\(\text{www.the}\beta\text{etafunction.com}\)
Home
Login
Questions: Algebra BusinessCalculus

Please login to create an exam or a quiz.

Algebra
Logarithms
New Random

Solve \(\displaystyle \log_{6}(x + 24)+\log_{6}(x + 235) = 5\).


Using logarithmic properties and expanding the argument gives \(\displaystyle \log_{6}(x^{2} + 259 x + 5640)=5\). Making both sides an exponent on the base gives \(\displaystyle x^{2} + 259 x + 5640=6^{5}\). Expanding and setting equal to zero gives \(\displaystyle x^{2} + 259 x - 2136=0\). Factoring gives \(\displaystyle \left(x - 8\right) \left(x + 267\right)=0\). Solving gives the two possible solutions \(\displaystyle x = -267\) and \(\displaystyle x = 8\). The domain of the original is \(\displaystyle \left(-24, \infty\right) \bigcap \left(-235, \infty\right)=\left(-24, \infty\right)\). Checking if each possible solution is in the domain gives: \(\displaystyle x = -267\) is not a solution. \(\displaystyle x=8\) is a solution.

Download \(\LaTeX\)

\begin{question}Solve $\log_{6}(x + 24)+\log_{6}(x + 235) = 5$. 
    \soln{9cm}{Using logarithmic properties and expanding the argument gives $\log_{6}(x^{2} + 259 x + 5640)=5$. Making both sides an exponent on the base gives $x^{2} + 259 x + 5640=6^{5}$. Expanding and setting equal to zero gives $x^{2} + 259 x - 2136=0$. Factoring gives $\left(x - 8\right) \left(x + 267\right)=0$. Solving gives the two possible solutions $x = -267$ and $x = 8$. The domain of the original is $\left(-24, \infty\right) \bigcap \left(-235, \infty\right)=\left(-24, \infty\right)$.  Checking if each possible solution is in the domain gives: $x = -267$ is not a solution. $x=8$ is a solution. }

\end{question}

Download Question and Solution Environment\(\LaTeX\)
\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}
HTML for Canvas
<p> <p>Solve  <img class="equation_image" title=" \displaystyle \log_{6}(x + 24)+\log_{6}(x + 235) = 5 " src="/equation_images/%20%5Cdisplaystyle%20%5Clog_%7B6%7D%28x%20%2B%2024%29%2B%5Clog_%7B6%7D%28x%20%2B%20235%29%20%3D%205%20" alt="LaTeX:  \displaystyle \log_{6}(x + 24)+\log_{6}(x + 235) = 5 " data-equation-content=" \displaystyle \log_{6}(x + 24)+\log_{6}(x + 235) = 5 " /> . </p> </p>
HTML for Canvas
<p> <p>Using logarithmic properties and expanding the argument gives  <img class="equation_image" title=" \displaystyle \log_{6}(x^{2} + 259 x + 5640)=5 " src="/equation_images/%20%5Cdisplaystyle%20%5Clog_%7B6%7D%28x%5E%7B2%7D%20%2B%20259%20x%20%2B%205640%29%3D5%20" alt="LaTeX:  \displaystyle \log_{6}(x^{2} + 259 x + 5640)=5 " data-equation-content=" \displaystyle \log_{6}(x^{2} + 259 x + 5640)=5 " /> . Making both sides an exponent on the base gives  <img class="equation_image" title=" \displaystyle x^{2} + 259 x + 5640=6^{5} " src="/equation_images/%20%5Cdisplaystyle%20x%5E%7B2%7D%20%2B%20259%20x%20%2B%205640%3D6%5E%7B5%7D%20" alt="LaTeX:  \displaystyle x^{2} + 259 x + 5640=6^{5} " data-equation-content=" \displaystyle x^{2} + 259 x + 5640=6^{5} " /> . Expanding and setting equal to zero gives  <img class="equation_image" title=" \displaystyle x^{2} + 259 x - 2136=0 " src="/equation_images/%20%5Cdisplaystyle%20x%5E%7B2%7D%20%2B%20259%20x%20-%202136%3D0%20" alt="LaTeX:  \displaystyle x^{2} + 259 x - 2136=0 " data-equation-content=" \displaystyle x^{2} + 259 x - 2136=0 " /> . Factoring gives  <img class="equation_image" title=" \displaystyle \left(x - 8\right) \left(x + 267\right)=0 " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28x%20-%208%5Cright%29%20%5Cleft%28x%20%2B%20267%5Cright%29%3D0%20" alt="LaTeX:  \displaystyle \left(x - 8\right) \left(x + 267\right)=0 " data-equation-content=" \displaystyle \left(x - 8\right) \left(x + 267\right)=0 " /> . Solving gives the two possible solutions  <img class="equation_image" title=" \displaystyle x = -267 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-267%20" alt="LaTeX:  \displaystyle x = -267 " data-equation-content=" \displaystyle x = -267 " />  and  <img class="equation_image" title=" \displaystyle x = 8 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%208%20" alt="LaTeX:  \displaystyle x = 8 " data-equation-content=" \displaystyle x = 8 " /> . The domain of the original is  <img class="equation_image" title=" \displaystyle \left(-24, \infty\right) \bigcap \left(-235, \infty\right)=\left(-24, \infty\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28-24%2C%20%5Cinfty%5Cright%29%20%5Cbigcap%20%5Cleft%28-235%2C%20%5Cinfty%5Cright%29%3D%5Cleft%28-24%2C%20%5Cinfty%5Cright%29%20" alt="LaTeX:  \displaystyle \left(-24, \infty\right) \bigcap \left(-235, \infty\right)=\left(-24, \infty\right) " data-equation-content=" \displaystyle \left(-24, \infty\right) \bigcap \left(-235, \infty\right)=\left(-24, \infty\right) " /> .  Checking if each possible solution is in the domain gives:  <img class="equation_image" title=" \displaystyle x = -267 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-267%20" alt="LaTeX:  \displaystyle x = -267 " data-equation-content=" \displaystyle x = -267 " />  is not a solution.  <img class="equation_image" title=" \displaystyle x=8 " src="/equation_images/%20%5Cdisplaystyle%20x%3D8%20" alt="LaTeX:  \displaystyle x=8 " data-equation-content=" \displaystyle x=8 " />  is a solution. </p> </p>