3.4 Logarithmic equations
From Förberedande kurs i matematik 1
Theory | Exercises |
Contents:
- Logarithmic equations
- Exponential equations
- Spurious roots
Learning outcomes:
After this section, you will have learned to:
- Solve equations, containing logarithmic or exponential expressions, that can be reduced to a linear or quadratic form.
- Deal with spurious roots, and know when they arise.
- Determine which of two logarithmic expressions is the largest by means of a comparison of bases argument.
Basic Equations
Equations involving logarithms can vary a lot. Here are two simple examples which we can solve straight away using the definition of the logarithm:
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In this section, we consider only logarithms to base 10 or natural logarithms (logarithms to base
Example 1
We solve the following equations for
10x=537 has a solutionx=lg537 .105x=537 gives5x=lg537 , i.e.x=51lg537 .3ex=5 . Multiplying both sides byex and division by 5 gives53=ex , which means thatx=ln53 .lgx=3 . The definition of logarithm then directly givesx=103=1000 .lg(2x−4)=2 . From the definition of logarithm 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 subtract 2 from both sides to get3ln2x=−1. Dividing both sides by 3 gives
ln2x=−31. Now, the definition of logarithm directly gives
2x=e−1 , so that3
x=21e−1 3=12e1
3.
In many practical applications of exponential growth or decay there appear equations of the type
where
Then by the laws of logarithms,
which gives the solution
Example 3
- Solve the equation
3x=20 .
Take logarithms of both sides to getlg3x=lg20. The left-hand side can be written as
lg3x=xlg3 givingx=lg3lg20( 2.727).
- Solve the equation
5000 .1.05x=10000
Divide both sides by 5000 to get1.05x=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 . Thenlg1.05
x=lg2lg1.05( 14.2).
Example 4
- Solve the equation
2x .3x=5
The left-hand side can be rewritten using the laws of indices to give2x 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 logarithms to get(2x+1)lg5 2xlg5+lg5=5xlg3=5xlg3.
Collecting
x on one side giveslg5 lg5=5xlg3−2xlg5=x(5lg3−2lg5).
The solution is then
x=lg55lg3−2lg5.
Some more complicated equations
Equations containing exponential or logarithmic expressions can sometimes be treated as linear or quadratic equations by considering "
Example 5
Solve the equation
Multiply both sides by
In this last step we have multiplied the equation by factors
Simplify both sides of the equation to get
Here we have used ex=e−x+x=e0=1
Taking logarithms then gives the answer:
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Example 6
Solve the equation
The term lnx=−lnx
We multiply both sides by =1
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Completing the square on the left-hand side we see that
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Then by taking roots,
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This means that the equation has two solutions
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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
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We solve the equation )
and taking the root. This gives
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 this 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
Completing the square on the left-hand side gives
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so that
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Since 3
1
3
0
3
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as the only solution of 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 some time studying logarithms.
Logarithms are not usually dealt with in detail in secondary school, but become important at university. This can cause problems.