Question 242310
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2}{3}\left(\ln\left(x^2-9\right)\ -\ \ln\left(x-3\right)\right)\ +\ \ln\left(x+4\right)]


The difference of the logs is the log of the quotient, that is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(x) - \log_b(y) = log_b\left(\frac{x}{y}\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2}{3}\left(\ln\left(\frac{x^2-9}{x-3}\right)\right)\ +\ \ln\left(x+4\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2}{3}\left(\ln\left(x+3\right)\right)\ +\ \ln\left(x+4\right)]


Use:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(x^n) = n\log_b(x)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\left(\sqrt[3]{(x+3)^2}\right)\ +\ \ln\left(x+4\right)]


The sum of the logs is the log of the product:


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\left(\left(\sqrt[3]{(x+3)^2}\right)\left(x+4\right)\right)]




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