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 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:
\displaystyle \begin{align*}
10^x = y\quad&\Leftrightarrow\quad x = \lg y\\
e^x = y\quad&\Leftrightarrow\quad x = \ln y \mbox{.}\\
\end{align*}
|
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 \displaystyle x:
- \displaystyle 10^x = 537\quad has a solution \displaystyle x = \lg 537.
- \displaystyle 10^{5x} = 537\quad gives \displaystyle 5x = \lg 537, i.e. \displaystyle x=\frac{1}{5} \lg 537.
- \displaystyle \frac{3}{e^x} = 5 \quad . Multiplication of both sides with \displaystyle e^x and division by 5 gives \displaystyle \tfrac{3}{5}=e^x , which means that \displaystyle x=\ln\tfrac{3}{5}.
- \displaystyle \lg x = 3 \quad. The definition gives directly \displaystyle x=10^3 = 1000.
- \displaystyle \lg(2x-4) = 2 \quad. From the definition we have \displaystyle 2x-4 = 10^2 = 100 and it follows that \displaystyle x = 52.
Example 2
- Solve the equation \displaystyle \,(\sqrt{10}\,)^x = 25.
Since \displaystyle \sqrt{10} = 10^{1/2} the left-hand side is equal to \displaystyle (\sqrt{10}\,)^x = (10^{1/2})^x = 10^{x/2} and the equation becomes
This equation has a solution \displaystyle \frac{x}{2} = \lg 25, ie. \displaystyle x = 2 \lg 25.\displaystyle 10^{x/2} = 25\,\mbox{.} - Solve the equation \displaystyle \,\frac{3 \ln 2x}{2} + 1 = \frac{1}{2}.
Multiply both sides by 2 and then subtract 2 from both sides to get\displaystyle 3 \ln 2x = -1\,\mbox{.} Dividing both sides by 3 gives
\displaystyle \ln 2x = -\frac{1}{3}\,\mbox{.} Now, the definition directly gives \displaystyle 2x = e^{-1/3}, so that
\displaystyle x = {\textstyle\frac{1}{2}} e^{-1/3} = \frac{1}{2e^{1/3}}\,\mbox{.}
In many practical applications of exponential growth or decline there appear equations of the type
| \displaystyle a^x = b\,\mbox{,} |
where \displaystyle a and \displaystyle b are positive numbers. These equations are best solved by taking the logarithm of both sides so that
| \displaystyle \lg a^x = \lg b |
Then by the law of logarithms,
| \displaystyle x \cdot \lg a = \lg b |
which gives the solution \displaystyle \ x = \displaystyle \frac{\lg b}{\lg a}.
Example 3
- Solve the equation \displaystyle \,3^x = 20.
Take logarithms of both sides to get\displaystyle \lg 3^x = \lg 20\,\mbox{.} The left-hand side can be written as \displaystyle \lg 3^x = x \cdot \lg 3 giving
\displaystyle x = \displaystyle \frac{\lg 20}{\lg 3} \quad ({}\approx 2\textrm{.}727)\,\mbox{.} - Solve the equation \displaystyle \ 5000 \cdot 1\textrm{.}05^x = 10\,000.
Divide both sides by 5000 to get\displaystyle 1\textrm{.}05^x = \displaystyle \frac{ 10\,000}{5\,000} = 2\,\mbox{.} This equation can be solved by taking the lg logarithm of both sides of and rewriting the left-hand side as \displaystyle \lg 1\textrm{.}05^x = x\cdot\lg 1\textrm{.}05. Then
\displaystyle x = \frac{\lg 2}{\lg 1\textrm{.}05} \quad ({}\approx 14\textrm{.}2)\,\mbox{.}
Example 4
- Solve the equation \displaystyle \ 2^x \cdot 3^x = 5.
The left-hand side can be rewritten using the laws of exponents giving \displaystyle 2^x\cdot 3^x=(2 \cdot 3)^x and the equation becomes\displaystyle 6^x = 5\,\mbox{.} This equation is solved in the usual way by taking logarithms giving
\displaystyle x = \frac{\lg 5}{\lg 6}\quad ({}\approx 0\textrm{.}898)\,\mbox{.} - Solve the equation \displaystyle \ 5^{2x + 1} = 3^{5x}.
Take logarithms of both sides and use the laws of logarithms to get\displaystyle \eqalign{(2x+1)\lg 5 &= 5x \cdot \lg 3\,\cr \Rightarrow 2x \cdot \lg 5 + \lg 5 &= 5x \cdot \lg 3\,\mbox{.}\cr} Collecting \displaystyle x to one side gives
\displaystyle \eqalign{\lg 5 &= 5x \cdot \lg 3 -2x \cdot \lg 5\,\cr \Rightarrow \lg 5 &= x\,(5 \lg 3 -2 \lg 5)\,\mbox{.}\cr} The solution is then
\displaystyle x = \frac{\lg 5}{5 \lg 3 -2 \lg 5}\,\mbox{.}
Some more complicated equations
Equations containing exponential or logarithmic expressions can sometimes be treated as first order or second order equations by considering "\displaystyle \ln x" or "\displaystyle e^x" as the unknown variable.
Example 5
Solve the equation \displaystyle \,\frac{6e^x}{3e^x+1}=\frac{5}{e^{-x}+2}.
Multiply both sides by \displaystyle 3e^x+1 and \displaystyle e^{-x}+2 to eliminate the denominators, so that
| \displaystyle 6e^x(e^{-x}+2) = 5(3e^x+1)\,\mbox{.} |
In this last step we have multiplied the equation by factors \displaystyle 3e^x+1 and \displaystyle e^{-x} +2. 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 to get
| \displaystyle 6+12e^x = 15e^x+5\,\mbox{.} |
Here we have used \displaystyle e^{-x} \cdot e^x = e^{-x + x} = e^0 = 1. If we treat \displaystyle e^x as the unknown variable, the equation is essentially a first order equation which has a solution
| \displaystyle e^x=\frac{1}{3}\,\mbox{.} |
Taking logarithms then gives the answer:
| \displaystyle x=\ln\frac{1}{3}= \ln 3^{-1} = -1 \cdot \ln 3 = -\ln 3\,\mbox{.} |
Example 6
Solve the equation \displaystyle \,\frac{1}{\ln x} + \ln\frac{1}{x} = 1.
The term \displaystyle \ln\frac{1}{x} can be written as \displaystyle \ln\frac{1}{x} = \ln x^{-1} = -1 \cdot \ln x = - \ln x and then the equation becomes
| \displaystyle \frac{1}{\ln x} - \ln x = 1\,\mbox{.} |
We multiply both sides by \displaystyle \ln x (which is different from zero when \displaystyle x \neq 1, and \displaystyle x = 1 is clearly not a solution) and this gives us a quadratic equation in \displaystyle \ln x:
| \displaystyle 1 - (\ln x)^2 = \ln x\, |
| \displaystyle \Rightarrow (\ln x)^2 + \ln x - 1 = 0\,\mbox{.} |
Completing the square on the left-hand side we see that
\displaystyle \begin{align*}
\textstyle (\ln x)^2 + \ln x -1
&= \bigl( \ln x + \frac{1}{2} \bigr)^2 - \bigl(\frac{1}{2} \bigr)^2 - 1\\
&= \bigl( \ln x + \frac{1}{2} \bigr)^2 - \frac{5}{4} \mbox{.}\\
\end{align*}
|
Then by taking roots,
\displaystyle
\ln x = -\frac{1}{2} \pm \frac{\sqrt{5}}{2} \,\mbox{.}
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This means that the equation has two solutions
\displaystyle
x= e^{(-1 + \sqrt{5})/2}
\quad \mbox{or} \quad
x= e^{-(1+\sqrt{5})/2}\,\mbox{.}
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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 \displaystyle e^{(\ldots)} can only have positive values. In other words we must be careful to make sure that our answer makes sense.
Example 7
Solve the equation \displaystyle \,\ln(4x^2 -2x) = \ln (1-2x).
For the equation to be satisfied the arguments \displaystyle 4x^2-2x and \displaystyle 1-2x must be be positive and equal i.e.
| \displaystyle 4x^2 - 2x = 1 - 2x\mbox{.}\,, | \displaystyle (*) |
We solve the equation \displaystyle (*) by moving all of the terms to one side
| \displaystyle 4x^2 - 1= 0 |
and taking the root. This gives
\displaystyle
\textstyle x= -\frac{1}{2}
\quad\mbox{or}\quad
x = \frac{1}{2} \; \mbox{.}
|
We now check if both sides of \displaystyle (*) are positive
- If \displaystyle x= -\tfrac{1}{2} then both are sides are equal to \displaystyle 4x^2 - 2x = 1-2x = 1-2 \cdot \bigl(-\tfrac{1}{2}\bigr) = 1+1 = 2 > 0.
- If \displaystyle x= \tfrac{1}{2} then both are sides are equal to \displaystyle 4x^2 - 2x = 1-2x = 1-2 \cdot \tfrac{1}{2} = 1-1 = 0 \not > 0.
So the logarithmic equation has only one solution \displaystyle x= -\frac{1}{2}.
Example 8
Solve the equation \displaystyle \,e^{2x} - e^{x} = \frac{1}{2}.
The first term can be written as \displaystyle e^{2x} = (e^x)^2. The whole equation is a quadratic with \displaystyle e^x as the unknown i.e.
| \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, so that we try and solve
| \displaystyle t^2 -t = \tfrac{1}{2}\,\mbox{.} |
Completing the square on the left-hand side gives
\displaystyle \begin{align*}
\textstyle \bigl(t-\frac{1}{2}\bigr)^2 - \bigl(\frac{1}{2}\bigr)^2
&= \frac{1}{2}\,\\
\Rightarrow \bigl(t-\frac{1}{2}\bigr)^2
&= \frac{3}{4}\,\mbox{.}\\
\end{align*}
|
so that
\displaystyle
t=\frac{1}{2} - \frac{\sqrt{3}}{2}
\quad\mbox{or}\quad
t=\frac{1}{2} + \frac{\sqrt{3}}{2} \, \mbox{.}
|
Since \displaystyle \sqrt3 > 1, \displaystyle \frac{1}{2}-\frac{1}{2}\sqrt3 <0. Therefore 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
\displaystyle
x = \ln \Bigl(\,\frac{1}{2}+\frac{\sqrt3}{2}\,\Bigr)
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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 some time studying logarithms.
Logarithms are not usually dealt with in detail in high school. Therefore, many college students tend to encounter problems when it comes to calculations with logarithms.
