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Solving logarithmic equations. Part 1.
Logarithmic equation is an equation in which the unknown is contained under the sign of the logarithm (in particular, in the base of the logarithm).
The simplest logarithmic equation has the form:
Solving any logarithmic equation involves a transition from logarithms to expressions under the sign of logarithms. However, this action expands the range of permissible values of the equation and can lead to the appearance of extraneous roots. To avoid the appearance of foreign roots, you can do one of three ways:
1. Make an equivalent transition from the original equation to a system including
depending on which inequality or simpler.
If the equation contains an unknown in the base of the logarithm:
then we go to the system:
2. Separately find the range of acceptable values of the equation, then solve the equation and check whether the solutions found satisfy the equation.
3. Solve the equation, and then check: substitute the found solutions into the original equation and check whether we get the correct equality.
A logarithmic equation of any level of complexity always ultimately reduces to the simplest logarithmic equation.
All logarithmic equations can be divided into four types:
1 . Equations that contain logarithms only to the first power. With the help of transformations and use, they are brought to the form
Example. Let's solve the equation:
Let's equate the expressions under the logarithm sign:
Let's check whether our root of the equation satisfies:
Yes, it satisfies.
Answer: x=5
2 . Equations that contain logarithms to powers other than 1 (particularly in the denominator of a fraction). Such equations can be solved using introducing a change of variable.
Example. Let's solve the equation:
Let's find the ODZ equation:
The equation contains logarithms squared, so it can be solved using a change of variable.
Important! Before introducing a replacement, you need to “pull apart” the logarithms that are part of the equation into “bricks”, using the properties of logarithms.
When “pulling apart” logarithms, it is important to use the properties of logarithms very carefully:
In addition, there is one more subtle point here, and in order to avoid a common mistake, we will use an intermediate equality: we will write the degree of the logarithm in this form:
Likewise,
Let's substitute the resulting expressions into the original equation. We get:
Now we see that the unknown is contained in the equation as part of . Let's introduce the replacement: . Since it can take any real value, we do not impose any restrictions on the variable.
Preparation for the final test in mathematics includes an important section - “Logarithms”. Tasks from this topic are necessarily contained in the Unified State Examination. Experience from past years shows that logarithmic equations caused difficulties for many schoolchildren. Therefore, students with different levels of training must understand how to find the correct answer and quickly cope with them.
Pass the certification test successfully using the Shkolkovo educational portal!
When preparing for the Unified State Exam, high school graduates need a reliable source that provides the most complete and accurate information for successfully solving test problems. However, a textbook is not always at hand, and searching for the necessary rules and formulas on the Internet often takes time.
The Shkolkovo educational portal allows you to prepare for the Unified State Exam anywhere at any time. Our website offers the most convenient approach to repeating and assimilating a large amount of information on logarithms, as well as with one and several unknowns. Start with easy equations. If you cope with them without difficulty, move on to more complex ones. If you have trouble solving a particular inequality, you can add it to your Favorites so you can return to it later.
You can find the necessary formulas to complete the task, repeat special cases and methods for calculating the root of a standard logarithmic equation by looking at the “Theoretical Help” section. Shkolkovo teachers collected, systematized and outlined everything necessary for successful completion materials in the simplest and most understandable form.
In order to easily cope with tasks of any complexity, on our portal you can familiarize yourself with the solution of some standard logarithmic equations. To do this, go to the “Catalogues” section. We have a large number of examples, including those with profile equations Unified State Exam level mathematics.
Students from schools throughout Russia can use our portal. To start classes, simply register in the system and start solving equations. To consolidate the results, we advise you to return to the Shkolkovo website daily.
Instructions
Write the given logarithmic expression. If the expression uses the logarithm of 10, then its notation is shortened and looks like this: lg b is the decimal logarithm. If the logarithm has the number e as its base, then write the expression: ln b – natural logarithm. It is understood that the result of any is the power to which the base number must be raised to obtain the number b.
When finding the sum of two functions, you simply need to differentiate them one by one and add the results: (u+v)" = u"+v";
When finding the derivative of the product of two functions, it is necessary to multiply the derivative of the first function by the second and add the derivative of the second function multiplied by the first function: (u*v)" = u"*v+v"*u;
In order to find the derivative of the quotient of two functions, it is necessary to subtract from the product of the derivative of the dividend multiplied by the divisor function the product of the derivative of the divisor multiplied by the function of the dividend, and divide all this by the divisor function squared. (u/v)" = (u"*v-v"*u)/v^2;
If a complex function is given, then it is necessary to multiply the derivative of the internal function and the derivative of the external one. Let y=u(v(x)), then y"(x)=y"(u)*v"(x).
Using the results obtained above, you can differentiate almost any function. So let's look at a few examples:
y=x^4, y"=4*x^(4-1)=4*x^3;
y=2*x^3*(e^x-x^2+6), y"=2*(3*x^2*(e^x-x^2+6)+x^3*(e^x-2 *x));
There are also problems involving calculating the derivative at a point. Let the function y=e^(x^2+6x+5) be given, you need to find the value of the function at the point x=1.
1) Find the derivative of the function: y"=e^(x^2-6x+5)*(2*x +6).
2) Calculate the value of the function at a given point y"(1)=8*e^0=8
Video on the topic
Learn the table of elementary derivatives. This will significantly save time.
Sources:
- derivative of a constant
So, what is the difference between an irrational equation and a rational one? If the unknown variable is under the sign square root, then the equation is considered irrational.
Instructions
The main method for solving such equations is the method of constructing both sides equations into a square. However. this is natural, the first thing you need to do is get rid of the sign. This method is not technically difficult, but sometimes it can lead to trouble. For example, the equation is v(2x-5)=v(4x-7). By squaring both sides you get 2x-5=4x-7. Solving such an equation is not difficult; x=1. But the number 1 will not be given equations. Why? Substitute one into the equation instead of the value of x. And the right and left sides will contain expressions that do not make sense, that is. This value is not valid for a square root. Therefore, 1 is an extraneous root, and therefore this equation has no roots.
So, an irrational equation is solved using the method of squaring both its sides. And having solved the equation, it is necessary to cut off extraneous roots. To do this, substitute the found roots into the original equation.
Consider another one.
2х+vх-3=0
Of course, this equation can be solved using the same equation as the previous one. Move Compounds equations, which do not have a square root, to the right side and then use the squaring method. solve the resulting rational equation and roots. But also another, more elegant one. Enter a new variable; vх=y. Accordingly, you will receive an equation of the form 2y2+y-3=0. That is, the usual quadratic equation. Find its roots; y1=1 and y2=-3/2. Next, solve two equations vх=1; vх=-3/2. The second equation has no roots; from the first we find that x=1. Don't forget to check the roots.
Solving identities is quite simple. To do this, it is necessary to carry out identical transformations until the set goal is achieved. Thus, with the help of simple arithmetic operations, the problem posed will be solved.
You will need
- - paper;
- - pen.
Instructions
The simplest of such transformations are algebraic abbreviated multiplications (such as the square of the sum (difference), difference of squares, sum (difference), cube of the sum (difference)). In addition, there are many and trigonometric formulas, which are essentially the same identities.
Indeed, the square of the sum of two terms is equal to the square of the first plus twice the product of the first by the second and plus the square of the second, that is, (a+b)^2= (a+b)(a+b)=a^2+ab +ba+b ^2=a^2+2ab+b^2.
Simplify both
General principles of the solution
Repeat from a textbook on mathematical analysis or higher mathematics what a definite integral is. As is known, the solution to a definite integral is a function whose derivative will give an integrand. This function is called antiderivative. Based on this principle, the main integrals are constructed.Determine by type integrand function, which of the table integrals fits in in this case. It is not always possible to determine this immediately. Often, the tabular form becomes noticeable only after several transformations to simplify the integrand.
Variable Replacement Method
If the integrand is a trigonometric function whose argument is a polynomial, then try using the change of variables method. In order to do this, replace the polynomial in the argument of the integrand with some new variable. Based on the relationship between the new and old variables, determine the new limits of integration. By differentiating this expression, find the new differential in . So you will get the new kind of the previous integral, close to or even corresponding to any tabular one.Solving integrals of the second kind
If the integral is an integral of the second kind, a vector form of the integrand, then you will need to use the rules for the transition from these integrals to scalar ones. One such rule is the Ostrogradsky-Gauss relation. This law allows us to move from the rotor flux of a certain vector function to the triple integral over the divergence of a given vector field.Substitution of integration limits
After finding the antiderivative, it is necessary to substitute the limits of integration. First, substitute the value of the upper limit into the expression for the antiderivative. You will get some number. Next, subtract from the resulting number another number obtained from the lower limit into the antiderivative. If one of the limits of integration is infinity, then when substituting it into the antiderivative function, it is necessary to go to the limit and find what the expression tends to.If the integral is two-dimensional or three-dimensional, then you will have to represent the limits of integration geometrically to understand how to evaluate the integral. Indeed, in the case of, say, a three-dimensional integral, the limits of integration can be entire planes that limit the volume being integrated.
Examples:
\(\log_(2)(x) = 32\)
\(\log_3x=\log_39\)
\(\log_3((x^2-3))=\log_3((2x))\)
\(\log_(x+1)((x^2+3x-7))=2\)
\(\lg^2((x+1))+10=11 \lg((x+1))\)
How to solve logarithmic equations:
When solving a logarithmic equation, you should strive to transform it to the form \(\log_a(f(x))=\log_a(g(x))\), and then make the transition to \(f(x)=g(x) \).
\(\log_a(f(x))=\log_a(g(x))\) \(⇒\) \(f(x)=g(x)\).
Example:\(\log_2(x-2)=3\)
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Solution: |
ODZ: |
Very important! This transition can only be made if:
You have written for the original equation, and at the end you will check whether those found are included in the ODZ. If this is not done, extra roots may appear, which means a wrong decision.
The number (or expression) on the left and right is the same;
The logarithms on the left and right are “pure”, that is, there should be no multiplications, divisions, etc. – only single logarithms on either side of the equal sign.
For example:
Note that Equations 3 and 4 can be easily solved by applying the necessary properties of logarithms.
Example . Solve the equation \(2\log_8x=\log_82.5+\log_810\)
Solution :
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Let's write the ODZ: \(x>0\). |
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\(2\log_8x=\log_82.5+\log_810\) ODZ: \(x>0\) |
On the left in front of the logarithm is the coefficient, on the right is the sum of the logarithms. This bothers us. Let's move the two to the exponent \(x\) according to the property: \(n \log_b(a)=\log_b(a^n)\). Let us represent the sum of logarithms as one logarithm according to the property: \(\log_ab+\log_ac=\log_a(bc)\) |
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\(\log_8(x^2)=\log_825\) |
We reduced the equation to the form \(\log_a(f(x))=\log_a(g(x))\) and wrote down the ODZ, which means we can move to the form \(f(x)=g(x)\ ). |
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Happened . We solve it and get the roots. |
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\(x_1=5\) \(x_2=-5\) |
We check whether the roots are suitable for ODZ. To do this, in \(x>0\) instead of \(x\) we substitute \(5\) and \(-5\). This operation can be performed orally. |
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\(5>0\), \(-5>0\) |
The first inequality is true, the second is not. This means that \(5\) is the root of the equation, but \(-5\) is not. We write down the answer. |
Answer : \(5\)
Example : Solve the equation \(\log^2_2(x)-3 \log_2(x)+2=0\)
Solution :
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Let's write the ODZ: \(x>0\). |
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\(\log^2_2(x)-3 \log_2(x)+2=0\) ODZ: \(x>0\) |
A typical equation solved using . Replace \(\log_2x\) with \(t\). |
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\(t=\log_2x\) |
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We received the usual one. We are looking for its roots. |
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\(t_1=2\) \(t_2=1\) |
Making a reverse replacement |
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\(\log_2(x)=2\) \(\log_2(x)=1\) |
We transform the right-hand sides, representing them as logarithms: \(2=2 \cdot 1=2 \log_22=\log_24\) and \(1=\log_22\) |
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\(\log_2(x)=\log_24\) \(\log_2(x)=\log_22 \) |
Now our equations are \(\log_a(f(x))=\log_a(g(x))\), and we can transition to \(f(x)=g(x)\). |
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\(x_1=4\) \(x_2=2\) |
We check the correspondence of the roots of the ODZ. To do this, substitute \(4\) and \(2\) into the inequality \(x>0\) instead of \(x\). |
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\(4>0\) \(2>0\) |
Both inequalities are true. This means that both \(4\) and \(2\) are roots of the equation. |
Answer : \(4\); \(2\).
