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Algebra
Logarithms
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Solve \(\displaystyle \log_{ 15 }(x + 20) + \log_{ 15 }(x + 230) = 3\)


Using the product rule for logarithms gives \(\displaystyle \log_{ 15 }(\left(x + 20\right) \left(x + 230\right)) \) and rewriting in exponential form gives \(\displaystyle \left(x + 20\right) \left(x + 230\right) = 3375\) expanding and setting the equation equal to zero gives \(\displaystyle x^{2} + 250 x + 1225 = 0\). Factoring gives \(\displaystyle \left(x + 5\right) \left(x + 245\right)=0\). This gives two possible solutions \(\displaystyle x=-245\) or \(\displaystyle x=-5\). \(\displaystyle x=-245\) is an extraneous solution. The only soution is \(\displaystyle x=-5\).

Download \(\LaTeX\)

\begin{question}Solve $\log_{ 15 }(x + 20) + \log_{ 15 }(x + 230) = 3$
    \soln{10cm}{Using the product rule for logarithms gives $\log_{ 15 }(\left(x + 20\right) \left(x + 230\right)) $ and rewriting in exponential form gives $\left(x + 20\right) \left(x + 230\right) = 3375$ expanding and setting the equation equal to zero gives $x^{2} + 250 x + 1225 = 0$.  Factoring gives $\left(x + 5\right) \left(x + 245\right)=0$. This gives two possible solutions $x=-245$ or $x=-5$. $x=-245$ 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 + 230) = 3 " 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%20230%29%20%3D%203%20" alt="LaTeX:  \displaystyle \log_{ 15 }(x + 20) + \log_{ 15 }(x + 230) = 3 " data-equation-content=" \displaystyle \log_{ 15 }(x + 20) + \log_{ 15 }(x + 230) = 3 " /> </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 + 230\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%20230%5Cright%29%29%20%20" alt="LaTeX:  \displaystyle \log_{ 15 }(\left(x + 20\right) \left(x + 230\right))  " data-equation-content=" \displaystyle \log_{ 15 }(\left(x + 20\right) \left(x + 230\right))  " />  and rewriting in exponential form gives  <img class="equation_image" title=" \displaystyle \left(x + 20\right) \left(x + 230\right) = 3375 " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28x%20%2B%2020%5Cright%29%20%5Cleft%28x%20%2B%20230%5Cright%29%20%3D%203375%20" alt="LaTeX:  \displaystyle \left(x + 20\right) \left(x + 230\right) = 3375 " data-equation-content=" \displaystyle \left(x + 20\right) \left(x + 230\right) = 3375 " />  expanding and setting the equation equal to zero gives  <img class="equation_image" title=" \displaystyle x^{2} + 250 x + 1225 = 0 " src="/equation_images/%20%5Cdisplaystyle%20x%5E%7B2%7D%20%2B%20250%20x%20%2B%201225%20%3D%200%20" alt="LaTeX:  \displaystyle x^{2} + 250 x + 1225 = 0 " data-equation-content=" \displaystyle x^{2} + 250 x + 1225 = 0 " /> .  Factoring gives  <img class="equation_image" title=" \displaystyle \left(x + 5\right) \left(x + 245\right)=0 " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28x%20%2B%205%5Cright%29%20%5Cleft%28x%20%2B%20245%5Cright%29%3D0%20" alt="LaTeX:  \displaystyle \left(x + 5\right) \left(x + 245\right)=0 " data-equation-content=" \displaystyle \left(x + 5\right) \left(x + 245\right)=0 " /> . This gives two possible solutions  <img class="equation_image" title=" \displaystyle x=-245 " src="/equation_images/%20%5Cdisplaystyle%20x%3D-245%20" alt="LaTeX:  \displaystyle x=-245 " data-equation-content=" \displaystyle x=-245 " />  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=-245 " src="/equation_images/%20%5Cdisplaystyle%20x%3D-245%20" alt="LaTeX:  \displaystyle x=-245 " data-equation-content=" \displaystyle x=-245 " />  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>