) = f 1 + f 2 = log a s1 + log as 2, as required.
  3. The logarithm of the quotient looks like this: log a (s 1 / s 2) = log a s 1 - log a s 2.
  4. A 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 regular postulates. Let's look at the proof.

Let log a b = t, it turns out a t = b. If both parts are raised to the power m: a tn = b n,

but since a tn = (a q) nt / q = b n, therefore 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 problems on the topic of logarithms are examples of equations and inequalities. They are found in almost all problem books, and are also included in the required part of math exams. To enter the university or take entrance examinations in mathematics, you need to know how to correctly solve such problems.

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 need to find out whether it is possible to simplify the expression or lead to a general view. Long logarithmic expressions can be simplified if their properties are used correctly. Let's get to know them soon.

When solving the logarithmic equations, it is necessary to determine what kind of logarithm is in front of us: an example of an expression may contain a natural logarithm or a decimal.

Here are examples of ln100, ln1026. Their solution is reduced to the fact that it is necessary to determine the degree to which the base 10 will be equal to 100 and 1026, respectively. For solutions of natural logarithms, one needs to apply logarithmic identities or their properties. Let's look at examples of solving logarithmic problems of various types.

How to use logarithm formulas: with examples and solutions

So, let's look at examples of using the basic theorems on logarithms.

  1. The property of the logarithm of the product can be used in tasks where it is necessary to decompose the 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.
  2. log 4 8 = log 2 2 2 3 = 3/2 log 2 2 = 1.5 - as you can see, using the fourth property of the degree of the logarithm, we managed to solve at first glance a complex and unsolvable expression. It is only necessary to factor the basis and then derive the degree from 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 implies an accurate and perfect knowledge of the topic "Natural Logarithms".

Examples and solutions to problems are taken from the official exams. Let's see how such tasks are solved.

Given log 2 (2x-1) = 4. Solution:
we rewrite the expression, simplifying it a bit 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.

  • All logarithms are best reduced to the same base so that the solution is not cumbersome and confusing.
  • The whole expression under the logarithm sign is indicated as positive, therefore, when the factor makes the exponent of the expression, which stands under the sign of the logarithm and as its basis, the expression remaining under the logarithm should be positive.

    Check if negative numbers or one are under the logarithm sign. This method is applicable to expressions of the form log b ⁡ (x) log b ⁡ (a) < displaystyle < frac < log _(x)> < log _(a) >>>. However, it is not suitable for some special cases:

    • The logarithm of a negative number is not defined for any reason (e.g. log ⁡ (- 3) < displaystyle log (-3)> or log 4 ⁡ (- 5) < displaystyle log _ <4> (-5)>) . In this case, write "no solution."
    • The logarithm of zero for any reason is also not defined. If you come across ln ⁡ (0) < displaystyle ln (0)>, write down "no solution".
    • The logarithm of a unit for any reason (log ⁡ (1) < displaystyle log (1)>) is always zero, since x 0 = 1 < displaystyle x ^ <0> = 1> for all values x . Write instead of logarithm 1 and do not use the method below.
    • If the logarithms have different bases, for example l o g 3 (x) l o g 4 (a) < displaystyle < frac (x)>(a)>>> , и не сводятся к целым числам, значение выражения нельзя найти вручную.

    Преобразуйте выражение в один логарифм. Если выражение не относится к приведенным выше особым случаям, его можно представить в виде одного логарифма. Используйте для этого следующую формулу: log b ⁡ (x) log b ⁡ (a) = log a ⁡ (x) <displaystyle <frac <log _(x)><log _(a)>>=log _(x)> .

    • Пример 1: рассмотрим выражение log ⁡ 16 log ⁡ 2 <displaystyle <frac <log <16>><log <2>>>> .
      Для начала представим выражение в виде одного логарифма с помощью приведенной выше формулы: log ⁡ 16 log ⁡ 2 = log 2 ⁡ (16) <displaystyle <frac <log <16>><log <2>>>=log _<2>(16)> .
    • Эта формула "замены основания" логарифма выводится из основных свойств логарифмов.

    • Пример 1 (продолжение): Перепишите в виде 2 ? = 16 <displaystyle 2^<?>=16> . Необходимо найти, какое число должно стоять вместо знака "?". Это можно сделать методом проб и ошибок:
      2 2 = 2 ∗ 2 = 4 < displaystyle 2 ^ <2> = 2 * 2 = 4>
      2 3 = 4 ∗ 2 = 8 < displaystyle 2 ^ <3> = 4 * 2 = 8>
      2 4 = 8 ∗ 2 = 16 < displaystyle 2 ^ <4> = 8 * 2 = 16>
      So, the desired number is 4: log 2 ⁡ (16) < displaystyle log _ <2> (16)> = 4 .

    Leave the answer in a logarithmic form if you are unable to simplify it. Many logarithms are very difficult to calculate manually. In this case, in order to get an accurate answer, you need a calculator. However, if you solve the task in the lesson, then the teacher will most likely satisfy the answer in a logarithmic form. Below, the method considered is used to solve a more complex example:

    • example 2: what is log 3 ⁡ (58) log 3 ⁡ (7) < displaystyle < frac < log _ <3> (58)> < log _ <3> (7) >>>?
    • We transform this expression into one logarithm: log 3 ⁡ (58) log 3 ⁡ (7) = log 7 ⁡ (58) < displaystyle < frac < log _ <3> (58)> < log _ <3> (7) >> = log _ <7> (58)>. Note that the common base 3 for both logarithms disappears, this is true for any base.
    • Rewrite the expression as 7? = 58 < displaystyle 7 ^ <?> = 58> and try to find the value ?:
      7 2 = 7 ∗ 7 = 49 < displaystyle 7 ^ <2> = 7 * 7 = 49>
      7 3 = 49 ∗ 7 = 343 < displaystyle 7 ^ <3> = 49 * 7 = 343>
      Since 58 is between the two numbers, it is not expressed as an integer.
    • We leave the answer in the logarithmic form: log 7 ⁡ (58) < displaystyle log _ <7> (58)>.

Comparison of logarithms (inequalities)

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

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

  • If a> 0, then f (x)> g (x)> 0
  • If 0 How to solve problems with logarithms: examples

Logarithm Jobs included in the exam 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. You can find all examples through site search.

What is the logarithm

Logarithms have always been considered a complex topic in a school math course. There are many different definitions of the logarithm, but most textbooks for some reason use the most complex and unsuccessful of them.

We will determine the logarithm simply and clearly. To do this, compile a table:

So, before us are powers of two.

Logarithms - properties, formulas, how to solve

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

And now - in fact, the definition of the logarithm:

on the basis of a from argument x, this is the degree to which the number a must be raised to obtain the number x.

Designation: log a x = b, where a is the base, x is the argument, b is actually what the logarithm is.

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

The operation of finding the logarithm of a number on a given basis is called. So, we supplement our table with a new line:

2 12 22 32 42 52 6
248163264
log 2 2 = 1log 2 4 = 2log 2 8 = 3log 2 16 = 4log 2 32 = 5log 2 64 = 6

Unfortunately, not all logarithms are considered so easy. For example, try to find log 2 5. The number 5 is not in the table, but the logic suggests that the logarithm will lie somewhere on the segment. Because 2 2 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:

  1. The argument and the base must always be greater than zero. This follows from determining the degree of a rational indicator, to which the definition of the logarithm is reduced.
  2. The base must be different from one, because the unit remains to any extent one. Because of this, the question "to what degree must a unit be raised to get a deuce" is meaningless. There is no such degree!

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

Note that there are no restrictions on the number b (the logarithm value). For example, the logarithm may well be negative: log 2 0.5 = −1, because 0.5 = 2 −1.

However, now we are considering only numerical expressions, where it is not required to know the logistic linear differential equation. All restrictions are already taken into account by the drafters of the tasks. But when the logarithmic equations and inequalities go, the requirements of ODZ will become mandatory. After all, the basis and argument can be quite non-weak constructions, which do not necessarily correspond to the above restrictions.

Now consider the general scheme for calculating logarithms. It consists of three steps:

  1. Represent the base a and the argument x as a power with the smallest possible base greater than one. Along the way, it’s better to get rid of decimal fractions,
  2. Solve the equation for variable b: x = a b,
  3. The resulting number b will be the answer.

That's all! If the logarithm turns out to be irrational, this will be seen already in the first step. The requirement that the base be more than one is very relevant: this reduces the likelihood of error and greatly simplifies calculations. Similarly with decimal fractions: if you immediately translate them into regular fractions, there will be many times fewer errors.

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

Task. Calculate the logarithm: log 5 25

  1. We represent the base and the argument as the degree of the five: 5 = 5 1, 25 = 5 2,

We compose and solve the equation:
log 5 25 = b ⇒ (5 1) b = 5 2 ⇒5 b = 5 2 ⇒ b = 2,

Task. Calculate the logarithm: log 4 64

  1. We represent the base and the argument as a power of two: 4 = 2 2, 64 = 2 6,
  2. We compose and solve the equation:
    log 4 64 = b ⇒ (2 2) b = 2 6 ⇒2 2b = 2 6 ⇒2b = 6 ⇒ b = 3,
  3. Received the answer: 3.

Task. Calculate the logarithm: log 16 1

  1. We represent the base and the argument as a power of two: 16 = 2 4, 1 = 2 0,
  2. We compose and solve the equation:
    log 16 1 = b ⇒ (2 4) b = 2 0 ⇒2 4b = 2 0 ⇒4b = 0 ⇒ b = 0,
  3. Received the answer: 0.

Task. Calculate the logarithm: log 7 14

  1. We represent the base and the argument as a power of seven: 7 = 7 1, 14 as a power of seven is not represented, since 7 1 The decimal logarithm

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

from argument x is the base 10 logarithm, i.e. the power to raise the number 10 to get the number x. Designation: log x.

For example, log 10 = 1, log 100 = 2, log 1000 = 3 - etc.

From now on, when a phrase like “Find lg 0.01” is found in a textbook, be aware that this is not a typo. This is the decimal logarithm. However, if you are unfamiliar with this notation, you can always rewrite it:
log x = log 10 x

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

Natural logarithm

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

from argument x is the base logarithm of e, i.e. the degree to which the number e must be raised to obtain 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 of it:
e = 2.718281828459 ...

We will not go deep into what this number is and why it is necessary. Just remember that e is the base of the natural logarithm:
ln x = log e x

Thus, ln e = 1, ln e 2 = 2, ln e 16 = 16 - and so on. 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 = 0.

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

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

How to represent a number as a logarithm?

We use the definition of the logarithm.

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

Thus, in order to represent a certain number c as a logarithm on the base of a, it is necessary to put a degree under the sign of the logarithm with the same base as the base of the logarithm, and write down the number c in the exponent:

In the form of a logarithm, you can imagine absolutely any number - positive, negative, integer, fractional, rational, irrational:

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

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

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

We have two numbers - 2 and 3. These numbers are the base and the end