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      <text>6.4.4.3 Logs in different bases</text>
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<h4 class="HELM_example">Example</h4>
<p>Solve for \(x\) if \(3\ln x+4\log_{10} x=1\).</p>
<h4 class="HELM_solution">Solution</h4>
<p>This has logs to two different bases. So we must first express each logarithm in terms of logs to the <i>same</i> base, \(e\) say Recall</p>
\[\log_{10} x=\dfrac{\ln x}{\ln 10}.\] So \(3\ln x+4\log x=1\) becomes \[ 3\ln x+4\dfrac{\ln x}{\ln 10}=1\qquad\text{or}\qquad \left(3+\dfrac{4}{\ln 10}\right)\ln x=1\] leading to \(\ln x=\dfrac{\ln 10}{3\ln 10 +4}=\dfrac{2.302585}{10.907755}=0.211096\)
and so \[ x=e^{0.211096}=1.2350311.\]]]></text>
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