\(\text{www.the}\beta\text{etafunction.com}\)
Home
Login
Questions: Algebra BusinessCalculus
Please login to create an exam or a quiz.
Solve \(\displaystyle \log_{ 4 }(x + 8) + \log_{ 4 }(x + 20) = 3\)
Using the product rule for logarithms gives \(\displaystyle \log_{ 4 }(\left(x + 8\right) \left(x + 20\right)) \) and rewriting in exponential form gives \(\displaystyle \left(x + 8\right) \left(x + 20\right) = 64\) expanding and setting the equation equal to zero gives \(\displaystyle x^{2} + 28 x + 96 = 0\). Factoring gives \(\displaystyle \left(x + 4\right) \left(x + 24\right)=0\). This gives two possible solutions \(\displaystyle x=-24\) or \(\displaystyle x=-4\). \(\displaystyle x=-24\) is an extraneous solution. The only soution is \(\displaystyle x=-4\).
\begin{question}Solve $\log_{ 4 }(x + 8) + \log_{ 4 }(x + 20) = 3$ \soln{10cm}{Using the product rule for logarithms gives $\log_{ 4 }(\left(x + 8\right) \left(x + 20\right)) $ and rewriting in exponential form gives $\left(x + 8\right) \left(x + 20\right) = 64$ expanding and setting the equation equal to zero gives $x^{2} + 28 x + 96 = 0$. Factoring gives $\left(x + 4\right) \left(x + 24\right)=0$. This gives two possible solutions $x=-24$ or $x=-4$. $x=-24$ is an extraneous solution. The only soution is $x=-4$. } \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 \log_{ 4 }(x + 8) + \log_{ 4 }(x + 20) = 3 " src="/equation_images/%20%5Cdisplaystyle%20%5Clog_%7B%204%20%7D%28x%20%2B%208%29%20%2B%20%5Clog_%7B%204%20%7D%28x%20%2B%2020%29%20%3D%203%20" alt="LaTeX: \displaystyle \log_{ 4 }(x + 8) + \log_{ 4 }(x + 20) = 3 " data-equation-content=" \displaystyle \log_{ 4 }(x + 8) + \log_{ 4 }(x + 20) = 3 " /> </p> </p>
<p> <p>Using the product rule for logarithms gives <img class="equation_image" title=" \displaystyle \log_{ 4 }(\left(x + 8\right) \left(x + 20\right)) " src="/equation_images/%20%5Cdisplaystyle%20%5Clog_%7B%204%20%7D%28%5Cleft%28x%20%2B%208%5Cright%29%20%5Cleft%28x%20%2B%2020%5Cright%29%29%20%20" alt="LaTeX: \displaystyle \log_{ 4 }(\left(x + 8\right) \left(x + 20\right)) " data-equation-content=" \displaystyle \log_{ 4 }(\left(x + 8\right) \left(x + 20\right)) " /> and rewriting in exponential form gives <img class="equation_image" title=" \displaystyle \left(x + 8\right) \left(x + 20\right) = 64 " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28x%20%2B%208%5Cright%29%20%5Cleft%28x%20%2B%2020%5Cright%29%20%3D%2064%20" alt="LaTeX: \displaystyle \left(x + 8\right) \left(x + 20\right) = 64 " data-equation-content=" \displaystyle \left(x + 8\right) \left(x + 20\right) = 64 " /> expanding and setting the equation equal to zero gives <img class="equation_image" title=" \displaystyle x^{2} + 28 x + 96 = 0 " src="/equation_images/%20%5Cdisplaystyle%20x%5E%7B2%7D%20%2B%2028%20x%20%2B%2096%20%3D%200%20" alt="LaTeX: \displaystyle x^{2} + 28 x + 96 = 0 " data-equation-content=" \displaystyle x^{2} + 28 x + 96 = 0 " /> . Factoring gives <img class="equation_image" title=" \displaystyle \left(x + 4\right) \left(x + 24\right)=0 " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28x%20%2B%204%5Cright%29%20%5Cleft%28x%20%2B%2024%5Cright%29%3D0%20" alt="LaTeX: \displaystyle \left(x + 4\right) \left(x + 24\right)=0 " data-equation-content=" \displaystyle \left(x + 4\right) \left(x + 24\right)=0 " /> . This gives two possible solutions <img class="equation_image" title=" \displaystyle x=-24 " src="/equation_images/%20%5Cdisplaystyle%20x%3D-24%20" alt="LaTeX: \displaystyle x=-24 " data-equation-content=" \displaystyle x=-24 " /> or <img class="equation_image" title=" \displaystyle x=-4 " src="/equation_images/%20%5Cdisplaystyle%20x%3D-4%20" alt="LaTeX: \displaystyle x=-4 " data-equation-content=" \displaystyle x=-4 " /> . <img class="equation_image" title=" \displaystyle x=-24 " src="/equation_images/%20%5Cdisplaystyle%20x%3D-24%20" alt="LaTeX: \displaystyle x=-24 " data-equation-content=" \displaystyle x=-24 " /> is an extraneous solution. The only soution is <img class="equation_image" title=" \displaystyle x=-4 " src="/equation_images/%20%5Cdisplaystyle%20x%3D-4%20" alt="LaTeX: \displaystyle x=-4 " data-equation-content=" \displaystyle x=-4 " /> . </p> </p>