Question 326404
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_3(x)\ =\ 2\ +\ \log_3(x\,-\,2)]


Add *[tex \Large -\log_3(x\,-\,2)] to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_3(x)\ -\ \log_3(x\,-\,2)\ =\ 2]


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)]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_3\left(\frac{x}{x\,-\,2}\right)\ =\ 2]


Use the definition of the logarithm function:


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


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2 = \log_3\left(\frac{x}{x\,-\,2}\right) \ \ \Rightarrow\ \ 3^2 = \frac{x}{x\,-\,2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9x\ -\ 18\ =\ x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8x\ =\ 18]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{9}{4}]



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