\(\text{www.the}\beta\text{etafunction.com}\)
Home
Login
Questions: Algebra BusinessCalculus
Please login to create an exam or a quiz.
Solve the equation \(\displaystyle \log_{9}(x + 59032)-\log_{9}(x + 712)=2\).
Using the quotient property of logarithms gives \(\displaystyle \log_{9}\frac{x + 59032}{x + 712} = 2\). Making both sides of the equation exponents on the base \(\displaystyle 9\) gives \(\displaystyle \frac{x + 59032}{x + 712}=81\). Clearing the fractions by multiplying by the LCD gives \(\displaystyle x + 59032=81 x + 57672\). Isolating \(\displaystyle x\) gives \(\displaystyle x = 17\).
\begin{question}Solve the equation $\log_{9}(x + 59032)-\log_{9}(x + 712)=2$. \soln{9cm}{Using the quotient property of logarithms gives $\log_{9}\frac{x + 59032}{x + 712} = 2$. Making both sides of the equation exponents on the base $9$ gives $\frac{x + 59032}{x + 712}=81$. Clearing the fractions by multiplying by the LCD gives $x + 59032=81 x + 57672$. Isolating $x$ gives $x = 17$. } \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 the equation <img class="equation_image" title=" \displaystyle \log_{9}(x + 59032)-\log_{9}(x + 712)=2 " src="/equation_images/%20%5Cdisplaystyle%20%5Clog_%7B9%7D%28x%20%2B%2059032%29-%5Clog_%7B9%7D%28x%20%2B%20712%29%3D2%20" alt="LaTeX: \displaystyle \log_{9}(x + 59032)-\log_{9}(x + 712)=2 " data-equation-content=" \displaystyle \log_{9}(x + 59032)-\log_{9}(x + 712)=2 " /> . </p> </p>
<p> <p>Using the quotient property of logarithms gives <img class="equation_image" title=" \displaystyle \log_{9}\frac{x + 59032}{x + 712} = 2 " src="/equation_images/%20%5Cdisplaystyle%20%5Clog_%7B9%7D%5Cfrac%7Bx%20%2B%2059032%7D%7Bx%20%2B%20712%7D%20%3D%202%20" alt="LaTeX: \displaystyle \log_{9}\frac{x + 59032}{x + 712} = 2 " data-equation-content=" \displaystyle \log_{9}\frac{x + 59032}{x + 712} = 2 " /> . Making both sides of the equation exponents on the base <img class="equation_image" title=" \displaystyle 9 " src="/equation_images/%20%5Cdisplaystyle%209%20" alt="LaTeX: \displaystyle 9 " data-equation-content=" \displaystyle 9 " /> gives <img class="equation_image" title=" \displaystyle \frac{x + 59032}{x + 712}=81 " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7Bx%20%2B%2059032%7D%7Bx%20%2B%20712%7D%3D81%20" alt="LaTeX: \displaystyle \frac{x + 59032}{x + 712}=81 " data-equation-content=" \displaystyle \frac{x + 59032}{x + 712}=81 " /> . Clearing the fractions by multiplying by the LCD gives <img class="equation_image" title=" \displaystyle x + 59032=81 x + 57672 " src="/equation_images/%20%5Cdisplaystyle%20x%20%2B%2059032%3D81%20x%20%2B%2057672%20" alt="LaTeX: \displaystyle x + 59032=81 x + 57672 " data-equation-content=" \displaystyle x + 59032=81 x + 57672 " /> . Isolating <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 = 17 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%2017%20" alt="LaTeX: \displaystyle x = 17 " data-equation-content=" \displaystyle x = 17 " /> . </p> </p>