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3.4 Logarithmic equations

From Förberedande kurs i matematik 1

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Current revision (17:08, 2 January 2009) (edit) (undo)
(Some changes to language, notation and terminology. NB: I've replaced "first and second order" w/ "linear and quadratic", throughout, but it should in any case have read "first and second degree".)
 
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After this section, you will have learned to:
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.
+
* 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.
* 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.
* Determine which of two logarithmic expressions is the largest by means of a comparison of bases argument.
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\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.
+
In this section, we consider only logarithms to base 10 or natural logarithms (logarithms to base <math>e</math>), though the methods can just as easily be applied in the case of logarithms with an arbitrary base.
<div class="exempel">
<div class="exempel">
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= \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>. Multiplying both sides by <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 of logarithm then directly gives<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 of logarithm 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>
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<br>
<br>
<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
{{Displayed math||<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 solution <math>\frac{x}{2} = \lg 25</math>, ie. <math>x = 2 \lg 25</math>.</li>
<li>Solve the equation <math>\,\frac{3 \ln 2x}{2} + 1 = \frac{1}{2}</math>.
<li>Solve the equation <math>\,\frac{3 \ln 2x}{2} + 1 = \frac{1}{2}</math>.
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Dividing both sides by 3 gives
Dividing both sides by 3 gives
{{Displayed math||<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>, so that
+
Now, the definition of logarithm directly gives <math>2x = e^{-1/3}</math>, so that
{{Displayed math||<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 decay there appear equations of the type
{{Displayed math||<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 so that
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
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{{Displayed math||<math>\lg a^x = \lg b</math>}}
{{Displayed math||<math>\lg a^x = \lg b</math>}}
-
Then by the law of logarithms,
+
Then by the laws of logarithms,
-
{{Displayed math||<math>x \cdot \lg a = \lg b</math>}}
+
{{Displayed math||<math>x \, \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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Take logarithms of both sides to get
Take logarithms of both sides to get
{{Displayed math||<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 \, \lg 3</math> giving
{{Displayed math||<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 \times 1\textrm{.}05^x = 10\,000</math>.
<br>
<br>
<br>
<br>
Divide both sides by 5000 to get
Divide both sides by 5000 to get
{{Displayed math||<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>. Then
+
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\times\lg 1\textrm{.}05</math>. Then
{{Displayed math||<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>
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<ol type="a">
<ol type="a">
-
<li>Solve the equation <math>\ 2^x \cdot 3^x = 5</math>.
+
<li>Solve the equation <math>\ 2^x \times 3^x = 5</math>.
<br>
<br>
<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 indices to give <math>2^x\times 3^x=(2 \times 3)^x</math> and the equation becomes
{{Displayed math||<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
{{Displayed math||<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>
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<br>
<br>
Take logarithms of both sides and use the laws of logarithms to get
Take logarithms of both sides and use the laws of logarithms to get
-
{{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>}}
+
{{Displayed math||<math>\eqalign{(2x+1)\lg 5 &= 5x \,\lg 3\,\cr \Rightarrow 2x \, \lg 5 + \lg 5 &= 5x \, \lg 3\,\mbox{.}\cr}</math>}}
-
Collecting <math>x</math> to one side gives
+
Collecting <math>x</math> on one side gives
-
{{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>}}
+
{{Displayed math||<math>\eqalign{\lg 5 &= 5x \, \lg 3 -2x \, \lg 5\,\cr \Rightarrow \lg 5 &= x\,(5 \lg 3 -2 \lg 5)\,\mbox{.}\cr}</math>}}
The solution is then
The solution is then
{{Displayed math||<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>
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== Some more complicated equations ==
== Some more complicated equations ==
-
Equations containing exponential or logarithmic expressions can sometimes be treated as first order or second order equations by considering "<math>\ln x</math>" or "<math>e^x</math>" as the unknown variable.
+
Equations containing exponential or logarithmic expressions can sometimes be treated as linear or quadratic equations by considering "<math>\ln x</math>" or "<math>e^x</math>" as the unknown variable.
<div class="exempel">
<div class="exempel">
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{{Displayed math||<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>}}
-
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.
+
In this last step we have multiplied the equation by factors <math>3e^x+1</math> and <math>e^{-x} +2</math>. Neither of these factors can possibly be zero, so we are safe; however, if either can be zero, there is a danger that spurious roots may thus be introduced.
Simplify both sides of the equation to get
Simplify both sides of the equation to get
{{Displayed math||<math>6+12e^x = 15e^x+5\,\mbox{.}</math>}}
{{Displayed math||<math>6+12e^x = 15e^x+5\,\mbox{.}</math>}}
-
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
+
Here we have used <math>e^{-x} \times 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 linear equation which has a solution
{{Displayed math||<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:
-
{{Displayed math||<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 \times \ln 3 = -\ln 3\,\mbox{.}</math>}}
</div>
</div>
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<br>
<br>
<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 \times \ln x = - \ln x</math>, and then the equation becomes
{{Displayed math||<math>\frac{1}{\ln x} - \ln x = 1\,\mbox{.}</math>}}
{{Displayed math||<math>\frac{1}{\ln x} - \ln x = 1\,\mbox{.}</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>:
+
We multiply both sides by <math>\ln x</math> (which is non-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>:
{{Displayed math||<math>1 - (\ln x)^2 = \ln x\,</math>}}
{{Displayed math||<math>1 - (\ln x)^2 = \ln x\,</math>}}
{{Displayed math||<math> \Rightarrow (\ln x)^2 + \ln x - 1 = 0\,\mbox{.}</math>}}
{{Displayed math||<math> \Rightarrow (\ln x)^2 + \ln x - 1 = 0\,\mbox{.}</math>}}
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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. We may generate spurious roots, so we must check that our answer makes sense.
+
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">
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We now check if both sides of <math>(*)</math> are positive
We now check if both sides of <math>(*)</math> are positive
-
* If <math>x= -\tfrac{1}{2}</math> then both are sides are equal to <math>4x^2 - 2x = 1-2x = 1-2 \cdot \bigl(-\tfrac{1}{2}\bigr) = 1+1 = 2 > 0</math>.
+
* If <math>x= -\tfrac{1}{2}</math> then both are sides are equal to <math>4x^2 - 2x = 1-2x = 1-2 \times \bigl(-\tfrac{1}{2}\bigr) = 1+1 = 2 > 0</math>.
-
* If <math>x= \tfrac{1}{2}</math> then both are sides are equal to <math>4x^2 - 2x = 1-2x = 1-2 \cdot \tfrac{1}{2} = 1-1 = 0 \not > 0</math>.
+
* If <math>x= \tfrac{1}{2}</math> then both are sides are equal to <math>4x^2 - 2x = 1-2x = 1-2 \times \tfrac{1}{2} = 1-1 = 0 \not > 0</math>.
-
So the logarithmic equation has only one solution <math>x= -\frac{1}{2}</math>.
+
So this logarithmic equation has only one solution <math>x= -\frac{1}{2}</math>.
</div>
</div>
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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>, <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
+
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 of the original equation because <math>e^x</math> is always positive. Taking logarithms finally gives
{{Displayed math||<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>}}
-
as the only solution to the equation.
+
as the only solution of the equation.
</div>
</div>
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'''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 secondary school, but become important at university. This can cause problems.
</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, 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:

10x=yex=yx=lgyx=lny.

In this section, we consider only logarithms to base 10 or natural logarithms (logarithms to base e), 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. Multiplying both sides by ex and division by 5 gives 53=ex, which means that x=ln53.
  4. lgx=3. The definition of logarithm then directly givesx=103=1000.
  5. lg(2x4)=2. From the definition of logarithm we have 2x4=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 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 of logarithm directly gives 2x=e13 , so that

    x=21e13=12e13.

In many practical applications of exponential growth or decay 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 laws 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 indices to give 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 on one side gives

    lg5lg5=5xlg32xlg5=x(5lg32lg5).

    The solution is then

    x=lg55lg32lg5.


Some more complicated equations

Equations containing exponential or logarithmic expressions can sometimes be treated as linear or quadratic equations by considering "lnx" or "ex" as the unknown variable.

Example 5

Solve the equation 6ex3ex+1=5ex+2.

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

6ex(ex+2)=5(3ex+1).

In this last step we have multiplied the equation by factors 3ex+1 and ex+2. Neither of these factors can possibly be zero, so we are safe; however, if either can be zero, there is a danger that spurious roots may thus be introduced.

Simplify both sides of the equation to get

6+12ex=15ex+5.

Here we have used exex=ex+x=e0=1. If we treat ex as the unknown variable, the equation is essentially a linear equation which has a solution

ex=31.

Taking logarithms then gives the answer:

x=ln31=ln31=1ln3=ln3.

Example 6

Solve the equation 1lnx+lnx1=1.

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

1lnxlnx=1.

We multiply both sides by lnx (which is non-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+lnx1=0.

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

(lnx)2+lnx1=lnx+2122121=lnx+21245.

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(4x22x)=ln(12x).

For the equation to be satisfied the arguments 4x22x and 12x must be be positive and equal i.e.

4x22x=12x. ()

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

4x21=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 4x22x=12x=1221=1+1=20 .
  • If x=21 then both are sides are equal to 4x22x=12x=1221=11=00.

So this logarithmic equation has only one solution x=21.

Example 8

Solve the equation e2xex=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)2ex=21.

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

t2t=21.

Completing the square on the left-hand side gives

t212212t212=21=43.

so that

t=2123ort=21+23. 

Since 31 , 212130 . Therefore it is only t=21+213  that provides a solution of the original equation because ex is always positive. Taking logarithms finally gives

x=ln21+23 

as the only solution of 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 secondary school, but become important at university. This can cause problems.