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Questions: Algebra BusinessCalculus
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Solve \(\displaystyle 2^{x} + 18=181\).
Isolating the exponential gives \(\displaystyle 2^{x}=163\). Taking the logarithm of both sides gives \(\displaystyle x = \log_{2}(163)\)
\begin{question}Solve $2^{x} + 18=181$. \soln{9cm}{Isolating the exponential gives $2^{x}=163$. Taking the logarithm of both sides gives $x = \log_{2}(163)$} \end{question}
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<p> <p>Solve <img class="equation_image" title=" \displaystyle 2^{x} + 18=181 " src="/equation_images/%20%5Cdisplaystyle%202%5E%7Bx%7D%20%2B%2018%3D181%20" alt="LaTeX: \displaystyle 2^{x} + 18=181 " data-equation-content=" \displaystyle 2^{x} + 18=181 " /> . </p> </p>
<p> <p>Isolating the exponential gives <img class="equation_image" title=" \displaystyle 2^{x}=163 " src="/equation_images/%20%5Cdisplaystyle%202%5E%7Bx%7D%3D163%20" alt="LaTeX: \displaystyle 2^{x}=163 " data-equation-content=" \displaystyle 2^{x}=163 " /> . Taking the logarithm of both sides gives <img class="equation_image" title=" \displaystyle x = \log_{2}(163) " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20%5Clog_%7B2%7D%28163%29%20" alt="LaTeX: \displaystyle x = \log_{2}(163) " data-equation-content=" \displaystyle x = \log_{2}(163) " /> </p> </p>