3.4 Logarithmic equations

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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,
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{{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\\
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<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>
Multiply both sides by 2 and then subtracting 2 from both sides
Multiply both sides by 2 and then subtracting 2 from both sides
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{{Fristående formel||<math> 3 \ln 2x = -1\,\mbox{.}</math>}}
+
{{Displayed math||<math> 3 \ln 2x = -1\,\mbox{.}</math>}}
Divide both sides by 3
Divide both sides by 3
-
{{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>, which means 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
-
{{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
and use the law of logarithms for powers
-
{{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>
Take logarithms of both sides
Take logarithms of both sides
-
{{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\textrm{.}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\textrm{.}05^x = 10\,000</math>.
<li>Solve the equation <math>\ 5000 \cdot 1\textrm{.}05^x = 10\,000</math>.
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<br>
<br>
Divide both sides by 5000
Divide both sides by 5000
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{{Fristående formel||<math>1\textrm{.}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\textrm{.}05^x = x\cdot\lg 1\textrm{.}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>,
-
{{Fristående formel||<math>x = \frac{\lg 2}{\lg 1\textrm{.}05} \quad ({}\approx 14\textrm{.}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>
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
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\textrm{.}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>.
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<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 <math>\lg a^b = b \cdot \lg a</math>
-
{{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\,\mbox{,}\cr 2x \cdot \lg 5 + \lg 5 &= 5x \cdot \lg 3\,\mbox{.}\cr}</math>}}
Collect <math>x</math> to one side
Collect <math>x</math> to one side
-
{{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\,\mbox{,}\cr \lg 5 &= x\,(5 \lg 3 -2 \lg 5)\,\mbox{.}\cr}</math>}}
The solution is
The solution is
-
{{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>
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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
-
{{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.
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
-
{{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
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
-
{{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>
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>}}
+
{{Displayed math||<math>1 - (\ln x)^2 = \ln x\,\mbox{,}</math>}}
-
{{Fristående formel||<math> (\ln x)^2 + \ln x - 1 = 0\,\mbox{.}</math>}}
+
{{Displayed math||<math> (\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
-
{{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\\
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We continue by taking the root giving
We continue by taking the root giving
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{{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{och} \quad
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<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 equal,
-
{{Fristående formel||<math>4x^2 - 2x = 1 - 2x\,,</math>|<math>(*)</math>}}
+
{{Displayed math||<math>4x^2 - 2x = 1 - 2x\,,</math>|<math>(*)</math>}}
and also be positive. We solve the equation <math>(*)</math> by moving all of the terms to one side
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>}}
+
{{Displayed math||<math>4x^2 - 1= 0</math>}}
and take the root. This gives that
and take the root. This gives that
-
{{Fristående formel||<math>
+
{{Displayed math||<math>
\textstyle x= -\frac{1}{2}
\textstyle x= -\frac{1}{2}
\quad\mbox{and}\quad
\quad\mbox{and}\quad
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<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>}}
+
{{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>,
-
{{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.
Complete the square for the left-hand side.
-
{{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}\,\mbox{,}\\
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which gives solutions
which gives solutions
-
{{Fristående formel||<math>
+
{{Displayed math||<math>
t=\frac{1}{2} - \frac{\sqrt{3}}{2}
t=\frac{1}{2} - \frac{\sqrt{3}}{2}
\quad\mbox{and}\quad
\quad\mbox{and}\quad
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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 finally gives that
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 finally gives that
-
{{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>}}

Revision as of 13:55, 10 September 2008

       Theory          Exercises      

Contents:

  • Logarithmic equations
  • Exponential equations
  • Spurious roots

Learning outcomes:

After this section, you will have learned to:

  • To solve equations that contain powers and by taking logarithms obtain an equation of the first degree.
  • 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.
  • To determine which of two logarithmic expressions is the largest by means of a comparison of bases / arguments.

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,

\displaystyle \begin{align*} 
   10^x = y\quad&\Leftrightarrow\quad x = \lg y\\
   e^x = y\quad&\Leftrightarrow\quad x = \ln y\\
 \end{align*}

(We consider only 10-logarithms or natural logarithms.)

Example 1

Solve the equations

  1. \displaystyle 10^x = 537\quad has a solution \displaystyle x = \lg 537.
  2. \displaystyle 10^{5x} = 537\quad gives \displaystyle 5x = \lg 537, i.e. \displaystyle x=\frac{1}{5} \lg 537.
  3. \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}.
  4. \displaystyle \lg x = 3 \quad The definition gives directly \displaystyle x=10^3 = 1000.
  5. \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

  1. 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
    \displaystyle 10^{x/2} = 25\,\mbox{.}
    This equation has a solution \displaystyle \frac{x}{2} = \lg 25, ie. \displaystyle x = 2 \lg 25.
  2. Solve the equation \displaystyle \,\frac{3 \ln 2x}{2} + 1 = \frac{1}{2}.

    Multiply both sides by 2 and then subtracting 2 from both sides
    \displaystyle 3 \ln 2x = -1\,\mbox{.}

    Divide both sides by 3

    \displaystyle \ln 2x = -\frac{1}{3}\,\mbox{.}

    Now, the definition directly gives \displaystyle 2x = e^{-1/3}, which means 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

\displaystyle \lg a^x = \lg b

and use the law of logarithms for powers

\displaystyle x \cdot \lg a = \lg b

which gives the solution \displaystyle \ x = \displaystyle \frac{\lg b}{\lg a}.

Example 3

  1. Solve the equation \displaystyle \,3^x = 20.

    Take logarithms of both sides
    \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{.}
  2. Solve the equation \displaystyle \ 5000 \cdot 1\textrm{.}05^x = 10\,000.

    Divide both sides by 5000
    \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,

    \displaystyle x = \frac{\lg 2}{\lg 1\textrm{.}05} \quad ({}\approx 14\textrm{.}2)\,\mbox{.}

Example 4

  1. 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{.}
  2. Solve the equation \displaystyle \ 5^{2x + 1} = 3^{5x}.

    Take logarithms of both sides and use the laws of logarithms \displaystyle \lg a^b = b \cdot \lg a
    \displaystyle \eqalign{(2x+1)\lg 5 &= 5x \cdot \lg 3\,\mbox{,}\cr 2x \cdot \lg 5 + \lg 5 &= 5x \cdot \lg 3\,\mbox{.}\cr}

    Collect \displaystyle x to one side

    \displaystyle \eqalign{\lg 5 &= 5x \cdot \lg 3 -2x \cdot \lg 5\,\mbox{,}\cr \lg 5 &= x\,(5 \lg 3 -2 \lg 5)\,\mbox{.}\cr}

    The solution is

    \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

\displaystyle 6e^x(e^{-x}+2) = 5(3e^x+1)\,\mbox{.}

Note that since \displaystyle e^x and \displaystyle e^{-x} are always positive regardless of the value of \displaystyle x, in this latest 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

\displaystyle 6+12e^x = 15e^x+5\,\mbox{,}

where we 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{,}

where we can consider \displaystyle \ln x as a new unknown. We multiply both sides by \displaystyle \ln x (which is different from zero when \displaystyle x \neq 1) and this gives us a quadratic equation in \displaystyle \ln x

\displaystyle 1 - (\ln x)^2 = \ln x\,\mbox{,}
\displaystyle (\ln x)^2 + \ln x - 1 = 0\,\mbox{.}

Completing the square on the left-hand side

\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}\\
 \end{align*}

We continue by taking the root giving

\displaystyle 
 \ln x = -\frac{1}{2} \pm \frac{\sqrt{5}}{2} \,\mbox{.}

This means that the equation has two solutions

\displaystyle 
 x= e^{(-1 + \sqrt{5})/2}
 \quad \mbox{och} \quad
 x= e^{-(1+\sqrt{5})/2}\,\mbox{.}


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. The risk is otherwise that you get spurious roots.

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 equal,

\displaystyle 4x^2 - 2x = 1 - 2x\,, \displaystyle (*)

and also be positive. We solve the equation \displaystyle (*) by moving all of the terms to one side

\displaystyle 4x^2 - 1= 0

and take the root. This gives that

\displaystyle 
 \textstyle x= -\frac{1}{2}
 \quad\mbox{and}\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

\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.


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 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.

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