(from Greek λόγος - “word”, “relation” and ἀριθμός - “number”) numbers b based on a(log α b) is called such a number c, And b= a c, that is, records log α b=c And b=ac are equivalent. The logarithm makes sense if a > 0, a ≠ 1, b > 0.
In other words logarithm numbers b based on A formulated as an exponent to which a number must be raised a to get the number b(logarithm exists only for positive numbers).
From this formulation it follows that the calculation x= log α b, is equivalent to solving the equation a x =b.
For example:
log 2 8 = 3 because 8 = 2 3 .
Let us emphasize that the indicated formulation of the logarithm makes it possible to immediately determine logarithm value, when the number under the logarithm sign acts as a certain power of the base. Indeed, the formulation of the logarithm makes it possible to justify that if b=a c, then the logarithm of the number b based on a equals With. It is also clear that the topic of logarithms is closely related to the topic powers of a number.
Calculating the logarithm is called logarithm. Logarithm is the mathematical operation of taking a logarithm. When taking logarithms, products of factors are transformed into sums of terms.
Potentiation is the inverse mathematical operation of logarithm. During potentiation, a given base is raised to the degree of expression over which potentiation is performed. In this case, the sums of terms are transformed into a product of factors.
Quite often, real logarithms are used with bases 2 (binary), Euler's number e ≈ 2.718 (natural logarithm) and 10 (decimal).
At this stage it is advisable to consider logarithm samples log 7 2 , ln √ 5, lg0.0001.
And the entries lg(-3), log -3 3.2, log -1 -4.3 do not make sense, since in the first of them a negative number is placed under the sign of the logarithm, in the second there is a negative number in the base, and in the third there is a negative number under the logarithm sign and unit at the base.
Conditions for determining the logarithm.
It is worth considering separately the conditions a > 0, a ≠ 1, b > 0.under which we get definition of logarithm. Let's consider why these restrictions were taken. An equality of the form x = log α will help us with this b, called the basic logarithmic identity, which directly follows from the definition of logarithm given above.
Let's take the condition a≠1. Since one to any power is equal to one, then the equality x=log α b can only exist when b=1, but log 1 1 will be any real number. To eliminate this ambiguity, we take a≠1.
Let us prove the necessity of the condition a>0. At a=0 according to the formulation of the logarithm can exist only when b=0. And accordingly then log 0 0 can be any non-zero real number, since zero to any non-zero power is zero. This ambiguity can be eliminated by the condition a≠0. And when a<0 we would have to reject the analysis of rational and irrational values of the logarithm, since a degree with a rational and irrational exponent is defined only for non-negative bases. It is for this reason that the condition is stipulated a>0.
And the last condition b>0 follows from inequality a>0, since x=log α b, and the value of the degree with a positive base a always positive.
Features of logarithms.
Logarithms characterized by distinctive features, which led to their widespread use to significantly facilitate painstaking calculations. When moving “into the world of logarithms,” multiplication is transformed into a much easier addition, division is transformed into subtraction, and exponentiation and root extraction are transformed, respectively, into multiplication and division by the exponent.
Formulation of logarithms and table of their values (for trigonometric functions) was first published in 1614 by the Scottish mathematician John Napier. Logarithmic tables, enlarged and detailed by other scientists, were widely used in scientific and engineering calculations, and remained relevant until the use of electronic calculators and computers.
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\(a^(b)=c\) \(\Leftrightarrow\) \(\log_(a)(c)=b\)
Let's explain it more simply. For example, \(\log_(2)(8)\) is equal to the power to which \(2\) must be raised to get \(8\). From this it is clear that \(\log_(2)(8)=3\).
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Examples: |
\(\log_(5)(25)=2\) |
because \(5^(2)=25\) |
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\(\log_(3)(81)=4\) |
because \(3^(4)=81\) |
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\(\log_(2)\)\(\frac(1)(32)\) \(=-5\) |
because \(2^(-5)=\)\(\frac(1)(32)\) |
Argument and base of logarithm
Any logarithm has the following “anatomy”:
The argument of a logarithm is usually written at its level, and the base is written in subscript closer to the logarithm sign. And this entry reads like this: “logarithm of twenty-five to base five.”
How to calculate logarithm?
To calculate the logarithm, you need to answer the question: to what power should the base be raised to get the argument?
For example, calculate the logarithm: a) \(\log_(4)(16)\) b) \(\log_(3)\)\(\frac(1)(3)\) c) \(\log_(\sqrt (5))(1)\) d) \(\log_(\sqrt(7))(\sqrt(7))\) e) \(\log_(3)(\sqrt(3))\)
a) To what power must \(4\) be raised to get \(16\)? Obviously the second one. That's why:
\(\log_(4)(16)=2\)
\(\log_(3)\)\(\frac(1)(3)\) \(=-1\)
c) To what power must \(\sqrt(5)\) be raised to get \(1\)? What power makes any number one? Zero, of course!
\(\log_(\sqrt(5))(1)=0\)
d) To what power must \(\sqrt(7)\) be raised to obtain \(\sqrt(7)\)? Firstly, any number to the first power is equal to itself.
\(\log_(\sqrt(7))(\sqrt(7))=1\)
e) To what power must \(3\) be raised to obtain \(\sqrt(3)\)? From we know that is a fractional power, which means square root is the power of \(\frac(1)(2)\) .
\(\log_(3)(\sqrt(3))=\)\(\frac(1)(2)\)
Example : Calculate logarithm \(\log_(4\sqrt(2))(8)\)
Solution :
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\(\log_(4\sqrt(2))(8)=x\) |
We need to find the value of the logarithm, let's denote it as x. Now let's use the definition of a logarithm: |
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\((4\sqrt(2))^(x)=8\) |
What connects \(4\sqrt(2)\) and \(8\)? Two, because both numbers can be represented by twos: |
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\(((2^(2)\cdot2^(\frac(1)(2))))^(x)=2^(3)\) |
On the left we use the properties of the degree: \(a^(m)\cdot a^(n)=a^(m+n)\) and \((a^(m))^(n)=a^(m\cdot n)\) |
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\(2^(\frac(5)(2)x)=2^(3)\) |
The bases are equal, we move on to equality of indicators |
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\(\frac(5x)(2)\) \(=3\) |
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Multiply both sides of the equation by \(\frac(2)(5)\) |
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The resulting root is the value of the logarithm |
Answer : \(\log_(4\sqrt(2))(8)=1,2\)
Why was the logarithm invented?
To understand this, let's solve the equation: \(3^(x)=9\). Just match \(x\) to make the equation work. Of course, \(x=2\).
Now solve the equation: \(3^(x)=8\).What is x equal to? That's the point.
The smartest ones will say: “X is a little less than two.” How exactly to write this number? To answer this question, the logarithm was invented. Thanks to him, the answer here can be written as \(x=\log_(3)(8)\).
I want to emphasize that \(\log_(3)(8)\), like any logarithm is just a number. Yes, it looks unusual, but it’s short. Because if we wanted to write it in the form decimal, then it would look like this: \(1.892789260714.....\)
Example : Solve the equation \(4^(5x-4)=10\)
Solution :
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\(4^(5x-4)=10\) |
\(4^(5x-4)\) and \(10\) cannot be brought to the same base. This means you can’t do without a logarithm. Let's use the definition of logarithm: |
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\(\log_(4)(10)=5x-4\) |
Let's flip the equation so that X is on the left |
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\(5x-4=\log_(4)(10)\) |
Before us. Let's move \(4\) to the right. And don't be afraid of the logarithm, treat it like an ordinary number. |
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\(5x=\log_(4)(10)+4\) |
Divide the equation by 5 |
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\(x=\)\(\frac(\log_(4)(10)+4)(5)\) |
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This is our root. Yes, it looks unusual, but they don’t choose the answer. |
Answer : \(\frac(\log_(4)(10)+4)(5)\)
Decimal and natural logarithms
As stated in the definition of a logarithm, its base can be any positive number except one \((a>0, a\neq1)\). And among all the possible bases, there are two that occur so often that a special short notation was invented for logarithms with them:
Natural logarithm: a logarithm whose base is Euler's number \(e\) (equal to approximately \(2.7182818…\)), and the logarithm is written as \(\ln(a)\).
That is, \(\ln(a)\) is the same as \(\log_(e)(a)\)
Decimal Logarithm: A logarithm whose base is 10 is written \(\lg(a)\).
That is, \(\lg(a)\) is the same as \(\log_(10)(a)\), where \(a\) is some number.
Basic logarithmic identity
Logarithms have many properties. One of them is called the “Basic Logarithmic Identity” and looks like this:
| \(a^(\log_(a)(c))=c\) |
This property follows directly from the definition. Let's see exactly how this formula came about.
Let us recall a short notation of the definition of logarithm:
if \(a^(b)=c\), then \(\log_(a)(c)=b\)
That is, \(b\) is the same as \(\log_(a)(c)\). Then we can write \(\log_(a)(c)\) instead of \(b\) in the formula \(a^(b)=c\). It turned out \(a^(\log_(a)(c))=c\) - the main logarithmic identity.
You can find other properties of logarithms. With their help, you can simplify and calculate the values of expressions with logarithms, which are difficult to calculate directly.
Example : Find the value of the expression \(36^(\log_(6)(5))\)
Solution :
Answer : \(25\)
How to write a number as a logarithm?
As mentioned above, any logarithm is just a number. The converse is also true: any number can be written as a logarithm. For example, we know that \(\log_(2)(4)\) is equal to two. Then you can write \(\log_(2)(4)\) instead of two.
But \(\log_(3)(9)\) is also equal to \(2\), which means we can also write \(2=\log_(3)(9)\) . Likewise with \(\log_(5)(25)\), and with \(\log_(9)(81)\), etc. That is, it turns out
\(2=\log_(2)(4)=\log_(3)(9)=\log_(4)(16)=\log_(5)(25)=\log_(6)(36)=\ log_(7)(49)...\)
Thus, if we need, we can write two as a logarithm with any base anywhere (even in an equation, even in an expression, even in an inequality) - we simply write the squared base as an argument.
It’s the same with the triple – it can be written as \(\log_(2)(8)\), or as \(\log_(3)(27)\), or as \(\log_(4)(64) \)... Here we write the base in the cube as an argument:
\(3=\log_(2)(8)=\log_(3)(27)=\log_(4)(64)=\log_(5)(125)=\log_(6)(216)=\ log_(7)(343)...\)
And with four:
\(4=\log_(2)(16)=\log_(3)(81)=\log_(4)(256)=\log_(5)(625)=\log_(6)(1296)=\ log_(7)(2401)...\)
And with minus one:
\(-1=\) \(\log_(2)\)\(\frac(1)(2)\) \(=\) \(\log_(3)\)\(\frac(1)( 3)\) \(=\) \(\log_(4)\)\(\frac(1)(4)\) \(=\) \(\log_(5)\)\(\frac(1 )(5)\) \(=\) \(\log_(6)\)\(\frac(1)(6)\) \(=\) \(\log_(7)\)\(\frac (1)(7)\) \(...\)
And with one third:
\(\frac(1)(3)\) \(=\log_(2)(\sqrt(2))=\log_(3)(\sqrt(3))=\log_(4)(\sqrt( 4))=\log_(5)(\sqrt(5))=\log_(6)(\sqrt(6))=\log_(7)(\sqrt(7))...\)
Any number \(a\) can be represented as a logarithm with base \(b\): \(a=\log_(b)(b^(a))\)
Example : Find the meaning of the expression \(\frac(\log_(2)(14))(1+\log_(2)(7))\)
Solution :
Answer : \(1\)
main properties.
- logax + logay = loga(x y);
- logax − logay = 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 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:
Of course, all these rules make sense if the ODZ of the logarithm is observed: a > 0, a ≠ 1, x >
Task. Find the meaning of the expression:
Transition to a new foundation
Let the logarithm logax be given. Then for any number c such that c > 0 and c ≠ 1, the equality is true:
Task. Find the meaning of the expression:
See also:
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 equal to 2.7 and twice the year of birth of Leo Nikolaevich Tolstoy.
Basic properties of logarithms
Knowing this rule, you will know and exact value exhibitors, and the date of birth of Leo Tolstoy.
Examples for logarithms
Logarithm expressions
Example 1.
A). x=10ac^2 (a>0,c>0).
Using properties 3.5 we calculate 
2.
3. 
4. 
Where 
.
Example 2. Find x if
Example 3. Let the value of logarithms be given
Calculate log(x) if
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: logax and logay. Then they can be added and subtracted, and:
- logax + logay = loga(x y);
- logax − logay = loga (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:
Since logarithms have the same bases, we use the sum formula:
log6 4 + log6 9 = log6 (4 9) = log6 36 = 2.
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 = 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, which are not calculated separately. But after the transformations, completely normal numbers are obtained. Many are built on this fact tests. 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
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. This is what is most often required.
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 = 6 log7 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 = 24; 49 = 72. 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.
Logarithm formulas. Logarithms examples solutions.
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: 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 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 logax 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 by deciding 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: log5 16 log2 25.
Note that the arguments of both logarithms contain exact powers. Let's take out the indicators: log5 16 = log5 24 = 4log5 2; log2 25 = log2 52 = 2log2 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: log9 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 log25 64 = log5 8 - simply took the square from the base and argument of the logarithm. Taking into account the rules for multiplying powers with the same base, 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.
- logaa = 1 is. Remember once and for all: the logarithm to any base a of that base itself is equal to one.
- loga 1 = 0 is. The base a can be anything, but if the argument contains one, the logarithm is equal to zero! Because a0 = 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.
See also:
The logarithm of b to base a denotes the expression. To calculate the logarithm means to find a power x () at which the equality is satisfied
Basic properties of the logarithm
It is necessary to know the above properties, since almost all problems and examples related to logarithms are solved on their basis. The rest of the exotic properties can be derived through mathematical manipulations with these formulas
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
When calculating the formula for the sum and difference of logarithms (3.4) you come across 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 most common logarithms are those in which the base is equal to ten, exponential or two.
The logarithm to base ten is usually called the decimal logarithm and is simply denoted by lg(x).
It is clear from the recording that the basics are not written in the recording. For example
Natural logarithm is a logarithm with an exponent as its basis (denoted by ln(x)).
The exponent is 2.718281828…. To remember the exponent, you can study the rule: the exponent is equal to 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 logarithm to base two is denoted by
The derivative of the logarithm of a function is equal to one divided by the variable
The integral or antiderivative logarithm is determined by the relationship
The given material is enough for you to solve a wide class of problems related to logarithms and logarithms. To help you understand the material, I will give just a few common examples from school curriculum and universities.
Examples for logarithms
Logarithm expressions
Example 1.
A). x=10ac^2 (a>0,c>0).
Using properties 3.5 we calculate
2.
By the property of difference of logarithms we have
3.
Using properties 3.5 we find
4. Where .
A seemingly complex expression is simplified to form using a number of rules
Finding logarithm values
Example 2. Find x if
Solution. For calculation, we apply to the last term 5 and 13 properties
We put it on record and mourn
Since the bases are equal, we equate the expressions
Logarithms. Entry level.
Let the value of logarithms be given
Calculate log(x) if
Solution: Let's take a logarithm of the variable to write the logarithm through the sum of its terms
This is just the beginning of our acquaintance with logarithms and their properties. Practice calculations, enrich your practical skills - you will soon need the knowledge you gain to solve logarithmic equations. Having studied the basic methods for solving such equations, we will expand your knowledge to another equally important topic - logarithmic inequalities...
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: logax and logay. Then they can be added and subtracted, and:
- logax + logay = loga(x y);
- logax − logay = loga (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 a logarithmic expression even when its individual parts are not considered (see the lesson “What is a logarithm”). Take a look at the examples and see:
Task. Find the value of the expression: log6 4 + log6 9.
Since logarithms have the same bases, we use the sum formula:
log6 4 + log6 9 = log6 (4 9) = log6 36 = 2.
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 = 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, which are not calculated separately. But after the transformations, completely normal numbers are obtained. Many tests are based on this fact. 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: log7 496.
Let's get rid of the degree in the argument using the first formula:
log7 496 = 6 log7 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 = 24; 49 = 72. 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: 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 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 logax 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 ordinary 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: log5 16 log2 25.
Note that the arguments of both logarithms contain exact powers. Let's take out the indicators: log5 16 = log5 24 = 4log5 2; log2 25 = log2 52 = 2log2 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: log9 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 log25 64 = log5 8 - simply took the square from the base and argument of the logarithm. Taking into account the rules for multiplying powers with the same base, 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.
- logaa = 1 is. Remember once and for all: the logarithm to any base a of that base itself is equal to one.
- loga 1 = 0 is. The base a can be anything, but if the argument contains one, the logarithm is equal to zero! Because a0 = 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.
The basic properties of the natural logarithm, graph, domain of definition, set of values, basic formulas, derivative, integral, expansion in power series and representation of the function ln x using complex numbers.
Definition
Natural logarithm is the function y = ln x, the inverse of the exponential, x = e y, and is the logarithm to the base of the number e: ln x = log e x.
The natural logarithm is widely used in mathematics because its derivative has the simplest form: (ln x)′ = 1/ x.
Based on definitions, the base of the natural logarithm is the number e:
e ≅ 2.718281828459045...;
.
Graph of the function y = ln x.
Graph of natural logarithm (functions y = ln x) is obtained from the exponential graph mirror image relative to the straight line y = x.
The natural logarithm is defined for positive values of the variable x. It increases monotonically in its domain of definition.
At x → 0 the limit of the natural logarithm is minus infinity (-∞).
As x → + ∞, the limit of the natural logarithm is plus infinity (+ ∞). For large x, the logarithm increases quite slowly. Any power function x a with a positive exponent a grows faster than the logarithm.
Properties of the natural logarithm
Domain of definition, set of values, extrema, increase, decrease
The natural logarithm is a monotonically increasing function, so it has no extrema. The main properties of the natural logarithm are presented in the table.
ln x values
ln 1 = 0
Basic formulas for natural logarithms
Formulas following from the definition of the inverse function:
The main property of logarithms and its consequences
Base replacement formula
Any logarithm can be expressed in terms of natural logarithms using the base substitution formula:
Proofs of these formulas are presented in the section "Logarithm".
Inverse function
The inverse of the natural logarithm is the exponent.
If , then
If, then.
Derivative ln x
Derivative of the natural logarithm:
.
Derivative of the natural logarithm of modulus x:
.
Derivative of nth order:
.
Deriving formulas > > >
Integral
The integral is calculated by integration by parts:
.
So,
Expressions using complex numbers
Consider the function of the complex variable z:
.
Let's express the complex variable z via module r and argument φ
:
.
Using the properties of the logarithm, we have:
.
Or
.
The argument φ is not uniquely defined. If you put
, where n is an integer,
it will be the same number for different n.
Therefore, the natural logarithm, as a function of a complex variable, is not a single-valued function.
Power series expansion
When the expansion takes place:
Used literature:
I.N. Bronstein, K.A. Semendyaev, Handbook of mathematics for engineers and college students, “Lan”, 2009.