Question 1166167
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2(3)^{(4x\,+\,5)}\ -\ 2\ =\ -1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2(3)^{(4x\,+\,5)}\ = 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3^{(4x\,+\,5)}\ =\ \frac{1}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\(3^{(4x\,+\,5)}\)\ =\ \ln\(\frac{1}{2}\)\ =\ -\ln(2)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (4x\,+\,5)\ln(3)\ =\ -\ln(2)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4x\ +\ 5\ =\ -\frac{\ln(2)}{\ln(3)}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4x\ =\ -\frac{\ln(2)}{\ln(3)}\ -\ 5]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{-\frac{\ln(2)}{\ln(3)}\,-\,5}{4}]


The rest is just arithmetic.


																
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">


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