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Algebra
Logarithms
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Solve \(\displaystyle \log_{ 12 }(x + 15) + \log_{ 12 }(x + 15) = 2\)

Using the product rule for logarithms gives \(\displaystyle \log_{ 12 }(\left(x + 15\right)^{2}) \) and rewriting in exponential form gives \(\displaystyle \left(x + 15\right)^{2} = 144\) expanding and setting the equation equal to zero gives \(\displaystyle x^{2} + 30 x + 81 = 0\). Factoring gives \(\displaystyle \left(x + 3\right) \left(x + 27\right)=0\). This gives two possible solutions \(\displaystyle x=-27\) or \(\displaystyle x=-3\). \(\displaystyle x=-27\) is an extraneous solution. The only soution is \(\displaystyle x=-3\).

Download \(\LaTeX\) 

\begin{question}Solve $\log_{ 12 }(x + 15) + \log_{ 12 }(x + 15) = 2$
    \soln{10cm}{Using the product rule for logarithms gives $\log_{ 12 }(\left(x + 15\right)^{2}) $ and rewriting in exponential form gives $\left(x + 15\right)^{2} = 144$ expanding and setting the equation equal to zero gives $x^{2} + 30 x + 81 = 0$.  Factoring gives $\left(x + 3\right) \left(x + 27\right)=0$. This gives two possible solutions $x=-27$ or $x=-3$. $x=-27$ is an extraneous solution.  The only soution is $x=-3$. }

\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_{ 12 }(x + 15) + \log_{ 12 }(x + 15) = 2 " src="/equation_images/%20%5Cdisplaystyle%20%5Clog_%7B%2012%20%7D%28x%20%2B%2015%29%20%2B%20%5Clog_%7B%2012%20%7D%28x%20%2B%2015%29%20%3D%202%20" alt="LaTeX:  \displaystyle \log_{ 12 }(x + 15) + \log_{ 12 }(x + 15) = 2 " data-equation-content=" \displaystyle \log_{ 12 }(x + 15) + \log_{ 12 }(x + 15) = 2 " /> </p> </p>
HTML for Canvas
<p> <p>Using the product rule for logarithms gives  <img class="equation_image" title=" \displaystyle \log_{ 12 }(\left(x + 15\right)^{2})  " src="/equation_images/%20%5Cdisplaystyle%20%5Clog_%7B%2012%20%7D%28%5Cleft%28x%20%2B%2015%5Cright%29%5E%7B2%7D%29%20%20" alt="LaTeX:  \displaystyle \log_{ 12 }(\left(x + 15\right)^{2})  " data-equation-content=" \displaystyle \log_{ 12 }(\left(x + 15\right)^{2})  " />  and rewriting in exponential form gives  <img class="equation_image" title=" \displaystyle \left(x + 15\right)^{2} = 144 " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28x%20%2B%2015%5Cright%29%5E%7B2%7D%20%3D%20144%20" alt="LaTeX:  \displaystyle \left(x + 15\right)^{2} = 144 " data-equation-content=" \displaystyle \left(x + 15\right)^{2} = 144 " />  expanding and setting the equation equal to zero gives  <img class="equation_image" title=" \displaystyle x^{2} + 30 x + 81 = 0 " src="/equation_images/%20%5Cdisplaystyle%20x%5E%7B2%7D%20%2B%2030%20x%20%2B%2081%20%3D%200%20" alt="LaTeX:  \displaystyle x^{2} + 30 x + 81 = 0 " data-equation-content=" \displaystyle x^{2} + 30 x + 81 = 0 " /> .  Factoring gives  <img class="equation_image" title=" \displaystyle \left(x + 3\right) \left(x + 27\right)=0 " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28x%20%2B%203%5Cright%29%20%5Cleft%28x%20%2B%2027%5Cright%29%3D0%20" alt="LaTeX:  \displaystyle \left(x + 3\right) \left(x + 27\right)=0 " data-equation-content=" \displaystyle \left(x + 3\right) \left(x + 27\right)=0 " /> . This gives two possible solutions  <img class="equation_image" title=" \displaystyle x=-27 " src="/equation_images/%20%5Cdisplaystyle%20x%3D-27%20" alt="LaTeX:  \displaystyle x=-27 " data-equation-content=" \displaystyle x=-27 " />  or  <img class="equation_image" title=" \displaystyle x=-3 " src="/equation_images/%20%5Cdisplaystyle%20x%3D-3%20" alt="LaTeX:  \displaystyle x=-3 " data-equation-content=" \displaystyle x=-3 " /> .  <img class="equation_image" title=" \displaystyle x=-27 " src="/equation_images/%20%5Cdisplaystyle%20x%3D-27%20" alt="LaTeX:  \displaystyle x=-27 " data-equation-content=" \displaystyle x=-27 " />  is an extraneous solution.  The only soution is  <img class="equation_image" title=" \displaystyle x=-3 " src="/equation_images/%20%5Cdisplaystyle%20x%3D-3%20" alt="LaTeX:  \displaystyle x=-3 " data-equation-content=" \displaystyle x=-3 " /> . </p> </p>