Question 1039009
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln(3x\ +\ 5)\ +\ \ln(2)\ =\ 2]


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln(6x\ +\ 10)\ =\ 2]


*[tex \Large b^{\log_b(x)}\ =\ x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6x\ +\ 10\ =\ e^2]


Solve for *[tex \Large x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{e^2\ -\ 10}{6}]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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