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  <question type="description">
    <name>
      <text>4.2.21 (Using logs to solve exponentials examples)</text>
    </name>
    <questiontext format="moodle_auto_format">
      <text><![CDATA[<p>To solve equations which involve logarithms or exponentials we need to use the index and logarithm laws, as well as the fact these two processes are inverses of each other. </p>
<br/>
<h3>Example</h3>

<p>Solve the following for the variable \(x\).</p>
<ol>
<li>\(3=10^x\)</li>
<li>\(10^{x/4}=\log 3\)</li>
<li>\(\dfrac{1}{17-e^x}=4\)</li>
</ol>

<h3>Solution</h3>
<ol>
 <li>Here we take logs (to base \(10\) because of the term \(10^x\)) of both sides to get
\[
\log 3=\log 10^x=x\log 10=x
\]
<nowrap>where we have used the general property that \(\log_aA^k=k\log_aA\) and the specific property that \(\log 10=1\). Hence \(x=\log 3\) or, in numerical form, \(x=0.47712\) to \(5\) d.p.</nowrap></li>
 <li>The approach used in (a) is used here. Take logs of both sides:
\[
\log(10^{x/4})=\log(\log3)
\]
that is, 
\[
\dfrac{x}{4}\log 10=\log (\log 3)=\log (0.4771212)=-0.3213712.
\]
So, since \(\log 10=1\), we have \(x=4(-0.3213712)=-1.28549\) to \(5\) d.p.</li>
 <li>Here we simplify the expression before taking logs.
\[
\dfrac{1}{17-e^x}=4\qquad\text{implies}\qquad 1=4(17-e^x)
\] 
or \(4e^x=4(17)-1=67\) so \(e^x=16.75\). Now taking natural logs of both sides (because of the presence of the \(e^x\) term) we have:
\[
\ln (e^x)=\ln (16.75)=2.8183983.\] But \(\ln (e^x)=x\ln e=x\) and so the solution to \(\dfrac{1}{17-e^x}=4\) is \(x=2.81840\) to \(5\) d.p.</li>
</ol>]]></text>
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