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Solve \(\displaystyle 3^{x + 2}=5^{x}\).
Taking the natural logarithm of both sides gives \(\displaystyle (x + 2)\ln(3)=x\ln(5)\). Distributing gives \(\displaystyle x \ln{\left(3 \right)} + 2 \ln{\left(3 \right)} = x \ln{\left(5 \right)}\). Moving all the \(\displaystyle x\) terms to the left hand side and all of the constants to the right side gives: \(\displaystyle - x \ln{\left(5 \right)} + x \ln{\left(3 \right)} = - 2 \ln{\left(3 \right)}\). Factoring out the \(\displaystyle x\) gives \(\displaystyle x \left(- \ln{\left(5 \right)} + \ln{\left(3 \right)}\right)=- 2 \ln{\left(3 \right)}\). Isolating \(\displaystyle x\) yeilds \(\displaystyle x = \frac{- \ln{\left(3 \right)}}{- \ln{\left(5 \right)} + \ln{\left(3 \right)}}\).
\begin{question}Solve $3^{x + 2}=5^{x}$.
\soln{9cm}{Taking the natural logarithm of both sides gives $(x + 2)\ln(3)=x\ln(5)$. Distributing gives $ x \ln{\left(3 \right)} + 2 \ln{\left(3 \right)} = x \ln{\left(5 \right)}$. Moving all the $x$ terms to the left hand side and all of the constants to the right side gives: $- x \ln{\left(5 \right)} + x \ln{\left(3 \right)} = - 2 \ln{\left(3 \right)}$. Factoring out the $x$ gives $x \left(- \ln{\left(5 \right)} + \ln{\left(3 \right)}\right)=- 2 \ln{\left(3 \right)}$. Isolating $x$ yeilds $x = \frac{- \ln{\left(3 \right)}}{- \ln{\left(5 \right)} + \ln{\left(3 \right)}}$. }
\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 3^{x + 2}=5^{x} " src="/equation_images/%20%5Cdisplaystyle%203%5E%7Bx%20%2B%202%7D%3D5%5E%7Bx%7D%20" alt="LaTeX: \displaystyle 3^{x + 2}=5^{x} " data-equation-content=" \displaystyle 3^{x + 2}=5^{x} " /> . </p> </p><p> <p>Taking the natural logarithm of both sides gives <img class="equation_image" title=" \displaystyle (x + 2)\ln(3)=x\ln(5) " src="/equation_images/%20%5Cdisplaystyle%20%28x%20%2B%202%29%5Cln%283%29%3Dx%5Cln%285%29%20" alt="LaTeX: \displaystyle (x + 2)\ln(3)=x\ln(5) " data-equation-content=" \displaystyle (x + 2)\ln(3)=x\ln(5) " /> . Distributing gives <img class="equation_image" title=" \displaystyle x \ln{\left(3 \right)} + 2 \ln{\left(3 \right)} = x \ln{\left(5 \right)} " src="/equation_images/%20%5Cdisplaystyle%20%20x%20%5Cln%7B%5Cleft%283%20%5Cright%29%7D%20%2B%202%20%5Cln%7B%5Cleft%283%20%5Cright%29%7D%20%3D%20x%20%5Cln%7B%5Cleft%285%20%5Cright%29%7D%20" alt="LaTeX: \displaystyle x \ln{\left(3 \right)} + 2 \ln{\left(3 \right)} = x \ln{\left(5 \right)} " data-equation-content=" \displaystyle x \ln{\left(3 \right)} + 2 \ln{\left(3 \right)} = x \ln{\left(5 \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(5 \right)} + x \ln{\left(3 \right)} = - 2 \ln{\left(3 \right)} " src="/equation_images/%20%5Cdisplaystyle%20-%20x%20%5Cln%7B%5Cleft%285%20%5Cright%29%7D%20%2B%20x%20%5Cln%7B%5Cleft%283%20%5Cright%29%7D%20%3D%20-%202%20%5Cln%7B%5Cleft%283%20%5Cright%29%7D%20" alt="LaTeX: \displaystyle - x \ln{\left(5 \right)} + x \ln{\left(3 \right)} = - 2 \ln{\left(3 \right)} " data-equation-content=" \displaystyle - x \ln{\left(5 \right)} + x \ln{\left(3 \right)} = - 2 \ln{\left(3 \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(5 \right)} + \ln{\left(3 \right)}\right)=- 2 \ln{\left(3 \right)} " src="/equation_images/%20%5Cdisplaystyle%20x%20%5Cleft%28-%20%5Cln%7B%5Cleft%285%20%5Cright%29%7D%20%2B%20%5Cln%7B%5Cleft%283%20%5Cright%29%7D%5Cright%29%3D-%202%20%5Cln%7B%5Cleft%283%20%5Cright%29%7D%20" alt="LaTeX: \displaystyle x \left(- \ln{\left(5 \right)} + \ln{\left(3 \right)}\right)=- 2 \ln{\left(3 \right)} " data-equation-content=" \displaystyle x \left(- \ln{\left(5 \right)} + \ln{\left(3 \right)}\right)=- 2 \ln{\left(3 \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(3 \right)}}{- \ln{\left(5 \right)} + \ln{\left(3 \right)}} " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20%5Cfrac%7B-%20%5Cln%7B%5Cleft%283%20%5Cright%29%7D%7D%7B-%20%5Cln%7B%5Cleft%285%20%5Cright%29%7D%20%2B%20%5Cln%7B%5Cleft%283%20%5Cright%29%7D%7D%20" alt="LaTeX: \displaystyle x = \frac{- \ln{\left(3 \right)}}{- \ln{\left(5 \right)} + \ln{\left(3 \right)}} " data-equation-content=" \displaystyle x = \frac{- \ln{\left(3 \right)}}{- \ln{\left(5 \right)} + \ln{\left(3 \right)}} " /> . </p> </p>