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Logarithms: Examples and Solutions


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Logarithms, like any numbers, can be added, subtracted and converted in every way. But since logarithms are not quite ordinary numbers, there are rules that are called basic properties.

You must 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.

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:

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:
log6 4 + log6 9 = log6 (4 · 9) = log6 36 = 2.

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

Task. Find the value of the expression: log3 135 - log3 5.

Again, the bases are the same, so we have:
log3 135 - log3 5 = log3 (135: 5) = log3 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:

  1. log a x n = n a x

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: log7 49 6 .

Let's get rid of the degree in the argument by the first formula:
log7 49 6 = 67 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:


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: log2 7. Since log2 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. Then for any number c such that c> 0 and c ≠ 1, the equality


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


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: log5 16 log2 25.

Note that the arguments of both logarithms contain exact degrees. We will take out indicators: log5 16 = log5 2 4 = 4log5 2, log2 25 = log2 5 2 = 2log2 5,

And now, “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 figured out the logarithms.

Task. Find the value of the expression: log9 100 lg 3.

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


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


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:

  1. n = log a a n

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. It is called:.

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 log25 64 = log5 8 - just made a square from the base and the argument of the logarithm. Given the rules of multiplication of degrees with the same base, we get:


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 this. 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 this. 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.

Definition in math

A logarithm is an expression of the following form: logab = 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" to which the base "a" must be raised in order to finally get the value "b". Let's analyze the logarithm with examples, let's say there is an expression log28. 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. The decimal logarithm is log 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 such a degree, raising to which the number ten, we get 100. This, of course, is a quadratic degree! 10 2 = 100.

Now let's imagine this expression as a logarithmic. We get log10100 = 2. When solving logarithms, all actions almost converge to find the degree to which you need to enter the base of the logarithm to get a 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 at base 3, which is four (log381 = 4). For negative degrees, the rules are the same: 2 -5 = 1/32 we write in the form of a logarithm, we get log2 (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 is given as follows: log2(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 (an example is the logarithm2x = √9) imply one or more specific numerical values ​​in the answer, while solving the inequality determines both the region of admissible values ​​and the break points of 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: logd(s1* s2) = logds1 + logds2. In this case, a prerequisite is: d, s1 and s2 > 0, and ≠ 1. You can give a proof for this formula of logarithms, with examples and a solution. Let logas1 = f1 and logas2 = f2then a f1 = s1, a f2 = s2. We get that s1* s2 = a f1 * a f2 = a f1 + f2 (properties of degrees), and then by definition: loga(s1* s2) = f1+ f2 = logas1 + logas2, Q.E.D.
  3. The logarithm of a private looks like this: loga(s1/s2) = logas1- logas2.
  4. A theorem in the form of a formula takes the following form: loga q b n = n / q logab.

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 logab = 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, loga q b n = (n * t) / t, then loga q b n = n / q logab. 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 decimal logarithms: 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, log24 + log2128 = log2(4 * 128) = log2512. The answer is 9.
  2. log48 = log2 2 2 3 = 3/2 log22 = 1.5 - as you can see, using the fourth property of the degree of the logarithm, it was possible 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 log2(2x-1) = 4. Solution:
rewrite the expression by simplifying it a bit log2(2x-1) = 2 2, by the definition of the logarithm we get that 2x-1 = 2 4, therefore 2x = 17, x = 8.5.

Below are a few recommendations, following which you can easily solve all equations containing expressions that are under the sign of the logarithm.

  • 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 logarithm and as its basis, the expression remaining under the logarithm should be positive.