3.4 Logarithmic equations
From Förberedande kurs i matematik 1
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{{Displayed math||<math>\begin{align*} | {{Displayed math||<math>\begin{align*} | ||
10^x = y\quad&\Leftrightarrow\quad x = \lg y\\ | 10^x = y\quad&\Leftrightarrow\quad x = \lg y\\ | ||
| - | e^x = y\quad&\Leftrightarrow\quad x = \ln y | + | e^x = y\quad&\Leftrightarrow\quad x = \ln y \\ |
| - | \end{align*}</math>}} | + | \end{align*}.</math>}} |
We consider only 10-logarithms or natural logarithms, though the methods can just as easily be applied in the case of logarithms with an arbitrary base. | We consider only 10-logarithms or natural logarithms, though the methods can just as easily be applied in the case of logarithms with an arbitrary base. | ||
Revision as of 21:28, 28 October 2008
| Theory | Exercises |
Contents:
- Logarithmic equations
- Exponential equations
- Spurious roots
Learning outcomes:
After this section, you will have learned to:
- Solve equations that contain logarithm or exponential expressions that can be reduced to first or second order equations.
- 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:
 x=lgyx=lny |
We consider only 10-logarithms or natural logarithms, though the methods can just as easily be applied in the case of logarithms with an arbitrary 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 . 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
2 ( and the equation becomes10)x=(1012)x=10x2
This equation has a solution10x 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 directly gives
2x=e−1 , so that3 x=21e−1 3=12e13.
In many practical applications of exponential growth or decline there appear equations of the type
where
Then by the law of logarithms,
 lga=lgb |
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=x givinglg3x=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 . Thenlg1.05x=lg2lg1.05( 14.2).
Example 4
- Solve the equation
2x .3x=5
The left-hand side can be rewritten using the laws of exponents giving2x and the equation becomes3x=(23)x6x=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.Collect
x to one sidelg5lg5=5x lg3−2xlg5,=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
| ln3=−ln3. |
Example 6
Solve the equation
The term lnx=−lnx
where we can consider 
=1
Completing the square on the left-hand side
 lnx+21 2−212−1=lnx+212−45 |
We continue by taking the root giving
 25. |
This means that the equation has two solutions
 5)2ochx=e−(1+5)2. |
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=00
So the logarithmic equation has only one solution
Example 8
Solve the equation
The first term can be written as
| \displaystyle (e^x)^2 - e^x = \tfrac{1}{2}\,\mbox{.} |
The equation can be a little easier to manage if we write \displaystyle t instead of \displaystyle e^x,
| \displaystyle t^2 -t = \tfrac{1}{2}\,\mbox{.} |
Complete the square for the left-hand side.
\displaystyle \begin{align*}
\textstyle \bigl(t-\frac{1}{2}\bigr)^2 - \bigl(\frac{1}{2}\bigr)^2
&= \frac{1}{2}\,\mbox{,}\\
\bigl(t-\frac{1}{2}\bigr)^2
&= \frac{3}{4}\,\mbox{,}\\
\end{align*}
|
which gives solutions
\displaystyle
t=\frac{1}{2} - \frac{\sqrt{3}}{2}
\quad\mbox{and}\quad
t=\frac{1}{2} + \frac{\sqrt{3}}{2} \, \mbox{.}
|
Since \displaystyle \sqrt3 > 1 then \displaystyle \frac{1}{2}-\frac{1}{2}\sqrt3 <0 and it is only \displaystyle t= \frac{1}{2}+\frac{1}{2}\sqrt3 that provides a solution to the original equation because \displaystyle e^x is always positive. Taking logarithms finally gives that
\displaystyle
x = \ln \Bigl(\,\frac{1}{2}+\frac{\sqrt3}{2}\,\Bigr)
|
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.
