Question 453996
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_3(x\ +\ 6)\ -\ \log_3(x)\ =\ 5]


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_3\left(\frac{x\ +\ 6}{x}\right)\ =\ 5]


Use the definition of logarithms:


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


to write


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5 = \log_3\left(\frac{x\ +\ 6}{x}\right) \ \ \Rightarrow\ \ 3^5 = \frac{x\ +\ 6}{x}]


Solve


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  3^5 = \frac{x\ +\ 6}{x}]


for *[tex \Large x]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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