Question 1167514
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The difference of the logs is the log of the quotient.


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


*[tex \Large n\cdot\log(x)\ =\ \log\(x^n\)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\cdot\log(x\,+\,2)\ -\ \log(x\,+\,5)\ =\ \log(2)\ +\ \log(x\,-\,2)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log\(\frac{(x\,+\,2)^2}{x\,+\,5}\)\ =\ \log(2x\,-\,4)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log(a)\ =\ \log(b)\ \Leftrightarrow\ a\ =\ b]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{(x\,+\,2)^2}{x\,+\,5}\ =\ 2x\,-\,4]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -x^2\ +\ 18x\ -\ 16\ =\ 0]


Solve the quadratic.  One of the roots will be extraneous (gives a negative argument for a couple of the log functions)

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


I > Ø
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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