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      <text>6.3.4.3 Inverse of taking logs</text>
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<h4 class="HELM_example">Example&nbsp;</h4><p>Simplify the expression \(10^{\log_{10} x}\).</p>
<h4 class="HELM_solution">Solution</h4>
<p>Let \(y=10^{\log_{10} x}\), then take logs (to base \(10\)) of both sides:
\[\log_{10} y =\log_{10} (10^{\log_{10} x}) =(\log_{10} x)\log_{10} 10,\]
where we have used: \(\log_{10} A^k=k\log_{10} A\). However, since we are using logs to base \(10\), 
\(\log_{10} 10=1\) and so \(\log_{10} y=\log_{10} x\), which implies \(y=x\).</p>
<p>Therefore, finally we conclude that \(10^{\log_{10} x}=x\).</p>
<p>This is an important result, true for logarithms of any base. It follows from the basic definition of the logarithm.</p>
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<h4>Key Point&nbsp;</h4>
<p>Raising to the power and taking logs are <b>inverse</b> operations:\[a^{\log_a x}=x,\\ \log_a(a^x) =x.\]
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