The logarithm of a positive number b to base a (a>0, a is not equal to 1) is a number c such that a c = b: log a b = c ⇔ a c = b (a > 0, a ≠ 1, b > 0)
Note that the logarithm of a non-positive number is undefined. In addition, the base of the logarithm must be a positive number that is not equal to 1. For example, if we square -2, we get the number 4, but this does not mean that the base -2 logarithm of 4 is equal to 2.
Basic logarithmic identity
a log a b = b (a > 0, a ≠ 1) (2)It is important that the scope of definition of the right and left sides of this formula is different. The left side is defined only for b>0, a>0 and a ≠ 1. The right side is defined for any b, and does not depend on a at all. Thus, the application of the basic logarithmic “identity” when solving equations and inequalities can lead to a change in the OD.
Two obvious consequences of the definition of logarithm
log a a = 1 (a > 0, a ≠ 1) (3)log a 1 = 0 (a > 0, a ≠ 1) (4)
Indeed, when raising the number a to the first power, we get the same number, and when raising it to the zero power, we get one.
Logarithm of the product and logarithm of the quotient
log a (b c) = log a b + log a c (a > 0, a ≠ 1, b > 0, c > 0) (5)Log a b c = log a b − log a c (a > 0, a ≠ 1, b > 0, c > 0) (6)
I would like to warn schoolchildren against thoughtlessly applying these formulas when solving logarithmic equations and inequalities. When using them “from left to right,” the ODZ narrows, and when moving from the sum or difference of logarithms to the logarithm of the product or quotient, the ODZ expands.
Indeed, the expression log a (f (x) g (x)) is defined in two cases: when both functions are strictly positive or when f (x) and g (x) are both less than zero.
Transforming this expression into the sum log a f (x) + log a g (x), we are forced to limit ourselves only to the case when f(x)>0 and g(x)>0. There is a narrowing of the range of acceptable values, and this is categorically unacceptable, since it can lead to a loss of solutions. A similar problem exists for formula (6).
The degree can be taken out of the sign of the logarithm
log a b p = p log a b (a > 0, a ≠ 1, b > 0) (7)And again I would like to urge caution. Consider the following example:
Log a (f (x) 2 = 2 log a f (x)
The left side of the equality is obviously defined for all values of f(x) except zero. The right side is only for f(x)>0! By taking the degree out of the logarithm, we again narrow the ODZ. The reverse procedure leads to an expansion of the range of acceptable values. All these remarks apply not only to power 2, but also to any even power.
Formula for moving to a new foundation
log a b = log c b log c a (a > 0, a ≠ 1, b > 0, c > 0, c ≠ 1) (8)That rare case when the ODZ does not change during transformation. If you have chosen base c wisely (positive and not equal to 1), the formula for moving to a new base is completely safe.
If we choose the number b as the new base c, we obtain an important special case of formula (8):
Log a b = 1 log b a (a > 0, a ≠ 1, b > 0, b ≠ 1) (9)
Some simple examples with logarithms
Example 1. Calculate: log2 + log50.
Solution. log2 + log50 = log100 = 2. We used the sum of logarithms formula (5) and the definition of the decimal logarithm.
Example 2. Calculate: lg125/lg5.
Solution. log125/log5 = log 5 125 = 3. We used the formula for moving to a new base (8).
Table of formulas related to logarithms
| a log a b = b (a > 0, a ≠ 1) |
| log a a = 1 (a > 0, a ≠ 1) |
| log a 1 = 0 (a > 0, a ≠ 1) |
| log a (b c) = log a b + log a c (a > 0, a ≠ 1, b > 0, c > 0) |
| log a b c = log a b − log a c (a > 0, a ≠ 1, b > 0, c > 0) |
| log a b p = p log a b (a > 0, a ≠ 1, b > 0) |
| log a b = log c b log c a (a > 0, a ≠ 1, b > 0, c > 0, c ≠ 1) |
| log a b = 1 log b a (a > 0, a ≠ 1, b > 0, b ≠ 1) |
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Maintaining your privacy is important to us. For this reason, we have developed a Privacy Policy that describes how we use and store your information. Please review our privacy practices and let us know if you have any questions.
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Logarithm of the number b (b > 0) to base a (a > 0, a ≠ 1)– exponent to which the number a must be raised to obtain b.
The base 10 logarithm of b can be written as log(b), and the logarithm to 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 > 0, a ≠ 1, x > 0 and y > 0.
Property 1. Logarithm of the product
Logarithm of the product equal to the sum logarithms:
log a (x ⋅ y) = log a x + log a y
Property 2. Logarithm of the quotient
Logarithm of the quotient equal to the difference of logarithms:
log a (x / y) = log a x – log a y
Property 3. Logarithm of power
Logarithm of degree equal to the product of the power and the logarithm:
If the base of the logarithm is in the power, then another formula applies:
Property 4. Logarithm of the root
This property can be obtained from the property of the logarithm of a power, since the nth root of the power is equal to the power of 1/n:
Formula for converting from a logarithm in one base to a logarithm in another base
This formula is also often used to solve various tasks to logarithms:
Special case:
Comparing logarithms (inequalities)
Let us 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 need to first look at the base of the logarithms a:
- If a > 0, then f(x) > g(x) > 0
- If 0< a < 1, то 0 < f(x) < g(x)
How to solve problems with logarithms: examples
Problems with logarithms included in the Unified State Exam in mathematics for grade 11 in task 5 and task 7, you can find tasks with solutions on our website in the appropriate sections. Also, tasks with logarithms are found in the math task bank. You can find all examples by searching the site.
What is a logarithm
Logarithms have always been considered a difficult topic in school course mathematics. There are many different definitions of logarithm, but for some reason most textbooks use the most complex and unsuccessful of them.
We will define the logarithm simply and clearly. To do this, let's create a table:
So, we have powers of two.
Logarithms - properties, formulas, how to solve
If you take the number from the bottom line, you can easily find the power to which you will have to raise 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:
the base a of the argument x is the power 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 what the logarithm is actually equal to.
For example, 2 3 = 8 ⇒log 2 8 = 3 (the base 2 logarithm of 8 is three because 2 3 = 8). With the same success, log 2 64 = 6, since 2 6 = 64.
The operation of finding the logarithm of a number to 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 = 1 | log 2 4 = 2 | log 2 8 = 3 | log 2 16 = 4 | log 2 32 = 5 | log 2 64 = 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 interval. Because 2 2< 5 < 2 3 , а чем more degree twos, the larger the number.
Such numbers are called irrational: the numbers after the decimal point can be written ad infinitum, 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 a logarithm is an expression with two variables (the base and the argument). At first, many people confuse where the basis is and where the argument is. To avoid annoying misunderstandings, just look at the picture:
Before us is nothing more than the definition of a logarithm. Remember: logarithm is a power, into which the base must be built in order to obtain an argument. It is the base that is raised to a power - it is highlighted in red in the picture. It turns out that the base is always at the bottom! I tell my students this wonderful rule at the very first lesson - and no confusion arises.
How to count logarithms
We've figured out the definition - all that remains is 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:
- The argument and the base must always be greater than zero. This follows from the definition of a degree by a rational exponent, to which the definition of a logarithm is reduced.
- The base must be different from one, since one to any degree still remains one. Because of this, the question “to what power must one be raised to get two” is meaningless. There is no such degree!
Such restrictions are called range of acceptable values(ODZ). 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 value of the logarithm). 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 VA of the logarithm. All restrictions have already been taken into account by the authors of the problems. But when logarithmic equations and inequalities come into play, DL requirements will become mandatory. After all, the basis and argument may contain 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:
- Express the base a and the argument x as a power with the minimum possible base greater than one. Along the way, it’s better to get rid of decimals;
- Solve the equation for variable b: x = a b ;
- The resulting number b will be the answer.
That's all! If the logarithm turns out to be irrational, this will be visible already in the first step. The requirement that the base be greater than one is very important: this reduces the likelihood of error and greatly simplifies the calculations. Same with decimals: if you immediately convert them to regular ones, there will be many fewer errors.
Let's see how this scheme works using specific examples:
Task. Calculate the logarithm: log 5 25
- Let's imagine the base and argument as a power of five: 5 = 5 1 ; 25 = 5 2 ;
- We received the answer: 2.
Let's create and solve the equation:
log 5 25 = b ⇒(5 1) b = 5 2 ⇒5 b = 5 2 ⇒ b = 2;
Task. Calculate the logarithm:
Task. Calculate the logarithm: log 4 64
- Let's imagine the base and argument as a power of two: 4 = 2 2 ; 64 = 2 6 ;
- Let's create and solve the equation:
log 4 64 = b ⇒(2 2) b = 2 6 ⇒2 2b = 2 6 ⇒2b = 6 ⇒ b = 3; - We received the answer: 3.
Task. Calculate the logarithm: log 16 1
- Let's imagine the base and argument as a power of two: 16 = 2 4 ; 1 = 2 0 ;
- Let's create and solve the equation:
log 16 1 = b ⇒(2 4) b = 2 0 ⇒2 4b = 2 0 ⇒4b = 0 ⇒ b = 0; - We received the answer: 0.
Task. Calculate the logarithm: log 7 14
- Let's imagine the base and argument as a power of seven: 7 = 7 1 ; 14 cannot be represented as a power of seven, since 7 1< 14 < 7 2 ;
- From the previous paragraph it follows that the logarithm does not count;
- The answer is no change: log 7 14.
A small note on the last example. How can you be sure that a number is not an exact power of another number? It's very simple - just break it down into prime factors. If the expansion has at least two different factors, the number is not an exact power.
Task. Find out whether the numbers are exact powers: 8; 48; 81; 35; 14.
8 = 2 · 2 · 2 = 2 3 - exact degree, because there is only one multiplier;
48 = 6 · 8 = 3 · 2 · 2 · 2 · 2 = 3 · 2 4 - is not an exact power, since there are two factors: 3 and 2;
81 = 9 · 9 = 3 · 3 · 3 · 3 = 3 4 - exact degree;
35 = 7 · 5 - again not an exact power;
14 = 7 · 2 - again not an exact degree;
Let us also note that we ourselves prime numbers are always exact degrees of themselves.
Decimal logarithm
Some logarithms are so common that they have a special name and symbol.
of the argument x is the logarithm to base 10, i.e. The power to which the number 10 must be raised to obtain the number x. Designation: lg x.
For example, log 10 = 1; lg 100 = 2; lg 1000 = 3 - etc.
From now on, when a phrase like “Find lg 0.01” appears in a textbook, know that this is not a typo. This is a 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 logarithms.
Natural logarithm
There is another logarithm that has its own designation. In some ways, it's even more important than decimal. We are talking about the natural logarithm.
of the argument x is the logarithm to base e, i.e. the power to which the number e must be raised to obtain the number x. Designation: ln x.
Many will ask: what is the number e? This is an irrational number, its exact value impossible to find and record. I will give only the first figures:
e = 2.718281828459…
We will not go into detail about what this number is and why it is needed. 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 - 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, for one: ln 1 = 0.
For natural logarithms all the rules that are true for ordinary logarithms are valid.
See also:
Logarithm. Properties of the logarithm (power of the logarithm).
How to represent a number as a logarithm?
We use the definition of logarithm.
A logarithm is an exponent to which the base must be raised to obtain the number under the logarithm sign.
Thus, in order to represent a certain number c as a logarithm to base a, you need to put a power with the same base as the base of the logarithm under the sign of the logarithm, and write this number c as the exponent:
Absolutely any number can be represented as a logarithm - positive, negative, integer, fractional, rational, irrational:
In order not to confuse a and c under stressful conditions of a test or exam, you can use the following memorization rule:
what is below goes down, what is above goes up.
For example, you need 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 should be written down, to the base of the power, and which – up, to the exponent.
The base 3 in the notation of a logarithm is at the bottom, which means that when we represent two as a logarithm to the base 3, we will also write 3 down to the base.
2 is higher than three. And in notation of the degree two we write above the three, that is, as an exponent:
Logarithms. First level.
Logarithms
Logarithm positive number b based on a, Where a > 0, a ≠ 1, is called the exponent to which the number must be raised a, To obtain b.
Definition of logarithm can be briefly written like this:
This equality is valid for b > 0, a > 0, a ≠ 1. It is usually called logarithmic identity.
The action of finding the logarithm of a number is called by logarithm.
Properties of logarithms:
Logarithm of the product:
Logarithm of the quotient:
Replacing the logarithm base:
Logarithm of degree:
Logarithm of the root:
Logarithm with power base:
Decimal and natural logarithms.
Decimal logarithm numbers call the logarithm of this number to base 10 and write lg b
Natural logarithm numbers are called the logarithm of that number to the base e, Where e- an irrational number approximately equal to 2.7. At the same time they write ln b.
Other notes on algebra and geometry
Basic properties of logarithms
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 main properties.
You definitely need to know these rules - without them, not a single serious logarithmic problem can be solved. In addition, there are very few of them - you can learn everything in one day. So let's get started.
Adding and subtracting 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 = log a (x y);
- log a x − log a y = log a (x: y).
So, the sum of logarithms is equal to the logarithm of the product, and the difference is equal to 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 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 logarithms have the same bases, 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, which are not calculated separately. But after the transformations, completely normal numbers are obtained. Many are built on this fact test papers. Yes, test-like expressions are offered in all seriousness (sometimes with virtually no changes) on the Unified State Examination.
Extracting the exponent from the logarithm
Now let's complicate the task a little. What if the base or argument of a logarithm is a power? Then the exponent of this degree can be taken out of the sign of the logarithm according to the following rules:
It is easy to see that the last rule follows the first two. But it’s better to remember it anyway - in some cases it will significantly reduce the amount of calculations.
Of course, all these rules make sense if the ODZ of the logarithm is observed: a > 0, a ≠ 1, x > 0. And one more thing: learn to apply all formulas not only from left to right, but also vice versa, i.e. You can enter the numbers before the logarithm sign into the logarithm itself.
How to solve logarithms
This is what is most often required.
Task. Find the value of the expression: log 7 49 6 .
Let's get rid of the degree in the argument using the first formula:
log 7 49 6 = 6 log 7 49 = 6 2 = 12
Task. Find the meaning of the expression:
Note that the denominator contains a logarithm, the base and argument of which are exact powers: 16 = 2 4 ; 49 = 7 2. We have:
I think the last example requires some clarification. Where have logarithms gone? Until the very last moment we work only with the denominator. We presented the base and argument of the logarithm standing there in the form of powers and took out the exponents - we got a “three-story” fraction.
Now let's look at the main fraction. The numerator and denominator contain 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 is what was done. The result was the answer: 2.
Transition to a new foundation
Speaking about the rules for adding and subtracting logarithms, I specifically emphasized that they only work with the same bases. What if the reasons are different? What if they are not exact powers of the same number?
Formulas for 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 > 0 and c ≠ 1, the equality is true:
In particular, if we set c = x, we get:
From the second formula it follows that the base and argument of the logarithm can be swapped, but in this case the entire expression is “turned over”, i.e. the logarithm appears in the denominator.
These formulas are rarely found in conventional numerical expressions. It is possible to evaluate how convenient they are only when solving logarithmic equations and inequalities.
However, there are problems that cannot be solved at all except by moving to a new foundation. Let's look at 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 powers. Let's take out the indicators: log 5 16 = log 5 2 4 = 4log 5 2; log 2 25 = log 2 5 2 = 2log 2 5;
Now let’s “reverse” the second logarithm:
Since the product does not change when rearranging factors, we calmly multiplied four and two, and then dealt with logarithms.
Task. Find the value of the expression: log 9 100 lg 3.
The base and argument of the first logarithm are exact powers. Let's write this down and get rid of the indicators:
Now let's get rid of the decimal logarithm by moving to a new base:
Basic logarithmic identity
Often in the solution process it is necessary to represent a number as a logarithm to a given base.
In this case, the following 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 a logarithm value.
The second formula is actually a paraphrased definition. That's what it's called: .
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: the result is the same number a. Read this paragraph carefully again - many people get stuck on it.
Like formulas for moving to a new base, the basic logarithmic identity is sometimes the only possible solution.
Task. Find the meaning of the expression:
Note that log 25 64 = log 5 8 - we simply took the square from the base and argument of the logarithm. Considering the rules for multiplying powers with the same basis, we get:
If anyone doesn’t know, this was a real task from the Unified State 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 constantly appear in problems and, surprisingly, create problems even for “advanced” students.
- log a a = 1 is. Remember once and for all: the logarithm to any base a of that base itself is equal to one.
- log a 1 = 0 is. The base a can be anything, but if the argument contains one, the logarithm is equal to zero! Because a 0 = 1 is a direct consequence of the definition.
That's all the properties. Be sure to practice putting them into practice! Download the cheat sheet at the beginning of the lesson, print it out, and solve the problems.
Today we will talk about logarithmic formulas and give indicative solution examples.
They themselves imply solution patterns according to the basic properties of logarithms. Before applying logarithmic formulas to solve, let us remind you of all the properties:
Now, based on these formulas (properties), we will show examples of solving logarithms.
Examples of solving logarithms based on formulas.
Logarithm a positive number b to base a (denoted by log a b) is an exponent to which a must be raised to obtain b, with b > 0, a > 0, and 1.
According to the definition, log a b = x, which is equivalent to a x = b, therefore log a a x = x.
Logarithms, examples:
log 2 8 = 3, because 2 3 = 8
log 7 49 = 2, because 7 2 = 49
log 5 1/5 = -1, because 5 -1 = 1/5
Decimal logarithm- this is an ordinary logarithm, the base of which is 10. It is denoted as lg.
log 10 100 = 2, because 10 2 = 100
Natural logarithm- also an ordinary logarithm, a logarithm, but with the base e (e = 2.71828... - an irrational number). Denoted as ln.
It is advisable to memorize the formulas or properties of logarithms, because we will need them later when solving logarithms, logarithmic equations and inequalities. Let's work through each formula again with examples.
- Basic logarithmic identity
a log a b = b8 2log 8 3 = (8 2log 8 3) 2 = 3 2 = 9
- The logarithm of the product is equal to the sum of the logarithms
log a (bc) = log a b + log a clog 3 8.1 + log 3 10 = log 3 (8.1*10) = log 3 81 = 4
- The logarithm of the quotient is equal to the difference of the logarithms
log a (b/c) = log a b - log a c9 log 5 50 /9 log 5 2 = 9 log 5 50- log 5 2 = 9 log 5 25 = 9 2 = 81
- Properties of the power of a logarithmic number and the base of the logarithm
Exponent of the logarithmic number log a b m = mlog a b
Exponent of the base of the logarithm log a n b =1/n*log a b
log a n b m = m/n*log a b,
if m = n, we get log a n b n = log a b
log 4 9 = log 2 2 3 2 = log 2 3
- Transition to a new foundation
log a b = log c b/log c a,if c = b, we get log b b = 1
then log a b = 1/log b a
log 0.8 3*log 3 1.25 = log 0.8 3*log 0.8 1.25/log 0.8 3 = log 0.8 1.25 = log 4/5 5/4 = -1
As you can see, the formulas for logarithms are not as complicated as they seem. Now, having looked at examples of solving logarithms, we can move on to logarithmic equations. We will look at 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 a different class of education and study abroad as an option.