3.4 Logarithmic equations
From Förberedande kurs i matematik 1
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- | {{Ej vald flik|[[3.4 Övningar| | + | {{Ej vald flik|[[3.4 Övningar|Exercises]]}} |
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== Basic Equations == | == Basic Equations == | ||
- | Equations where logarithms appear can vary a lot. Here are some examples where the solution is given almost immediately by the definition of a logarithm, that is | + | Equations where logarithms appear can vary a lot. Here are some examples where the solution is given almost immediately by the definition of a logarithm, that is, |
{{Fristående formel||<math>\begin{align*} | {{Fristående formel||<math>\begin{align*} | ||
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''' Example 1''' | ''' Example 1''' | ||
- | + | Solve the equations | |
<ol type="a"> | <ol type="a"> | ||
<li><math>10^x = 537\quad</math> has a solution <math>x = \lg 537</math>.</li> | <li><math>10^x = 537\quad</math> has a solution <math>x = \lg 537</math>.</li> | ||
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<br> | <br> | ||
<br> | <br> | ||
- | Multiply both sides by <math>3e^x+1</math> | + | Multiply both sides by <math>3e^x+1</math> and <math>e^{-x}+2</math> to eliminate the denominators |
{{Fristående formel||<math>6e^x(e^{-x}+2) = 5(3e^x+1)\,\mbox{.}</math>}} | {{Fristående formel||<math>6e^x(e^{-x}+2) = 5(3e^x+1)\,\mbox{.}</math>}} | ||
- | Note that since <math>e^x</math> and <math>e^{-x}</math> are always positive regardless of the value of <math>x</math> in this latest step we have | + | Note that since <math>e^x</math> and <math>e^{-x}</math> are always positive regardless of the value of <math>x</math> in this latest step we have multiplied the equation by factors <math>3e^x+1</math> and <math>e^{-x} +2</math>. Both of these factors are different from zero, so this step cannot introduce new (spurious) roots of the equation. |
Simplify both sides of the equation | Simplify both sides of the equation | ||
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The term <math>\ln\frac{1}{x}</math> can be written as <math>\ln\frac{1}{x} = \ln x^{-1} = -1 \cdot \ln x = - \ln x</math> and then the equation becomes | The term <math>\ln\frac{1}{x}</math> can be written as <math>\ln\frac{1}{x} = \ln x^{-1} = -1 \cdot \ln x = - \ln x</math> and then the equation becomes | ||
{{Fristående formel||<math>\frac{1}{\ln x} - \ln x = 1\,\mbox{,}</math>}} | {{Fristående formel||<math>\frac{1}{\ln x} - \ln x = 1\,\mbox{,}</math>}} | ||
- | where we can consider <math>\ln x</math> as a new unknown. We multiply both sides | + | where we can consider <math>\ln x</math> as a new unknown. We multiply both sides by <math>\ln x</math> (which is different from zero when <math>x \neq 1</math>) and this gives us a quadratic equation in <math>\ln x</math> |
{{Fristående formel||<math>1 - (\ln x)^2 = \ln x\,\mbox{,}</math>}} | {{Fristående formel||<math>1 - (\ln x)^2 = \ln x\,\mbox{,}</math>}} | ||
{{Fristående formel||<math> (\ln x)^2 + \ln x - 1 = 0\,\mbox{.}</math>}} | {{Fristående formel||<math> (\ln x)^2 + \ln x - 1 = 0\,\mbox{.}</math>}} | ||
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{{Fristående formel||<math>4x^2 - 2x = 1 - 2x\,,</math>|<math>(*)</math>}} | {{Fristående formel||<math>4x^2 - 2x = 1 - 2x\,,</math>|<math>(*)</math>}} | ||
- | and also positive. We solve the equation <math>(*)</math> by moving | + | and also be positive. We solve the equation <math>(*)</math> by moving all of the terms to one side |
{{Fristående formel||<math>4x^2 - 1= 0</math>}} | {{Fristående formel||<math>4x^2 - 1= 0</math>}} | ||
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<br> | <br> | ||
<br> | <br> | ||
- | The first term can be written as <math>e^{2x} = (e^x)^2</math>. The whole equation is a quadratic with <math>e^x</math>as the unknown | + | The first term can be written as <math>e^{2x} = (e^x)^2</math>. The whole equation is a quadratic with <math>e^x</math> as the unknown |
{{Fristående formel||<math>(e^x)^2 - e^x = \tfrac{1}{2}\,\mbox{.}</math>}} | {{Fristående formel||<math>(e^x)^2 - e^x = \tfrac{1}{2}\,\mbox{.}</math>}} | ||
Revision as of 16:58, 17 July 2008
Theory | Exercises |
Innehåll:
- Logarithmic Equations
- Exponential equations
- Spurious roots
- To solve equations that contain powers and by taking logarithms obtain an equation of the first degree.
- To determining which of two logarithmic expressions is the largest by means of a comparison of bases / arguments.
Learning outcomes::
After this section, you will have learned to:
- Solve equations that contain logarithm or exponential expressions and which can be reduced to first or second order equations.
- Deal with spurious roots, and know when they arise.
Basic Equations
Equations where logarithms appear can vary a lot. Here are some examples where the solution is given almost immediately by the definition of a logarithm, that is,
![]() ![]() |
(We consider only 10-logarithms or natural logarithms.)
Example 1
Solve the equations
10x=537 has a solutionx=lg537 .105x=537 gives5x=lg537 , i.e.x=51lg537 .3ex=5 Multiplication of both sides withex and division by 5 gives53=ex , which means thatx=ln53 .lgx=3 The definition gives directlyx=103=1000 .lg(2x−4)=2 From the definition we have2x−4=102=100 and it follows thatx=52 .
Example 2
- Solve the equation
( .10)x=25
Since the left-hand side is equal to10=101
2
( and the equation becomes10)x=(101
2)x=10x
2
10x 2=25.
x2=lg25 , ie.x=2lg25 . - Solve the equation
23ln2x+1=21 .
Multiply both sides by 2 and then subtracting 2 from both sides3ln2x=−1. Divide both sides by 3
ln2x=−31. Now, the definition directly gives
2x=e−1 , which means that3
x=21e−1 3=12e1
3.
In many practical applications of exponential growth or decline there appear equations of the type
where
and use the law of logarithms for powers
![]() |
which gives the solution
Example 3
- Solve the equation
3x=20 .
Take logarithms of both sideslg3x=lg20. The left-hand side can be written as
lg3x=x givinglg3
x=lg3lg20( 2
727).
- Solve the equation
5000 .1
05x=10000
Divide both sides by 50001 05x=500010000=2.
This equation can be solved by taking the lg logarithm of both sides of and rewriting the left-hand side as
lg1 ,05x=x
lg1
05
x=lg2lg1 05(
14
2).
Example 4
- Solve the equation
2x .3x=5
The left-hand side can be rewritten using the laws of powers giving2x and the equation becomes3x=(2
3)x
6x=5. This equation is solved in the usual way by taking logarithms giving
x=lg6lg5( 0
898).
- Solve the equation
52x+1=35x .
Take logarithms of both sides and use the laws of logarithmslgab=b lga
(2x+1)lg52x lg5+lg5=5x
lg3,=5x
lg3.
Collect
x to one sidelg5lg5=5x lg3−2x
lg5,=x(5lg3−2lg5).
The solution is
x=lg55lg3−2lg5.
Some more complicated equations
Equations containing exponential or logarithmic expressions can sometimes be treated as first order or second order equations by considering "
Example 5
Solve the equation
Multiply both sides by
Note that since
Simplify both sides of the equation
where we used ex=e−x+x=e0=1
Taking logarithms then gives the answer
![]() |
Example 6
Solve the equation
The term lnx=−lnx
where we can consider =1
Completing the square on the left-hand side
![]() ![]() ![]() ![]() ![]() ![]() |
We continue by taking the root giving
![]() ![]() |
This means that the equation has two solutions
![]() ![]() ![]() ![]() |
Spurious roots
When you solve equations you should also bear in mind that the arguments of logarithms have to be positive and that terms of the type
Example 7
Solve the equation
For the equation to be satisfied the arguments
![]() | ![]() |
and also be positive. We solve the equation )
and take the root. This gives that
We now check if both sides of )
- If
x=−21 then both are sides are equal to4x2−2x=1−2x=1−2 .−21
=1+1=2
0
- If
x=21 then both are sides are equal to4x2−2x=1−2x=1−2 .21=1−1=0
0
So the logarithmic equation has only one solution
Example 8
Solve the equation
The first term can be written as
The equation can be a little easier to manage if we write
Complete the square for the left-hand side.
![]() ![]() ![]() ![]() ![]() ![]() |
which gives solutions
![]() ![]() |
Since 3
1
3
0
3
![]() ![]() ![]() |
as the only solution to the equation.
Study advice
The basic and final tests
After you have read the text and worked through the exercises, you should do the basic and final tests to pass this section. You can find the link to the tests in your student lounge.
Keep in mind that:
You may need to spend much time studying logarithms. Logarithms usually are dealt with summarily in high school. Therefore, many college students tend to encounter problems when it comes to calculations with logarithms.