Question 237827
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The sum of the logs is the log of the product.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(x) + \log_b(y) = \log_b(xy)]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln(x) + \ln(x\ +\ 12) = \ln(x^2\ +\ 12x)\ =\ 1]


But we also know that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y = \log_b(x) \ \ \Rightarrow\ \ b^y = x]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 12\ =\ e^1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 12\ -\ e\ =\ 0]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \frac{-12 \pm sqrt{144\ +\ 4e}}{2}\ =\ -6\ \pm\ \sqrt{36\ +\ e}]




John
*[tex \LARGE e^{i\pi} + 1 = 0]
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