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Questions: Algebra BusinessCalculus
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Solve \(\displaystyle 6^{x} + 27=75\).
Isolating the exponential gives \(\displaystyle 6^{x}=48\). Taking the logarithm of both sides gives \(\displaystyle x = \log_{6}(48)\)
\begin{question}Solve $6^{x} + 27=75$.
\soln{9cm}{Isolating the exponential gives $6^{x}=48$. Taking the logarithm of both sides gives $x = \log_{6}(48)$}
\end{question}
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\begin{document}\begin{question}(10pts) The question goes here!
\soln{9cm}{The solution goes here.}
\end{question}\end{document}<p> <p>Solve <img class="equation_image" title=" \displaystyle 6^{x} + 27=75 " src="/equation_images/%20%5Cdisplaystyle%206%5E%7Bx%7D%20%2B%2027%3D75%20" alt="LaTeX: \displaystyle 6^{x} + 27=75 " data-equation-content=" \displaystyle 6^{x} + 27=75 " /> . </p> </p><p> <p>Isolating the exponential gives <img class="equation_image" title=" \displaystyle 6^{x}=48 " src="/equation_images/%20%5Cdisplaystyle%206%5E%7Bx%7D%3D48%20" alt="LaTeX: \displaystyle 6^{x}=48 " data-equation-content=" \displaystyle 6^{x}=48 " /> . Taking the logarithm of both sides gives <img class="equation_image" title=" \displaystyle x = \log_{6}(48) " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20%5Clog_%7B6%7D%2848%29%20" alt="LaTeX: \displaystyle x = \log_{6}(48) " data-equation-content=" \displaystyle x = \log_{6}(48) " /> </p> </p>