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3.4 Logarithmic equations

From Förberedande kurs i matematik 1

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{{Vald flik|[[3.4 Logaritmekvationer|Theory]]}}
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{{Selected tab|[[3.4 Logarithmic equations|Theory]]}}
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{{Ej vald flik|[[3.4 Övningar|Exercises]]}}
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{{Not selected tab|[[3.4 Exercises|Exercises]]}}
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{{Info|
{{Info|
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'''Innehåll:'''
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'''Contents:'''
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* Logarithmic Equations
+
* Logarithmic equations
-
* Exponential equations
+
* Exponential equations
-
* Spurious roots
+
* 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.
+
}}
}}
{{Info|
{{Info|
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'''Learning outcomes::'''
+
'''Learning outcomes:'''
After this section, you will have learned to:
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.
+
* 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.
* 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 ==
== 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 involving logarithms can vary a lot. Here are two simple examples which we can solve straight away using the definition of the logarithm:
-
{{Fristående formel||<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 \mbox{.}\\
\end{align*}</math>}}
\end{align*}</math>}}
-
(We consider only 10-logarithms or natural logarithms.)
+
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.
<div class="exempel">
<div class="exempel">
''' Example 1'''
''' Example 1'''
-
Solve the equations
+
We solve the following equations for <math>x</math>:
<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>
Line 44: Line 43:
= \lg 537</math>, i.e. <math>x=\frac{1}{5} \lg 537</math>.</li>
= \lg 537</math>, i.e. <math>x=\frac{1}{5} \lg 537</math>.</li>
<li><math>\frac{3}{e^x} = 5 \quad
<li><math>\frac{3}{e^x} = 5 \quad
-
</math> Multiplication of both sides with <math>e^x
+
</math>. Multiplication of both sides with <math>e^x
</math> and division by 5 gives <math>\tfrac{3}{5}=e^x
</math> and division by 5 gives <math>\tfrac{3}{5}=e^x
</math>, which means that <math>x=\ln\tfrac{3}{5}</math>.</li>
</math>, which means that <math>x=\ln\tfrac{3}{5}</math>.</li>
-
<li><math>\lg x = 3 \quad</math> The definition gives directly <math>
+
<li><math>\lg x = 3 \quad</math>. The definition gives directly <math>
x=10^3 = 1000</math>.</li>
x=10^3 = 1000</math>.</li>
-
<li><math>\lg(2x-4) = 2 \quad</math> From the definition we have <math>
+
<li><math>\lg(2x-4) = 2 \quad</math>. From the definition we have <math>
2x-4 = 10^2 = 100</math> and it follows that <math>x = 52</math>.</li>
2x-4 = 10^2 = 100</math> and it follows that <math>x = 52</math>.</li>
</ol>
</ol>
Line 62: Line 61:
<br>
<br>
Since <math>\sqrt{10} = 10^{1/2}</math> the left-hand side is equal to <math>(\sqrt{10}\,)^x = (10^{1/2})^x = 10^{x/2}</math> and the equation becomes
Since <math>\sqrt{10} = 10^{1/2}</math> the left-hand side is equal to <math>(\sqrt{10}\,)^x = (10^{1/2})^x = 10^{x/2}</math> and the equation becomes
-
{{Fristående formel||<math>10^{x/2} = 25\,\mbox{.}</math>}}
+
{{Displayed math||<math>10^{x/2} = 25\,\mbox{.}</math>}}
This equation has a solution <math>\frac{x}{2} = \lg 25</math>, ie. <math>x = 2 \lg 25</math>.</li>
This equation has a solution <math>\frac{x}{2} = \lg 25</math>, ie. <math>x = 2 \lg 25</math>.</li>
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<br>
<br>
<br>
<br>
-
Multiply both sides by 2 and then subtracting 2 from both sides
+
Multiply both sides by 2 and then subtract 2 from both sides to get
-
{{Fristående formel||<math> 3 \ln 2x = -1\,\mbox{.}</math>}}
+
{{Displayed math||<math> 3 \ln 2x = -1\,\mbox{.}</math>}}
-
Divide both sides by 3
+
Dividing both sides by 3 gives
-
{{Fristående formel||<math> \ln 2x = -\frac{1}{3}\,\mbox{.}</math>}}
+
{{Displayed math||<math> \ln 2x = -\frac{1}{3}\,\mbox{.}</math>}}
-
Now, the definition directly gives <math>2x = e^{-1/3}</math>, which means that
+
Now, the definition directly gives <math>2x = e^{-1/3}</math>, so that
-
{{Fristående formel||<math> x = {\textstyle\frac{1}{2}} e^{-1/3} = \frac{1}{2e^{1/3}}\,\mbox{.} </math>}}</li>
+
{{Displayed math||<math> x = {\textstyle\frac{1}{2}} e^{-1/3} = \frac{1}{2e^{1/3}}\,\mbox{.} </math>}}</li>
</ol>
</ol>
</div>
</div>
In many practical applications of exponential growth or decline there appear equations of the type
In many practical applications of exponential growth or decline there appear equations of the type
-
{{Fristående formel||<math>a^x = b\,\mbox{,}</math>}}
+
{{Displayed math||<math>a^x = b\,\mbox{,}</math>}}
-
where <math>a</math> and <math>b</math> are positive numbers. These equations are best solved by taking the logarithm of both sides
+
where <math>a</math> and <math>b</math> are positive numbers. These equations are best solved by taking the logarithm of both sides so that
-
{{Fristående formel||<math>\lg a^x = \lg b</math>}}
+
{{Displayed math||<math>\lg a^x = \lg b</math>}}
-
and use the law of logarithms for powers
+
Then by the law of logarithms,
-
{{Fristående formel||<math>x \cdot \lg a = \lg b</math>}}
+
{{Displayed math||<math>x \cdot \lg a = \lg b</math>}}
which gives the solution <math>\ x = \displaystyle \frac{\lg b}{\lg a}</math>.
which gives the solution <math>\ x = \displaystyle \frac{\lg b}{\lg a}</math>.
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<br>
<br>
<br>
<br>
-
Take logarithms of both sides
+
Take logarithms of both sides to get
-
{{Fristående formel||<math>\lg 3^x = \lg 20\,\mbox{.}</math>}}
+
{{Displayed math||<math>\lg 3^x = \lg 20\,\mbox{.}</math>}}
The left-hand side can be written as <math>\lg 3^x = x \cdot \lg 3</math> giving
The left-hand side can be written as <math>\lg 3^x = x \cdot \lg 3</math> giving
-
{{Fristående formel||<math>x = \displaystyle \frac{\lg 20}{\lg 3} \quad ({}\approx 2{,}727)\,\mbox{.}</math>}}</li>
+
{{Displayed math||<math>x = \displaystyle \frac{\lg 20}{\lg 3} \quad ({}\approx 2\textrm{.}727)\,\mbox{.}</math>}}</li>
-
<li>Solve the equation <math>\ 5000 \cdot 1{,}05^x = 10\,000</math>.
+
<li>Solve the equation <math>\ 5000 \cdot 1\textrm{.}05^x = 10\,000</math>.
<br>
<br>
<br>
<br>
-
Divide both sides by 5000
+
Divide both sides by 5000 to get
-
{{Fristående formel||<math>1{,}05^x = \displaystyle \frac{ 10\,000}{5\,000} = 2\,\mbox{.}</math>}}
+
{{Displayed math||<math>1\textrm{.}05^x = \displaystyle \frac{ 10\,000}{5\,000} = 2\,\mbox{.}</math>}}
-
This equation can be solved by taking the lg logarithm of both sides of and rewriting the left-hand side as <math>\lg 1{,}05^x = x\cdot\lg 1{,}05</math>,
+
This equation can be solved by taking the lg logarithm of both sides of and rewriting the left-hand side as <math>\lg 1\textrm{.}05^x = x\cdot\lg 1\textrm{.}05</math>. Then
-
{{Fristående formel||<math>x = \frac{\lg 2}{\lg 1{,}05} \quad ({}\approx 14{,}2)\,\mbox{.}</math>}}</li>
+
{{Displayed math||<math>x = \frac{\lg 2}{\lg 1\textrm{.}05} \quad ({}\approx 14\textrm{.}2)\,\mbox{.}</math>}}</li>
</ol>
</ol>
</div>
</div>
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<br>
<br>
<br>
<br>
-
The left-hand side can be rewritten using the laws of powers giving <math>2^x\cdot 3^x=(2 \cdot 3)^x</math> and the equation becomes
+
The left-hand side can be rewritten using the laws of exponents giving <math>2^x\cdot 3^x=(2 \cdot 3)^x</math> and the equation becomes
-
{{Fristående formel||<math>6^x = 5\,\mbox{.}</math>}}
+
{{Displayed math||<math>6^x = 5\,\mbox{.}</math>}}
This equation is solved in the usual way by taking logarithms giving
This equation is solved in the usual way by taking logarithms giving
-
{{Fristående formel||<math>x = \frac{\lg 5}{\lg 6}\quad ({}\approx 0{,}898)\,\mbox{.}</math>}}</li>
+
{{Displayed math||<math>x = \frac{\lg 5}{\lg 6}\quad ({}\approx 0\textrm{.}898)\,\mbox{.}</math>}}</li>
<li>Solve the equation <math>\ 5^{2x + 1} = 3^{5x}</math>.
<li>Solve the equation <math>\ 5^{2x + 1} = 3^{5x}</math>.
<br>
<br>
<br>
<br>
-
Take logarithms of both sides and use the laws of logarithms <math>\lg a^b = b \cdot \lg a</math>
+
Take logarithms of both sides and use the laws of logarithms to get
-
{{Fristående formel||<math>\eqalign{(2x+1)\lg 5 &= 5x \cdot \lg 3\,\mbox{,}\cr 2x \cdot \lg 5 + \lg 5 &= 5x \cdot \lg 3\,\mbox{.}\cr}</math>}}
+
{{Displayed math||<math>\eqalign{(2x+1)\lg 5 &= 5x \cdot \lg 3\,\cr \Rightarrow 2x \cdot \lg 5 + \lg 5 &= 5x \cdot \lg 3\,\mbox{.}\cr}</math>}}
-
Collect <math>x</math> to one side
+
Collecting <math>x</math> to one side gives
-
{{Fristående formel||<math>\eqalign{\lg 5 &= 5x \cdot \lg 3 -2x \cdot \lg 5\,\mbox{,}\cr \lg 5 &= x\,(5 \lg 3 -2 \lg 5)\,\mbox{.}\cr}</math>}}
+
{{Displayed math||<math>\eqalign{\lg 5 &= 5x \cdot \lg 3 -2x \cdot \lg 5\,\cr \Rightarrow \lg 5 &= x\,(5 \lg 3 -2 \lg 5)\,\mbox{.}\cr}</math>}}
-
The solution is
+
The solution is then
-
{{Fristående formel||<math>x = \frac{\lg 5}{5 \lg 3 -2 \lg 5}\,\mbox{.}</math>}}</li>
+
{{Displayed math||<math>x = \frac{\lg 5}{5 \lg 3 -2 \lg 5}\,\mbox{.}</math>}}</li>
</ol>
</ol>
Line 147: Line 146:
<br>
<br>
<br>
<br>
-
Multiply both sides by <math>3e^x+1</math> and <math>e^{-x}+2</math> to eliminate the denominators
+
Multiply both sides by <math>3e^x+1</math> and <math>e^{-x}+2</math> to eliminate the denominators, so that
-
{{Fristående formel||<math>6e^x(e^{-x}+2) = 5(3e^x+1)\,\mbox{.}</math>}}
+
{{Displayed math||<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 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.
+
In this last 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 to get
-
{{Fristående formel||<math>6+12e^x = 15e^x+5\,\mbox{,}</math>}}
+
{{Displayed math||<math>6+12e^x = 15e^x+5\,\mbox{.}</math>}}
-
where we used <math>e^{-x} \cdot e^x = e^{-x + x} = e^0 = 1</math>. If we treat <math>e^x</math> as the unknown variable, the equation is essentially a first order equation which has a solution
+
Here we have used <math>e^{-x} \cdot e^x = e^{-x + x} = e^0 = 1</math>. If we treat <math>e^x</math> as the unknown variable, the equation is essentially a first order equation which has a solution
-
{{Fristående formel||<math>e^x=\frac{1}{3}\,\mbox{.}</math>}}
+
{{Displayed math||<math>e^x=\frac{1}{3}\,\mbox{.}</math>}}
-
Taking logarithms then gives the answer
+
Taking logarithms then gives the answer:
-
{{Fristående formel||<math>x=\ln\frac{1}{3}= \ln 3^{-1} = -1 \cdot \ln 3 = -\ln 3\,\mbox{.}</math>}}
+
{{Displayed math||<math>x=\ln\frac{1}{3}= \ln 3^{-1} = -1 \cdot \ln 3 = -\ln 3\,\mbox{.}</math>}}
</div>
</div>
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<br>
<br>
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>}}
+
{{Displayed math||<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 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>
+
We multiply both sides by <math>\ln x</math> (which is different from zero when <math>x \neq 1</math>, and <math>x = 1</math> is clearly not a solution) and this gives us a quadratic equation in <math>\ln x</math>:
-
{{Fristående formel||<math>1 - (\ln x)^2 = \ln x\,\mbox{,}</math>}}
+
{{Displayed math||<math>1 - (\ln x)^2 = \ln x\,</math>}}
-
{{Fristående formel||<math> (\ln x)^2 + \ln x - 1 = 0\,\mbox{.}</math>}}
+
{{Displayed math||<math> \Rightarrow (\ln x)^2 + \ln x - 1 = 0\,\mbox{.}</math>}}
-
Completing the square on the left-hand side
+
Completing the square on the left-hand side we see that
-
{{Fristående formel||<math>\begin{align*}
+
{{Displayed math||<math>\begin{align*}
\textstyle (\ln x)^2 + \ln x -1
\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 - \bigl(\frac{1}{2} \bigr)^2 - 1\\
-
&= \bigl( \ln x + \frac{1}{2} \bigr)^2 - \frac{5}{4}\\
+
&= \bigl( \ln x + \frac{1}{2} \bigr)^2 - \frac{5}{4} \mbox{.}\\
\end{align*}</math>}}
\end{align*}</math>}}
-
We continue by taking the root giving
+
Then by taking roots,
-
{{Fristående formel||<math>
+
{{Displayed math||<math>
\ln x = -\frac{1}{2} \pm \frac{\sqrt{5}}{2} \,\mbox{.}</math>}}
\ln x = -\frac{1}{2} \pm \frac{\sqrt{5}}{2} \,\mbox{.}</math>}}
This means that the equation has two solutions
This means that the equation has two solutions
-
{{Fristående formel||<math>
+
{{Displayed math||<math>
x= e^{(-1 + \sqrt{5})/2}
x= e^{(-1 + \sqrt{5})/2}
-
\quad \mbox{och} \quad
+
\quad \mbox{or} \quad
x= e^{-(1+\sqrt{5})/2}\,\mbox{.}</math>}}
x= e^{-(1+\sqrt{5})/2}\,\mbox{.}</math>}}
</div>
</div>
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== Spurious roots ==
== 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 <math>e^{(\ldots)}</math> can only have positive values. The risk is otherwise that you get 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 <math>e^{(\ldots)}</math> can only have positive values. In other words we must be careful to make sure that our answer makes sense.
<div class="exempel">
<div class="exempel">
Line 206: Line 205:
<br>
<br>
<br>
<br>
-
For the equation to be satisfied the arguments <math>4x^2-2x</math> and <math>1-2x</math> must be equal,
+
For the equation to be satisfied the arguments <math>4x^2-2x</math> and <math>1-2x</math> must be be positive and equal i.e.
-
{{Fristående formel||<math>4x^2 - 2x = 1 - 2x\,,</math>|<math>(*)</math>}}
+
{{Displayed math||<math>4x^2 - 2x = 1 - 2x\mbox{.}\,,</math>|<math>(*)</math>}}
-
and also be positive. We solve the equation <math>(*)</math> by moving all of the terms to one side
+
We solve the equation <math>(*)</math> by moving all of the terms to one side
-
{{Fristående formel||<math>4x^2 - 1= 0</math>}}
+
{{Displayed math||<math>4x^2 - 1= 0</math>}}
-
and take the root. This gives that
+
and taking the root. This gives
-
{{Fristående formel||<math>
+
{{Displayed math||<math>
\textstyle x= -\frac{1}{2}
\textstyle x= -\frac{1}{2}
-
\quad\mbox{och}\quad
+
\quad\mbox{or}\quad
x = \frac{1}{2} \; \mbox{.}</math>}}
x = \frac{1}{2} \; \mbox{.}</math>}}
Line 233: Line 232:
<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 i.e.
-
{{Fristående formel||<math>(e^x)^2 - e^x = \tfrac{1}{2}\,\mbox{.}</math>}}
+
{{Displayed math||<math>(e^x)^2 - e^x = \tfrac{1}{2}\,\mbox{.}</math>}}
-
The equation can be a little easier to manage if we write <math>t</math> instead of <math>e^x</math>,
+
The equation can be a little easier to manage if we write <math>t</math> instead of <math>e^x</math>, so that we try and solve
-
{{Fristående formel||<math>t^2 -t = \tfrac{1}{2}\,\mbox{.}</math>}}
+
{{Displayed math||<math>t^2 -t = \tfrac{1}{2}\,\mbox{.}</math>}}
-
Complete the square for the left-hand side.
+
Completing the square on the left-hand side gives
-
{{Fristående formel||<math>\begin{align*}
+
{{Displayed math||<math>\begin{align*}
\textstyle \bigl(t-\frac{1}{2}\bigr)^2 - \bigl(\frac{1}{2}\bigr)^2
\textstyle \bigl(t-\frac{1}{2}\bigr)^2 - \bigl(\frac{1}{2}\bigr)^2
-
&= \frac{1}{2}\,\mbox{,}\\
+
&= \frac{1}{2}\,\\
-
\bigl(t-\frac{1}{2}\bigr)^2
+
\Rightarrow \bigl(t-\frac{1}{2}\bigr)^2
-
&= \frac{3}{4}\,\mbox{,}\\
+
&= \frac{3}{4}\,\mbox{.}\\
\end{align*}</math>}}
\end{align*}</math>}}
-
which gives solutions
+
so that
-
{{Fristående formel||<math>
+
{{Displayed math||<math>
t=\frac{1}{2} - \frac{\sqrt{3}}{2}
t=\frac{1}{2} - \frac{\sqrt{3}}{2}
-
\quad\mbox{och}\quad
+
\quad\mbox{or}\quad
t=\frac{1}{2} + \frac{\sqrt{3}}{2} \, \mbox{.}</math>}}
t=\frac{1}{2} + \frac{\sqrt{3}}{2} \, \mbox{.}</math>}}
-
Since <math>\sqrt3 > 1</math> then <math>\frac{1}{2}-\frac{1}{2}\sqrt3 <0</math> and it is only <math>t= \frac{1}{2}+\frac{1}{2}\sqrt3</math> that provides a solution to the original equation because <math>e^x</math> is always positive. Taking logarithms gives finally that
+
Since <math>\sqrt3 > 1</math>, <math>\frac{1}{2}-\frac{1}{2}\sqrt3 <0</math>. Therefore it is only <math>t= \frac{1}{2}+\frac{1}{2}\sqrt3</math> that provides a solution to the original equation because <math>e^x</math> is always positive. Taking logarithms finally gives
-
{{Fristående formel||<math>
+
{{Displayed math||<math>
x = \ln \Bigl(\,\frac{1}{2}+\frac{\sqrt3}{2}\,\Bigr)</math>}}
x = \ln \Bigl(\,\frac{1}{2}+\frac{\sqrt3}{2}\,\Bigr)</math>}}
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-
[[3.4 Övningar|Exercises]]
+
[[3.4 Exercises|Exercises]]
<div class="inforuta" style="width:580px;">
<div class="inforuta" style="width:580px;">
Line 277: Line 276:
'''Keep in mind that:'''
'''Keep in mind that:'''
-
You may need to spend much time studying logarithms.
+
You may need to spend some 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.
+
 
 +
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.
</div>
</div>

Current revision

       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:

10x=yex=yx=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 x:

  1. 10x=537 has a solution x=lg537.
  2. 105x=537 gives 5x=lg537, i.e. x=51lg537.
  3. 3ex=5. Multiplication of both sides with ex and division by 5 gives 53=ex, which means that x=ln53.
  4. lgx=3. The definition gives directly x=103=1000.
  5. lg(2x−4)=2. From the definition we have 2x−4=102=100 and it follows that x=52.

Example 2

  1. Solve the equation (10)x=25 .

    Since 10=1012  the left-hand side is equal to (10)x=(1012)x=10x2  and the equation becomes
    10x2=25. 
    This equation has a solution x2=lg25, ie. x=2lg25.
  2. Solve the equation 23ln2x+1=21.

    Multiply both sides by 2 and then subtract 2 from both sides to get
    3ln2x=−1.

    Dividing both sides by 3 gives

    ln2x=−31.

    Now, the definition directly gives 2x=e−13 , so that

    x=21e−13=12e13.

In many practical applications of exponential growth or decline there appear equations of the type

ax=b,

where a and b are positive numbers. These equations are best solved by taking the logarithm of both sides so that

lgax=lgb

Then by the law of logarithms,

xlga=lgb

which gives the solution  x=lgalgb.

Example 3

  1. Solve the equation 3x=20.

    Take logarithms of both sides to get
    lg3x=lg20.

    The left-hand side can be written as lg3x=xlg3 giving

    x=lg3lg20(2.727).
  2. Solve the equation  50001.05x=10000.

    Divide both sides by 5000 to get
    1.05x=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=xlg1.05. Then

    x=lg2lg1.05(14.2).

Example 4

  1. Solve the equation  2x3x=5.

    The left-hand side can be rewritten using the laws of exponents giving 2x3x=(23)x and the equation becomes
    6x=5.

    This equation is solved in the usual way by taking logarithms giving

    x=lg6lg5(0.898).
  2. Solve the equation  52x+1=35x.

    Take logarithms of both sides and use the laws of logarithms to get
    (2x+1)lg52xlg5+lg5=5xlg3=5xlg3.

    Collecting x to one side gives

    lg5lg5=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 first order or second order equations by considering "lnx" or "ex" as the unknown variable.

Example 5

Solve the equation 6ex3ex+1=5e−x+2.

Multiply both sides by 3ex+1 and e−x+2 to eliminate the denominators, so that

6ex(e−x+2)=5(3ex+1).

In this last step we have multiplied the equation by factors 3ex+1 and 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

6+12ex=15ex+5.

Here we have used e−xex=e−x+x=e0=1. If we treat ex as the unknown variable, the equation is essentially a first order equation which has a solution

ex=31.

Taking logarithms then gives the answer:

x=ln31=ln3−1=−1ln3=−ln3.

Example 6

Solve the equation 1lnx+lnx1=1.

The term lnx1 can be written as lnx1=lnx−1=−1lnx=−lnx and then the equation becomes

1lnx−lnx=1.

We multiply both sides by lnx (which is different from zero when x=1, and x=1 is clearly not a solution) and this gives us a quadratic equation in lnx:

1−(lnx)2=lnx
(lnx)2+lnx−1=0.

Completing the square on the left-hand side we see that

(lnx)2+lnx−1=lnx+212−212−1=lnx+212−45.

Then by taking roots,

lnx=−2125. 

This means that the equation has two solutions

x=e(−1+5)2orx=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 e(...) 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 ln(4x2−2x)=ln(1−2x).

For the equation to be satisfied the arguments 4x2−2x and 1−2x must be be positive and equal i.e.

4x2−2x=1−2x. ()

We solve the equation () by moving all of the terms to one side

4x2−1=0

and taking the root. This gives

x=−21orx=21.

We now check if both sides of () are positive

  • If x=−21 then both are sides are equal to 4x2−2x=1−2x=1−2−21=1+1=20 .
  • If x=21 then both are sides are equal to 4x2−2x=1−2x=1−221=1−1=00.

So the logarithmic equation has only one solution x=−21.

Example 8

Solve the equation e2x−ex=21.

The first term can be written as e2x=(ex)2. The whole equation is a quadratic with ex as the unknown i.e.

(ex)2−ex=21.

The equation can be a little easier to manage if we write t instead of ex, so that we try and solve

t2−t=21.

Completing the square on the left-hand side gives

t−212−212t−212=21=43.

so that

t=21−23ort=21+23. 

Since 31 , 21−2130 . Therefore it is only t=21+213  that provides a solution to the original equation because ex is always positive. Taking logarithms finally gives

x=ln21+23 

as the only solution to the equation.


Exercises

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.

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