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Algebra
Logarithms
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Solve \(\displaystyle \log_{ 15 }(x + 20) + \log_{ 15 }(x + 3380) = 4\)

Using the product rule for logarithms gives \(\displaystyle \log_{ 15 }(\left(x + 20\right) \left(x + 3380\right)) \) and rewriting in exponential form gives \(\displaystyle \left(x + 20\right) \left(x + 3380\right) = 50625\) expanding and setting the equation equal to zero gives \(\displaystyle x^{2} + 3400 x + 16975 = 0\). Factoring gives \(\displaystyle \left(x + 5\right) \left(x + 3395\right)=0\). This gives two possible solutions \(\displaystyle x=-3395\) or \(\displaystyle x=-5\). \(\displaystyle x=-3395\) is an extraneous solution. The only soution is \(\displaystyle x=-5\).

Download \(\LaTeX\) 

\begin{question}Solve $\log_{ 15 }(x + 20) + \log_{ 15 }(x + 3380) = 4$
    \soln{10cm}{Using the product rule for logarithms gives $\log_{ 15 }(\left(x + 20\right) \left(x + 3380\right)) $ and rewriting in exponential form gives $\left(x + 20\right) \left(x + 3380\right) = 50625$ expanding and setting the equation equal to zero gives $x^{2} + 3400 x + 16975 = 0$.  Factoring gives $\left(x + 5\right) \left(x + 3395\right)=0$. This gives two possible solutions $x=-3395$ or $x=-5$. $x=-3395$ 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 + 3380) = 4 " 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%203380%29%20%3D%204%20" alt="LaTeX:  \displaystyle \log_{ 15 }(x + 20) + \log_{ 15 }(x + 3380) = 4 " data-equation-content=" \displaystyle \log_{ 15 }(x + 20) + \log_{ 15 }(x + 3380) = 4 " /> </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) \left(x + 3380\right))  " src="/equation_images/%20%5Cdisplaystyle%20%5Clog_%7B%2015%20%7D%28%5Cleft%28x%20%2B%2020%5Cright%29%20%5Cleft%28x%20%2B%203380%5Cright%29%29%20%20" alt="LaTeX:  \displaystyle \log_{ 15 }(\left(x + 20\right) \left(x + 3380\right))  " data-equation-content=" \displaystyle \log_{ 15 }(\left(x + 20\right) \left(x + 3380\right))  " />  and rewriting in exponential form gives  <img class="equation_image" title=" \displaystyle \left(x + 20\right) \left(x + 3380\right) = 50625 " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28x%20%2B%2020%5Cright%29%20%5Cleft%28x%20%2B%203380%5Cright%29%20%3D%2050625%20" alt="LaTeX:  \displaystyle \left(x + 20\right) \left(x + 3380\right) = 50625 " data-equation-content=" \displaystyle \left(x + 20\right) \left(x + 3380\right) = 50625 " />  expanding and setting the equation equal to zero gives  <img class="equation_image" title=" \displaystyle x^{2} + 3400 x + 16975 = 0 " src="/equation_images/%20%5Cdisplaystyle%20x%5E%7B2%7D%20%2B%203400%20x%20%2B%2016975%20%3D%200%20" alt="LaTeX:  \displaystyle x^{2} + 3400 x + 16975 = 0 " data-equation-content=" \displaystyle x^{2} + 3400 x + 16975 = 0 " /> .  Factoring gives  <img class="equation_image" title=" \displaystyle \left(x + 5\right) \left(x + 3395\right)=0 " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28x%20%2B%205%5Cright%29%20%5Cleft%28x%20%2B%203395%5Cright%29%3D0%20" alt="LaTeX:  \displaystyle \left(x + 5\right) \left(x + 3395\right)=0 " data-equation-content=" \displaystyle \left(x + 5\right) \left(x + 3395\right)=0 " /> . This gives two possible solutions  <img class="equation_image" title=" \displaystyle x=-3395 " src="/equation_images/%20%5Cdisplaystyle%20x%3D-3395%20" alt="LaTeX:  \displaystyle x=-3395 " data-equation-content=" \displaystyle x=-3395 " />  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=-3395 " src="/equation_images/%20%5Cdisplaystyle%20x%3D-3395%20" alt="LaTeX:  \displaystyle x=-3395 " data-equation-content=" \displaystyle x=-3395 " />  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>