\(\text{www.the}\beta\text{etafunction.com}\)
Home
Login
Questions: Algebra BusinessCalculus
Please login to create an exam or a quiz.
Solve the equation \(\displaystyle \log_{7}(x + 16802)-\log_{7}(x + 2396)=1\).
Using the quotient property of logarithms gives \(\displaystyle \log_{7}\frac{x + 16802}{x + 2396} = 1\). Making both sides of the equation exponents on the base \(\displaystyle 7\) gives \(\displaystyle \frac{x + 16802}{x + 2396}=7\). Clearing the fractions by multiplying by the LCD gives \(\displaystyle x + 16802=7 x + 16772\). Isolating \(\displaystyle x\) gives \(\displaystyle x = 5\).
\begin{question}Solve the equation $\log_{7}(x + 16802)-\log_{7}(x + 2396)=1$.
\soln{9cm}{Using the quotient property of logarithms gives $\log_{7}\frac{x + 16802}{x + 2396} = 1$. Making both sides of the equation exponents on the base $7$ gives $\frac{x + 16802}{x + 2396}=7$. Clearing the fractions by multiplying by the LCD gives $x + 16802=7 x + 16772$. Isolating $x$ gives $x = 5$. }
\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_{7}(x + 16802)-\log_{7}(x + 2396)=1 " src="/equation_images/%20%5Cdisplaystyle%20%5Clog_%7B7%7D%28x%20%2B%2016802%29-%5Clog_%7B7%7D%28x%20%2B%202396%29%3D1%20" alt="LaTeX: \displaystyle \log_{7}(x + 16802)-\log_{7}(x + 2396)=1 " data-equation-content=" \displaystyle \log_{7}(x + 16802)-\log_{7}(x + 2396)=1 " /> . </p> </p><p> <p>Using the quotient property of logarithms gives <img class="equation_image" title=" \displaystyle \log_{7}\frac{x + 16802}{x + 2396} = 1 " src="/equation_images/%20%5Cdisplaystyle%20%5Clog_%7B7%7D%5Cfrac%7Bx%20%2B%2016802%7D%7Bx%20%2B%202396%7D%20%3D%201%20" alt="LaTeX: \displaystyle \log_{7}\frac{x + 16802}{x + 2396} = 1 " data-equation-content=" \displaystyle \log_{7}\frac{x + 16802}{x + 2396} = 1 " /> . Making both sides of the equation exponents on the base <img class="equation_image" title=" \displaystyle 7 " src="/equation_images/%20%5Cdisplaystyle%207%20" alt="LaTeX: \displaystyle 7 " data-equation-content=" \displaystyle 7 " /> gives <img class="equation_image" title=" \displaystyle \frac{x + 16802}{x + 2396}=7 " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7Bx%20%2B%2016802%7D%7Bx%20%2B%202396%7D%3D7%20" alt="LaTeX: \displaystyle \frac{x + 16802}{x + 2396}=7 " data-equation-content=" \displaystyle \frac{x + 16802}{x + 2396}=7 " /> . Clearing the fractions by multiplying by the LCD gives <img class="equation_image" title=" \displaystyle x + 16802=7 x + 16772 " src="/equation_images/%20%5Cdisplaystyle%20x%20%2B%2016802%3D7%20x%20%2B%2016772%20" alt="LaTeX: \displaystyle x + 16802=7 x + 16772 " data-equation-content=" \displaystyle x + 16802=7 x + 16772 " /> . 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 = 5 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%205%20" alt="LaTeX: \displaystyle x = 5 " data-equation-content=" \displaystyle x = 5 " /> . </p> </p>