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Solve \(\displaystyle 5^{x + 6}=6^{x}\).
Taking the natural logarithm of both sides gives \(\displaystyle (x + 6)\ln(5)=x\ln(6)\). Distributing gives \(\displaystyle x \ln{\left(5 \right)} + 6 \ln{\left(5 \right)} = x \ln{\left(6 \right)}\). Moving all the \(\displaystyle x\) terms to the left hand side and all of the constants to the right side gives: \(\displaystyle - x \ln{\left(6 \right)} + x \ln{\left(5 \right)} = - 6 \ln{\left(5 \right)}\). Factoring out the \(\displaystyle x\) gives \(\displaystyle x \left(- \ln{\left(6 \right)} + \ln{\left(5 \right)}\right)=- 6 \ln{\left(5 \right)}\). Isolating \(\displaystyle x\) yeilds \(\displaystyle x = \frac{- \ln{\left(5 \right)}}{- \ln{\left(6 \right)} + \ln{\left(5 \right)}}\).
\begin{question}Solve $5^{x + 6}=6^{x}$. \soln{9cm}{Taking the natural logarithm of both sides gives $(x + 6)\ln(5)=x\ln(6)$. Distributing gives $ x \ln{\left(5 \right)} + 6 \ln{\left(5 \right)} = x \ln{\left(6 \right)}$. Moving all the $x$ terms to the left hand side and all of the constants to the right side gives: $- x \ln{\left(6 \right)} + x \ln{\left(5 \right)} = - 6 \ln{\left(5 \right)}$. Factoring out the $x$ gives $x \left(- \ln{\left(6 \right)} + \ln{\left(5 \right)}\right)=- 6 \ln{\left(5 \right)}$. Isolating $x$ yeilds $x = \frac{- \ln{\left(5 \right)}}{- \ln{\left(6 \right)} + \ln{\left(5 \right)}}$. } \end{question}
\documentclass{article} \usepackage{tikz} \usepackage{amsmath} \usepackage[margin=2cm]{geometry} \usepackage{tcolorbox} \newcounter{ExamNumber} \newcounter{questioncount} \stepcounter{questioncount} \newenvironment{question}{{\noindent\bfseries Question \arabic{questioncount}.}}{\stepcounter{questioncount}} \renewcommand{\labelenumi}{{\bfseries (\alph{enumi})}} \newif\ifShowSolution \newcommand{\soln}[2]{% \ifShowSolution% \noindent\begin{tcolorbox}[colframe=blue,title=Solution]#2\end{tcolorbox}\else% \vspace{#1}% \fi% }% \newcommand{\hideifShowSolution}[1]{% \ifShowSolution% % \else% #1% \fi% }% \everymath{\displaystyle} \ShowSolutiontrue \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 5^{x + 6}=6^{x} " src="/equation_images/%20%5Cdisplaystyle%205%5E%7Bx%20%2B%206%7D%3D6%5E%7Bx%7D%20" alt="LaTeX: \displaystyle 5^{x + 6}=6^{x} " data-equation-content=" \displaystyle 5^{x + 6}=6^{x} " /> . </p> </p>
<p> <p>Taking the natural logarithm of both sides gives <img class="equation_image" title=" \displaystyle (x + 6)\ln(5)=x\ln(6) " src="/equation_images/%20%5Cdisplaystyle%20%28x%20%2B%206%29%5Cln%285%29%3Dx%5Cln%286%29%20" alt="LaTeX: \displaystyle (x + 6)\ln(5)=x\ln(6) " data-equation-content=" \displaystyle (x + 6)\ln(5)=x\ln(6) " /> . Distributing gives <img class="equation_image" title=" \displaystyle x \ln{\left(5 \right)} + 6 \ln{\left(5 \right)} = x \ln{\left(6 \right)} " src="/equation_images/%20%5Cdisplaystyle%20%20x%20%5Cln%7B%5Cleft%285%20%5Cright%29%7D%20%2B%206%20%5Cln%7B%5Cleft%285%20%5Cright%29%7D%20%3D%20x%20%5Cln%7B%5Cleft%286%20%5Cright%29%7D%20" alt="LaTeX: \displaystyle x \ln{\left(5 \right)} + 6 \ln{\left(5 \right)} = x \ln{\left(6 \right)} " data-equation-content=" \displaystyle x \ln{\left(5 \right)} + 6 \ln{\left(5 \right)} = x \ln{\left(6 \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(6 \right)} + x \ln{\left(5 \right)} = - 6 \ln{\left(5 \right)} " src="/equation_images/%20%5Cdisplaystyle%20-%20x%20%5Cln%7B%5Cleft%286%20%5Cright%29%7D%20%2B%20x%20%5Cln%7B%5Cleft%285%20%5Cright%29%7D%20%3D%20-%206%20%5Cln%7B%5Cleft%285%20%5Cright%29%7D%20" alt="LaTeX: \displaystyle - x \ln{\left(6 \right)} + x \ln{\left(5 \right)} = - 6 \ln{\left(5 \right)} " data-equation-content=" \displaystyle - x \ln{\left(6 \right)} + x \ln{\left(5 \right)} = - 6 \ln{\left(5 \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(6 \right)} + \ln{\left(5 \right)}\right)=- 6 \ln{\left(5 \right)} " src="/equation_images/%20%5Cdisplaystyle%20x%20%5Cleft%28-%20%5Cln%7B%5Cleft%286%20%5Cright%29%7D%20%2B%20%5Cln%7B%5Cleft%285%20%5Cright%29%7D%5Cright%29%3D-%206%20%5Cln%7B%5Cleft%285%20%5Cright%29%7D%20" alt="LaTeX: \displaystyle x \left(- \ln{\left(6 \right)} + \ln{\left(5 \right)}\right)=- 6 \ln{\left(5 \right)} " data-equation-content=" \displaystyle x \left(- \ln{\left(6 \right)} + \ln{\left(5 \right)}\right)=- 6 \ln{\left(5 \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(5 \right)}}{- \ln{\left(6 \right)} + \ln{\left(5 \right)}} " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20%5Cfrac%7B-%20%5Cln%7B%5Cleft%285%20%5Cright%29%7D%7D%7B-%20%5Cln%7B%5Cleft%286%20%5Cright%29%7D%20%2B%20%5Cln%7B%5Cleft%285%20%5Cright%29%7D%7D%20" alt="LaTeX: \displaystyle x = \frac{- \ln{\left(5 \right)}}{- \ln{\left(6 \right)} + \ln{\left(5 \right)}} " data-equation-content=" \displaystyle x = \frac{- \ln{\left(5 \right)}}{- \ln{\left(6 \right)} + \ln{\left(5 \right)}} " /> . </p> </p>