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Solve \(\displaystyle \log_{8}(x + 1016)+\log_{8}(x + 24) = 5\).

Using logarithmic properties and expanding the argument gives \(\displaystyle \log_{8}(x^{2} + 1040 x + 24384)=5\). Making both sides an exponent on the base gives \(\displaystyle x^{2} + 1040 x + 24384=8^{5}\). Expanding and setting equal to zero gives \(\displaystyle x^{2} + 1040 x - 8384=0\). Factoring gives \(\displaystyle \left(x - 8\right) \left(x + 1048\right)=0\). Solving gives the two possible solutions \(\displaystyle x = -1048\) and \(\displaystyle x = 8\). The domain of the original is \(\displaystyle \left(-1016, \infty\right) \bigcap \left(-24, \infty\right)=\left(-24, \infty\right)\). Checking if each possible solution is in the domain gives: \(\displaystyle x = -1048\) is not a solution. \(\displaystyle x=8\) is a solution.

Download \(\LaTeX\) 

\begin{question}Solve $\log_{8}(x + 1016)+\log_{8}(x + 24) = 5$. 
    \soln{9cm}{Using logarithmic properties and expanding the argument gives $\log_{8}(x^{2} + 1040 x + 24384)=5$. Making both sides an exponent on the base gives $x^{2} + 1040 x + 24384=8^{5}$. Expanding and setting equal to zero gives $x^{2} + 1040 x - 8384=0$. Factoring gives $\left(x - 8\right) \left(x + 1048\right)=0$. Solving gives the two possible solutions $x = -1048$ and $x = 8$. The domain of the original is $\left(-1016, \infty\right) \bigcap \left(-24, \infty\right)=\left(-24, \infty\right)$.  Checking if each possible solution is in the domain gives: $x = -1048$ is not a solution. $x=8$ is a solution. }

\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_{8}(x + 1016)+\log_{8}(x + 24) = 5 " src="/equation_images/%20%5Cdisplaystyle%20%5Clog_%7B8%7D%28x%20%2B%201016%29%2B%5Clog_%7B8%7D%28x%20%2B%2024%29%20%3D%205%20" alt="LaTeX:  \displaystyle \log_{8}(x + 1016)+\log_{8}(x + 24) = 5 " data-equation-content=" \displaystyle \log_{8}(x + 1016)+\log_{8}(x + 24) = 5 " /> . </p> </p>
HTML for Canvas
<p> <p>Using logarithmic properties and expanding the argument gives  <img class="equation_image" title=" \displaystyle \log_{8}(x^{2} + 1040 x + 24384)=5 " src="/equation_images/%20%5Cdisplaystyle%20%5Clog_%7B8%7D%28x%5E%7B2%7D%20%2B%201040%20x%20%2B%2024384%29%3D5%20" alt="LaTeX:  \displaystyle \log_{8}(x^{2} + 1040 x + 24384)=5 " data-equation-content=" \displaystyle \log_{8}(x^{2} + 1040 x + 24384)=5 " /> . Making both sides an exponent on the base gives  <img class="equation_image" title=" \displaystyle x^{2} + 1040 x + 24384=8^{5} " src="/equation_images/%20%5Cdisplaystyle%20x%5E%7B2%7D%20%2B%201040%20x%20%2B%2024384%3D8%5E%7B5%7D%20" alt="LaTeX:  \displaystyle x^{2} + 1040 x + 24384=8^{5} " data-equation-content=" \displaystyle x^{2} + 1040 x + 24384=8^{5} " /> . Expanding and setting equal to zero gives  <img class="equation_image" title=" \displaystyle x^{2} + 1040 x - 8384=0 " src="/equation_images/%20%5Cdisplaystyle%20x%5E%7B2%7D%20%2B%201040%20x%20-%208384%3D0%20" alt="LaTeX:  \displaystyle x^{2} + 1040 x - 8384=0 " data-equation-content=" \displaystyle x^{2} + 1040 x - 8384=0 " /> . Factoring gives  <img class="equation_image" title=" \displaystyle \left(x - 8\right) \left(x + 1048\right)=0 " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28x%20-%208%5Cright%29%20%5Cleft%28x%20%2B%201048%5Cright%29%3D0%20" alt="LaTeX:  \displaystyle \left(x - 8\right) \left(x + 1048\right)=0 " data-equation-content=" \displaystyle \left(x - 8\right) \left(x + 1048\right)=0 " /> . Solving gives the two possible solutions  <img class="equation_image" title=" \displaystyle x = -1048 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-1048%20" alt="LaTeX:  \displaystyle x = -1048 " data-equation-content=" \displaystyle x = -1048 " />  and  <img class="equation_image" title=" \displaystyle x = 8 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%208%20" alt="LaTeX:  \displaystyle x = 8 " data-equation-content=" \displaystyle x = 8 " /> . The domain of the original is  <img class="equation_image" title=" \displaystyle \left(-1016, \infty\right) \bigcap \left(-24, \infty\right)=\left(-24, \infty\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28-1016%2C%20%5Cinfty%5Cright%29%20%5Cbigcap%20%5Cleft%28-24%2C%20%5Cinfty%5Cright%29%3D%5Cleft%28-24%2C%20%5Cinfty%5Cright%29%20" alt="LaTeX:  \displaystyle \left(-1016, \infty\right) \bigcap \left(-24, \infty\right)=\left(-24, \infty\right) " data-equation-content=" \displaystyle \left(-1016, \infty\right) \bigcap \left(-24, \infty\right)=\left(-24, \infty\right) " /> .  Checking if each possible solution is in the domain gives:  <img class="equation_image" title=" \displaystyle x = -1048 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-1048%20" alt="LaTeX:  \displaystyle x = -1048 " data-equation-content=" \displaystyle x = -1048 " />  is not a solution.  <img class="equation_image" title=" \displaystyle x=8 " src="/equation_images/%20%5Cdisplaystyle%20x%3D8%20" alt="LaTeX:  \displaystyle x=8 " data-equation-content=" \displaystyle x=8 " />  is a solution. </p> </p>