Question 901583
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The sum of the logs is the log of the product.  The difference of the logs is the log of the quotient.


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3\log(x\,+\,1)\ +\ 5\log(9x\,+\,1)\ -\ \log(x\,+\,1)\ =\ \log\frac{(x\ +\ 1)^3(9x\ +\ 1)^5}{(x\ +\ 1)}\ =\ \log\left((x\ +\ 1)^2(9x\ +\ 1)^5\right)]





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