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Taking the natural logarithm of both sides gives \(\displaystyle (x + 1)\ln(8)=x\ln(3)\). Distributing gives \(\displaystyle 3 x \ln{\left(2 \right)} + 3 \ln{\left(2 \right)} = x \ln{\left(3 \right)}\). Moving all the \(\displaystyle x\) terms to the left hand side and all of the constants to the right side gives: \(\displaystyle - x \ln{\left(3 \right)} + x \ln{\left(8 \right)} = - \ln{\left(8 \right)}\). Factoring out the \(\displaystyle x\) gives \(\displaystyle x \left(- \ln{\left(3 \right)} + \ln{\left(8 \right)}\right)=- \ln{\left(8 \right)}\). Isolating \(\displaystyle x\) yeilds \(\displaystyle x = \frac{- \ln{\left(8 \right)}}{- \ln{\left(3 \right)} + \ln{\left(8 \right)}}\).

Download \(\LaTeX\)

\begin{question}Solve $8^{x + 1}=3^{x}$. 
    \soln{9cm}{Taking the natural logarithm of both sides gives $(x + 1)\ln(8)=x\ln(3)$. Distributing gives $ 3 x \ln{\left(2 \right)} + 3 \ln{\left(2 \right)} = x \ln{\left(3 \right)}$.  Moving all the $x$ terms to the left hand side and all of the constants to the right side gives: $- x \ln{\left(3 \right)} + x \ln{\left(8 \right)} = - \ln{\left(8 \right)}$. Factoring out the $x$ gives $x \left(- \ln{\left(3 \right)} + \ln{\left(8 \right)}\right)=- \ln{\left(8 \right)}$.  Isolating $x$ yeilds $x = \frac{- \ln{\left(8 \right)}}{- \ln{\left(3 \right)} + \ln{\left(8 \right)}}$. }

\end{question}

Download Question and Solution Environment\(\LaTeX\)

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\begin{document}\begin{question}(10pts) The question goes here!
    \soln{9cm}{The solution goes here.}

\end{question}\end{document}

HTML for Canvas

<p> <p>Solve  <img class="equation_image" title=" \displaystyle 8^{x + 1}=3^{x} " src="/equation_images/%20%5Cdisplaystyle%208%5E%7Bx%20%2B%201%7D%3D3%5E%7Bx%7D%20" alt="LaTeX:  \displaystyle 8^{x + 1}=3^{x} " data-equation-content=" \displaystyle 8^{x + 1}=3^{x} " /> . </p> </p>

HTML for Canvas

<p> <p>Taking the natural logarithm of both sides gives  <img class="equation_image" title=" \displaystyle (x + 1)\ln(8)=x\ln(3) " src="/equation_images/%20%5Cdisplaystyle%20%28x%20%2B%201%29%5Cln%288%29%3Dx%5Cln%283%29%20" alt="LaTeX:  \displaystyle (x + 1)\ln(8)=x\ln(3) " data-equation-content=" \displaystyle (x + 1)\ln(8)=x\ln(3) " /> . Distributing gives  <img class="equation_image" title=" \displaystyle  3 x \ln{\left(2 \right)} + 3 \ln{\left(2 \right)} = x \ln{\left(3 \right)} " src="/equation_images/%20%5Cdisplaystyle%20%203%20x%20%5Cln%7B%5Cleft%282%20%5Cright%29%7D%20%2B%203%20%5Cln%7B%5Cleft%282%20%5Cright%29%7D%20%3D%20x%20%5Cln%7B%5Cleft%283%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle  3 x \ln{\left(2 \right)} + 3 \ln{\left(2 \right)} = x \ln{\left(3 \right)} " data-equation-content=" \displaystyle  3 x \ln{\left(2 \right)} + 3 \ln{\left(2 \right)} = x \ln{\left(3 \right)} " /> .  Moving all the  <img class="equation_image" title=" \displaystyle x " src="/equation_images/%20%5Cdisplaystyle%20x%20" alt="LaTeX:  \displaystyle x " data-equation-content=" \displaystyle x " />  terms to the left hand side and all of the constants to the right side gives:  <img class="equation_image" title=" \displaystyle - x \ln{\left(3 \right)} + x \ln{\left(8 \right)} = - \ln{\left(8 \right)} " src="/equation_images/%20%5Cdisplaystyle%20-%20x%20%5Cln%7B%5Cleft%283%20%5Cright%29%7D%20%2B%20x%20%5Cln%7B%5Cleft%288%20%5Cright%29%7D%20%3D%20-%20%5Cln%7B%5Cleft%288%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle - x \ln{\left(3 \right)} + x \ln{\left(8 \right)} = - \ln{\left(8 \right)} " data-equation-content=" \displaystyle - x \ln{\left(3 \right)} + x \ln{\left(8 \right)} = - \ln{\left(8 \right)} " /> . Factoring out the  <img class="equation_image" title=" \displaystyle x " src="/equation_images/%20%5Cdisplaystyle%20x%20" alt="LaTeX:  \displaystyle x " data-equation-content=" \displaystyle x " />  gives  <img class="equation_image" title=" \displaystyle x \left(- \ln{\left(3 \right)} + \ln{\left(8 \right)}\right)=- \ln{\left(8 \right)} " src="/equation_images/%20%5Cdisplaystyle%20x%20%5Cleft%28-%20%5Cln%7B%5Cleft%283%20%5Cright%29%7D%20%2B%20%5Cln%7B%5Cleft%288%20%5Cright%29%7D%5Cright%29%3D-%20%5Cln%7B%5Cleft%288%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle x \left(- \ln{\left(3 \right)} + \ln{\left(8 \right)}\right)=- \ln{\left(8 \right)} " data-equation-content=" \displaystyle x \left(- \ln{\left(3 \right)} + \ln{\left(8 \right)}\right)=- \ln{\left(8 \right)} " /> .  Isolating  <img class="equation_image" title=" \displaystyle x " src="/equation_images/%20%5Cdisplaystyle%20x%20" alt="LaTeX:  \displaystyle x " data-equation-content=" \displaystyle x " />  yeilds  <img class="equation_image" title=" \displaystyle x = \frac{- \ln{\left(8 \right)}}{- \ln{\left(3 \right)} + \ln{\left(8 \right)}} " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20%5Cfrac%7B-%20%5Cln%7B%5Cleft%288%20%5Cright%29%7D%7D%7B-%20%5Cln%7B%5Cleft%283%20%5Cright%29%7D%20%2B%20%5Cln%7B%5Cleft%288%20%5Cright%29%7D%7D%20" alt="LaTeX:  \displaystyle x = \frac{- \ln{\left(8 \right)}}{- \ln{\left(3 \right)} + \ln{\left(8 \right)}} " data-equation-content=" \displaystyle x = \frac{- \ln{\left(8 \right)}}{- \ln{\left(3 \right)} + \ln{\left(8 \right)}} " /> . </p> </p>