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Solve \(\displaystyle \log_{6}(x + 240)+\log_{6}(x + 29) = 5\).

Using logarithmic properties and expanding the argument gives \(\displaystyle \log_{6}(x^{2} + 269 x + 6960)=5\). Making both sides an exponent on the base gives \(\displaystyle x^{2} + 269 x + 6960=6^{5}\). Expanding and setting equal to zero gives \(\displaystyle x^{2} + 269 x - 816=0\). Factoring gives \(\displaystyle \left(x - 3\right) \left(x + 272\right)=0\). Solving gives the two possible solutions \(\displaystyle x = -272\) and \(\displaystyle x = 3\). The domain of the original is \(\displaystyle \left(-240, \infty\right) \bigcap \left(-29, \infty\right)=\left(-29, \infty\right)\). Checking if each possible solution is in the domain gives: \(\displaystyle x = -272\) is not a solution. \(\displaystyle x=3\) is a solution.

Download \(\LaTeX\) 

\begin{question}Solve $\log_{6}(x + 240)+\log_{6}(x + 29) = 5$. 
    \soln{9cm}{Using logarithmic properties and expanding the argument gives $\log_{6}(x^{2} + 269 x + 6960)=5$. Making both sides an exponent on the base gives $x^{2} + 269 x + 6960=6^{5}$. Expanding and setting equal to zero gives $x^{2} + 269 x - 816=0$. Factoring gives $\left(x - 3\right) \left(x + 272\right)=0$. Solving gives the two possible solutions $x = -272$ and $x = 3$. The domain of the original is $\left(-240, \infty\right) \bigcap \left(-29, \infty\right)=\left(-29, \infty\right)$.  Checking if each possible solution is in the domain gives: $x = -272$ is not a solution. $x=3$ 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_{6}(x + 240)+\log_{6}(x + 29) = 5 " src="/equation_images/%20%5Cdisplaystyle%20%5Clog_%7B6%7D%28x%20%2B%20240%29%2B%5Clog_%7B6%7D%28x%20%2B%2029%29%20%3D%205%20" alt="LaTeX:  \displaystyle \log_{6}(x + 240)+\log_{6}(x + 29) = 5 " data-equation-content=" \displaystyle \log_{6}(x + 240)+\log_{6}(x + 29) = 5 " /> . </p> </p>
HTML for Canvas
<p> <p>Using logarithmic properties and expanding the argument gives  <img class="equation_image" title=" \displaystyle \log_{6}(x^{2} + 269 x + 6960)=5 " src="/equation_images/%20%5Cdisplaystyle%20%5Clog_%7B6%7D%28x%5E%7B2%7D%20%2B%20269%20x%20%2B%206960%29%3D5%20" alt="LaTeX:  \displaystyle \log_{6}(x^{2} + 269 x + 6960)=5 " data-equation-content=" \displaystyle \log_{6}(x^{2} + 269 x + 6960)=5 " /> . Making both sides an exponent on the base gives  <img class="equation_image" title=" \displaystyle x^{2} + 269 x + 6960=6^{5} " src="/equation_images/%20%5Cdisplaystyle%20x%5E%7B2%7D%20%2B%20269%20x%20%2B%206960%3D6%5E%7B5%7D%20" alt="LaTeX:  \displaystyle x^{2} + 269 x + 6960=6^{5} " data-equation-content=" \displaystyle x^{2} + 269 x + 6960=6^{5} " /> . Expanding and setting equal to zero gives  <img class="equation_image" title=" \displaystyle x^{2} + 269 x - 816=0 " src="/equation_images/%20%5Cdisplaystyle%20x%5E%7B2%7D%20%2B%20269%20x%20-%20816%3D0%20" alt="LaTeX:  \displaystyle x^{2} + 269 x - 816=0 " data-equation-content=" \displaystyle x^{2} + 269 x - 816=0 " /> . Factoring gives  <img class="equation_image" title=" \displaystyle \left(x - 3\right) \left(x + 272\right)=0 " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28x%20-%203%5Cright%29%20%5Cleft%28x%20%2B%20272%5Cright%29%3D0%20" alt="LaTeX:  \displaystyle \left(x - 3\right) \left(x + 272\right)=0 " data-equation-content=" \displaystyle \left(x - 3\right) \left(x + 272\right)=0 " /> . Solving gives the two possible solutions  <img class="equation_image" title=" \displaystyle x = -272 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-272%20" alt="LaTeX:  \displaystyle x = -272 " data-equation-content=" \displaystyle x = -272 " />  and  <img class="equation_image" title=" \displaystyle x = 3 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%203%20" alt="LaTeX:  \displaystyle x = 3 " data-equation-content=" \displaystyle x = 3 " /> . The domain of the original is  <img class="equation_image" title=" \displaystyle \left(-240, \infty\right) \bigcap \left(-29, \infty\right)=\left(-29, \infty\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28-240%2C%20%5Cinfty%5Cright%29%20%5Cbigcap%20%5Cleft%28-29%2C%20%5Cinfty%5Cright%29%3D%5Cleft%28-29%2C%20%5Cinfty%5Cright%29%20" alt="LaTeX:  \displaystyle \left(-240, \infty\right) \bigcap \left(-29, \infty\right)=\left(-29, \infty\right) " data-equation-content=" \displaystyle \left(-240, \infty\right) \bigcap \left(-29, \infty\right)=\left(-29, \infty\right) " /> .  Checking if each possible solution is in the domain gives:  <img class="equation_image" title=" \displaystyle x = -272 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-272%20" alt="LaTeX:  \displaystyle x = -272 " data-equation-content=" \displaystyle x = -272 " />  is not a solution.  <img class="equation_image" title=" \displaystyle x=3 " src="/equation_images/%20%5Cdisplaystyle%20x%3D3%20" alt="LaTeX:  \displaystyle x=3 " data-equation-content=" \displaystyle x=3 " />  is a solution. </p> </p>