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Questions: Algebra BusinessCalculus
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Solve \(\displaystyle \log_{ 15 }(x + 18) + \log_{ 15 }(x + 228) = 3\)
Using the product rule for logarithms gives \(\displaystyle \log_{ 15 }(\left(x + 18\right) \left(x + 228\right)) \) and rewriting in exponential form gives \(\displaystyle \left(x + 18\right) \left(x + 228\right) = 3375\) expanding and setting the equation equal to zero gives \(\displaystyle x^{2} + 246 x + 729 = 0\). Factoring gives \(\displaystyle \left(x + 3\right) \left(x + 243\right)=0\). This gives two possible solutions \(\displaystyle x=-243\) or \(\displaystyle x=-3\). \(\displaystyle x=-243\) is an extraneous solution. The only soution is \(\displaystyle x=-3\).
\begin{question}Solve $\log_{ 15 }(x + 18) + \log_{ 15 }(x + 228) = 3$
\soln{10cm}{Using the product rule for logarithms gives $\log_{ 15 }(\left(x + 18\right) \left(x + 228\right)) $ and rewriting in exponential form gives $\left(x + 18\right) \left(x + 228\right) = 3375$ expanding and setting the equation equal to zero gives $x^{2} + 246 x + 729 = 0$. Factoring gives $\left(x + 3\right) \left(x + 243\right)=0$. This gives two possible solutions $x=-243$ or $x=-3$. $x=-243$ is an extraneous solution. The only soution is $x=-3$. }
\end{question}
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\begin{document}\begin{question}(10pts) The question goes here!
\soln{9cm}{The solution goes here.}
\end{question}\end{document}<p> <p>Solve <img class="equation_image" title=" \displaystyle \log_{ 15 }(x + 18) + \log_{ 15 }(x + 228) = 3 " src="/equation_images/%20%5Cdisplaystyle%20%5Clog_%7B%2015%20%7D%28x%20%2B%2018%29%20%2B%20%5Clog_%7B%2015%20%7D%28x%20%2B%20228%29%20%3D%203%20" alt="LaTeX: \displaystyle \log_{ 15 }(x + 18) + \log_{ 15 }(x + 228) = 3 " data-equation-content=" \displaystyle \log_{ 15 }(x + 18) + \log_{ 15 }(x + 228) = 3 " /> </p> </p><p> <p>Using the product rule for logarithms gives <img class="equation_image" title=" \displaystyle \log_{ 15 }(\left(x + 18\right) \left(x + 228\right)) " src="/equation_images/%20%5Cdisplaystyle%20%5Clog_%7B%2015%20%7D%28%5Cleft%28x%20%2B%2018%5Cright%29%20%5Cleft%28x%20%2B%20228%5Cright%29%29%20%20" alt="LaTeX: \displaystyle \log_{ 15 }(\left(x + 18\right) \left(x + 228\right)) " data-equation-content=" \displaystyle \log_{ 15 }(\left(x + 18\right) \left(x + 228\right)) " /> and rewriting in exponential form gives <img class="equation_image" title=" \displaystyle \left(x + 18\right) \left(x + 228\right) = 3375 " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28x%20%2B%2018%5Cright%29%20%5Cleft%28x%20%2B%20228%5Cright%29%20%3D%203375%20" alt="LaTeX: \displaystyle \left(x + 18\right) \left(x + 228\right) = 3375 " data-equation-content=" \displaystyle \left(x + 18\right) \left(x + 228\right) = 3375 " /> expanding and setting the equation equal to zero gives <img class="equation_image" title=" \displaystyle x^{2} + 246 x + 729 = 0 " src="/equation_images/%20%5Cdisplaystyle%20x%5E%7B2%7D%20%2B%20246%20x%20%2B%20729%20%3D%200%20" alt="LaTeX: \displaystyle x^{2} + 246 x + 729 = 0 " data-equation-content=" \displaystyle x^{2} + 246 x + 729 = 0 " /> . Factoring gives <img class="equation_image" title=" \displaystyle \left(x + 3\right) \left(x + 243\right)=0 " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28x%20%2B%203%5Cright%29%20%5Cleft%28x%20%2B%20243%5Cright%29%3D0%20" alt="LaTeX: \displaystyle \left(x + 3\right) \left(x + 243\right)=0 " data-equation-content=" \displaystyle \left(x + 3\right) \left(x + 243\right)=0 " /> . This gives two possible solutions <img class="equation_image" title=" \displaystyle x=-243 " src="/equation_images/%20%5Cdisplaystyle%20x%3D-243%20" alt="LaTeX: \displaystyle x=-243 " data-equation-content=" \displaystyle x=-243 " /> 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=-243 " src="/equation_images/%20%5Cdisplaystyle%20x%3D-243%20" alt="LaTeX: \displaystyle x=-243 " data-equation-content=" \displaystyle x=-243 " /> 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>