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Algebra
Logarithms
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Solve \(\displaystyle \log_{ 15 }(x + 20) + \log_{ 15 }(x + 20) = 2\)

Using the product rule for logarithms gives \(\displaystyle \log_{ 15 }(\left(x + 20\right)^{2}) \) and rewriting in exponential form gives \(\displaystyle \left(x + 20\right)^{2} = 225\) expanding and setting the equation equal to zero gives \(\displaystyle x^{2} + 40 x + 175 = 0\). Factoring gives \(\displaystyle \left(x + 5\right) \left(x + 35\right)=0\). This gives two possible solutions \(\displaystyle x=-35\) or \(\displaystyle x=-5\). \(\displaystyle x=-35\) is an extraneous solution. The only soution is \(\displaystyle x=-5\).

Download \(\LaTeX\) 

\begin{question}Solve $\log_{ 15 }(x + 20) + \log_{ 15 }(x + 20) = 2$
    \soln{10cm}{Using the product rule for logarithms gives $\log_{ 15 }(\left(x + 20\right)^{2}) $ and rewriting in exponential form gives $\left(x + 20\right)^{2} = 225$ expanding and setting the equation equal to zero gives $x^{2} + 40 x + 175 = 0$.  Factoring gives $\left(x + 5\right) \left(x + 35\right)=0$. This gives two possible solutions $x=-35$ or $x=-5$. $x=-35$ is an extraneous solution.  The only soution is $x=-5$. }

\end{question}

Download Question and Solution Environment\(\LaTeX\)
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HTML for Canvas 
<p> <p>Solve  <img class="equation_image" title=" \displaystyle \log_{ 15 }(x + 20) + \log_{ 15 }(x + 20) = 2 " src="/equation_images/%20%5Cdisplaystyle%20%5Clog_%7B%2015%20%7D%28x%20%2B%2020%29%20%2B%20%5Clog_%7B%2015%20%7D%28x%20%2B%2020%29%20%3D%202%20" alt="LaTeX:  \displaystyle \log_{ 15 }(x + 20) + \log_{ 15 }(x + 20) = 2 " data-equation-content=" \displaystyle \log_{ 15 }(x + 20) + \log_{ 15 }(x + 20) = 2 " /> </p> </p>
HTML for Canvas
<p> <p>Using the product rule for logarithms gives  <img class="equation_image" title=" \displaystyle \log_{ 15 }(\left(x + 20\right)^{2})  " src="/equation_images/%20%5Cdisplaystyle%20%5Clog_%7B%2015%20%7D%28%5Cleft%28x%20%2B%2020%5Cright%29%5E%7B2%7D%29%20%20" alt="LaTeX:  \displaystyle \log_{ 15 }(\left(x + 20\right)^{2})  " data-equation-content=" \displaystyle \log_{ 15 }(\left(x + 20\right)^{2})  " />  and rewriting in exponential form gives  <img class="equation_image" title=" \displaystyle \left(x + 20\right)^{2} = 225 " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28x%20%2B%2020%5Cright%29%5E%7B2%7D%20%3D%20225%20" alt="LaTeX:  \displaystyle \left(x + 20\right)^{2} = 225 " data-equation-content=" \displaystyle \left(x + 20\right)^{2} = 225 " />  expanding and setting the equation equal to zero gives  <img class="equation_image" title=" \displaystyle x^{2} + 40 x + 175 = 0 " src="/equation_images/%20%5Cdisplaystyle%20x%5E%7B2%7D%20%2B%2040%20x%20%2B%20175%20%3D%200%20" alt="LaTeX:  \displaystyle x^{2} + 40 x + 175 = 0 " data-equation-content=" \displaystyle x^{2} + 40 x + 175 = 0 " /> .  Factoring gives  <img class="equation_image" title=" \displaystyle \left(x + 5\right) \left(x + 35\right)=0 " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28x%20%2B%205%5Cright%29%20%5Cleft%28x%20%2B%2035%5Cright%29%3D0%20" alt="LaTeX:  \displaystyle \left(x + 5\right) \left(x + 35\right)=0 " data-equation-content=" \displaystyle \left(x + 5\right) \left(x + 35\right)=0 " /> . This gives two possible solutions  <img class="equation_image" title=" \displaystyle x=-35 " src="/equation_images/%20%5Cdisplaystyle%20x%3D-35%20" alt="LaTeX:  \displaystyle x=-35 " data-equation-content=" \displaystyle x=-35 " />  or  <img class="equation_image" title=" \displaystyle x=-5 " src="/equation_images/%20%5Cdisplaystyle%20x%3D-5%20" alt="LaTeX:  \displaystyle x=-5 " data-equation-content=" \displaystyle x=-5 " /> .  <img class="equation_image" title=" \displaystyle x=-35 " src="/equation_images/%20%5Cdisplaystyle%20x%3D-35%20" alt="LaTeX:  \displaystyle x=-35 " data-equation-content=" \displaystyle x=-35 " />  is an extraneous solution.  The only soution is  <img class="equation_image" title=" \displaystyle x=-5 " src="/equation_images/%20%5Cdisplaystyle%20x%3D-5%20" alt="LaTeX:  \displaystyle x=-5 " data-equation-content=" \displaystyle x=-5 " /> . </p> </p>