Question 667600
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log(3x\ +\ 8)\ -\ \log(x\ -\ 2)\ =\ 1]


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log\left(\frac{3x\ +\ 8}{x\ -\ 2}\right)\ =\ 1]


Use the definition of the logarithm function and the convention that an unspecified base is assumed to be base 10.


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{3x\ +\ 8}{x\ -\ 2}\ =\ 10^1\ =\ 10]


Solve 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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