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Solve \(\displaystyle \log_{12}(x + 25)+\log_{12}(x + 62) = 3\).
Using logarithmic properties and expanding the argument gives \(\displaystyle \log_{12}(x^{2} + 87 x + 1550)=3\). Making both sides an exponent on the base gives \(\displaystyle x^{2} + 87 x + 1550=12^{3}\). Expanding and setting equal to zero gives \(\displaystyle x^{2} + 87 x - 178=0\). Factoring gives \(\displaystyle \left(x - 2\right) \left(x + 89\right)=0\). Solving gives the two possible solutions \(\displaystyle x = -89\) and \(\displaystyle x = 2\). The domain of the original is \(\displaystyle \left(-25, \infty\right) \bigcap \left(-62, \infty\right)=\left(-25, \infty\right)\). Checking if each possible solution is in the domain gives: \(\displaystyle x = -89\) is not a solution. \(\displaystyle x=2\) is a solution.
\begin{question}Solve $\log_{12}(x + 25)+\log_{12}(x + 62) = 3$. \soln{9cm}{Using logarithmic properties and expanding the argument gives $\log_{12}(x^{2} + 87 x + 1550)=3$. Making both sides an exponent on the base gives $x^{2} + 87 x + 1550=12^{3}$. Expanding and setting equal to zero gives $x^{2} + 87 x - 178=0$. Factoring gives $\left(x - 2\right) \left(x + 89\right)=0$. Solving gives the two possible solutions $x = -89$ and $x = 2$. The domain of the original is $\left(-25, \infty\right) \bigcap \left(-62, \infty\right)=\left(-25, \infty\right)$. Checking if each possible solution is in the domain gives: $x = -89$ is not a solution. $x=2$ is a solution. } \end{question}
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<p> <p>Solve <img class="equation_image" title=" \displaystyle \log_{12}(x + 25)+\log_{12}(x + 62) = 3 " src="/equation_images/%20%5Cdisplaystyle%20%5Clog_%7B12%7D%28x%20%2B%2025%29%2B%5Clog_%7B12%7D%28x%20%2B%2062%29%20%3D%203%20" alt="LaTeX: \displaystyle \log_{12}(x + 25)+\log_{12}(x + 62) = 3 " data-equation-content=" \displaystyle \log_{12}(x + 25)+\log_{12}(x + 62) = 3 " /> . </p> </p>
<p> <p>Using logarithmic properties and expanding the argument gives <img class="equation_image" title=" \displaystyle \log_{12}(x^{2} + 87 x + 1550)=3 " src="/equation_images/%20%5Cdisplaystyle%20%5Clog_%7B12%7D%28x%5E%7B2%7D%20%2B%2087%20x%20%2B%201550%29%3D3%20" alt="LaTeX: \displaystyle \log_{12}(x^{2} + 87 x + 1550)=3 " data-equation-content=" \displaystyle \log_{12}(x^{2} + 87 x + 1550)=3 " /> . Making both sides an exponent on the base gives <img class="equation_image" title=" \displaystyle x^{2} + 87 x + 1550=12^{3} " src="/equation_images/%20%5Cdisplaystyle%20x%5E%7B2%7D%20%2B%2087%20x%20%2B%201550%3D12%5E%7B3%7D%20" alt="LaTeX: \displaystyle x^{2} + 87 x + 1550=12^{3} " data-equation-content=" \displaystyle x^{2} + 87 x + 1550=12^{3} " /> . Expanding and setting equal to zero gives <img class="equation_image" title=" \displaystyle x^{2} + 87 x - 178=0 " src="/equation_images/%20%5Cdisplaystyle%20x%5E%7B2%7D%20%2B%2087%20x%20-%20178%3D0%20" alt="LaTeX: \displaystyle x^{2} + 87 x - 178=0 " data-equation-content=" \displaystyle x^{2} + 87 x - 178=0 " /> . Factoring gives <img class="equation_image" title=" \displaystyle \left(x - 2\right) \left(x + 89\right)=0 " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28x%20-%202%5Cright%29%20%5Cleft%28x%20%2B%2089%5Cright%29%3D0%20" alt="LaTeX: \displaystyle \left(x - 2\right) \left(x + 89\right)=0 " data-equation-content=" \displaystyle \left(x - 2\right) \left(x + 89\right)=0 " /> . Solving gives the two possible solutions <img class="equation_image" title=" \displaystyle x = -89 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-89%20" alt="LaTeX: \displaystyle x = -89 " data-equation-content=" \displaystyle x = -89 " /> 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(-25, \infty\right) \bigcap \left(-62, \infty\right)=\left(-25, \infty\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28-25%2C%20%5Cinfty%5Cright%29%20%5Cbigcap%20%5Cleft%28-62%2C%20%5Cinfty%5Cright%29%3D%5Cleft%28-25%2C%20%5Cinfty%5Cright%29%20" alt="LaTeX: \displaystyle \left(-25, \infty\right) \bigcap \left(-62, \infty\right)=\left(-25, \infty\right) " data-equation-content=" \displaystyle \left(-25, \infty\right) \bigcap \left(-62, \infty\right)=\left(-25, \infty\right) " /> . Checking if each possible solution is in the domain gives: <img class="equation_image" title=" \displaystyle x = -89 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-89%20" alt="LaTeX: \displaystyle x = -89 " data-equation-content=" \displaystyle x = -89 " /> 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>