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Solve \(\displaystyle \log_{10}(x + 123)+\log_{10}(x + 6) = 3\).

Using logarithmic properties and expanding the argument gives \(\displaystyle \log_{10}(x^{2} + 129 x + 738)=3\). Making both sides an exponent on the base gives \(\displaystyle x^{2} + 129 x + 738=10^{3}\). Expanding and setting equal to zero gives \(\displaystyle x^{2} + 129 x - 262=0\). Factoring gives \(\displaystyle \left(x - 2\right) \left(x + 131\right)=0\). Solving gives the two possible solutions \(\displaystyle x = -131\) and \(\displaystyle x = 2\). The domain of the original is \(\displaystyle \left(-123, \infty\right) \bigcap \left(-6, \infty\right)=\left(-6, \infty\right)\). Checking if each possible solution is in the domain gives: \(\displaystyle x = -131\) is not a solution. \(\displaystyle x=2\) is a solution.

Download \(\LaTeX\) 

\begin{question}Solve $\log_{10}(x + 123)+\log_{10}(x + 6) = 3$. 
    \soln{9cm}{Using logarithmic properties and expanding the argument gives $\log_{10}(x^{2} + 129 x + 738)=3$. Making both sides an exponent on the base gives $x^{2} + 129 x + 738=10^{3}$. Expanding and setting equal to zero gives $x^{2} + 129 x - 262=0$. Factoring gives $\left(x - 2\right) \left(x + 131\right)=0$. Solving gives the two possible solutions $x = -131$ and $x = 2$. The domain of the original is $\left(-123, \infty\right) \bigcap \left(-6, \infty\right)=\left(-6, \infty\right)$.  Checking if each possible solution is in the domain gives: $x = -131$ is not a solution. $x=2$ 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_{10}(x + 123)+\log_{10}(x + 6) = 3 " src="/equation_images/%20%5Cdisplaystyle%20%5Clog_%7B10%7D%28x%20%2B%20123%29%2B%5Clog_%7B10%7D%28x%20%2B%206%29%20%3D%203%20" alt="LaTeX:  \displaystyle \log_{10}(x + 123)+\log_{10}(x + 6) = 3 " data-equation-content=" \displaystyle \log_{10}(x + 123)+\log_{10}(x + 6) = 3 " /> . </p> </p>
HTML for Canvas
<p> <p>Using logarithmic properties and expanding the argument gives  <img class="equation_image" title=" \displaystyle \log_{10}(x^{2} + 129 x + 738)=3 " src="/equation_images/%20%5Cdisplaystyle%20%5Clog_%7B10%7D%28x%5E%7B2%7D%20%2B%20129%20x%20%2B%20738%29%3D3%20" alt="LaTeX:  \displaystyle \log_{10}(x^{2} + 129 x + 738)=3 " data-equation-content=" \displaystyle \log_{10}(x^{2} + 129 x + 738)=3 " /> . Making both sides an exponent on the base gives  <img class="equation_image" title=" \displaystyle x^{2} + 129 x + 738=10^{3} " src="/equation_images/%20%5Cdisplaystyle%20x%5E%7B2%7D%20%2B%20129%20x%20%2B%20738%3D10%5E%7B3%7D%20" alt="LaTeX:  \displaystyle x^{2} + 129 x + 738=10^{3} " data-equation-content=" \displaystyle x^{2} + 129 x + 738=10^{3} " /> . Expanding and setting equal to zero gives  <img class="equation_image" title=" \displaystyle x^{2} + 129 x - 262=0 " src="/equation_images/%20%5Cdisplaystyle%20x%5E%7B2%7D%20%2B%20129%20x%20-%20262%3D0%20" alt="LaTeX:  \displaystyle x^{2} + 129 x - 262=0 " data-equation-content=" \displaystyle x^{2} + 129 x - 262=0 " /> . Factoring gives  <img class="equation_image" title=" \displaystyle \left(x - 2\right) \left(x + 131\right)=0 " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28x%20-%202%5Cright%29%20%5Cleft%28x%20%2B%20131%5Cright%29%3D0%20" alt="LaTeX:  \displaystyle \left(x - 2\right) \left(x + 131\right)=0 " data-equation-content=" \displaystyle \left(x - 2\right) \left(x + 131\right)=0 " /> . Solving gives the two possible solutions  <img class="equation_image" title=" \displaystyle x = -131 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-131%20" alt="LaTeX:  \displaystyle x = -131 " data-equation-content=" \displaystyle x = -131 " />  and  <img class="equation_image" title=" \displaystyle x = 2 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%202%20" alt="LaTeX:  \displaystyle x = 2 " data-equation-content=" \displaystyle x = 2 " /> . The domain of the original is  <img class="equation_image" title=" \displaystyle \left(-123, \infty\right) \bigcap \left(-6, \infty\right)=\left(-6, \infty\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28-123%2C%20%5Cinfty%5Cright%29%20%5Cbigcap%20%5Cleft%28-6%2C%20%5Cinfty%5Cright%29%3D%5Cleft%28-6%2C%20%5Cinfty%5Cright%29%20" alt="LaTeX:  \displaystyle \left(-123, \infty\right) \bigcap \left(-6, \infty\right)=\left(-6, \infty\right) " data-equation-content=" \displaystyle \left(-123, \infty\right) \bigcap \left(-6, \infty\right)=\left(-6, \infty\right) " /> .  Checking if each possible solution is in the domain gives:  <img class="equation_image" title=" \displaystyle x = -131 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-131%20" alt="LaTeX:  \displaystyle x = -131 " data-equation-content=" \displaystyle x = -131 " />  is not a solution.  <img class="equation_image" title=" \displaystyle x=2 " src="/equation_images/%20%5Cdisplaystyle%20x%3D2%20" alt="LaTeX:  \displaystyle x=2 " data-equation-content=" \displaystyle x=2 " />  is a solution. </p> </p>