Question 428620
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The easy way to do this one is to rely on the relationship between the two logarithm arguments.  Note that *[tex \Large \frac{1}{9}\ =\ \left(\frac{1}{3}\right)^2]


Given:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4\,\cdot\,\ln\left(\frac{1}{3}\right)\ -\ 6\,\cdot\,\ln\left(\frac{1}{9}\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4\,\cdot\,\ln\left(\frac{1}{3}\right)\ -\ 6\,\cdot\,\ln\left(\frac{1}{3}\right)^2]


Use:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(x^n)\ =\ n\log_b(x)]


To write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4\,\cdot\,\ln\left(\frac{1}{3}\right)\ -\ 12\,\cdot\,\ln\left(\frac{1}{3}\right)]


Combine like terms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -8\,\cdot\,\ln\left(\frac{1}{3}\right)]


Use


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(x^n)\ =\ n\log_b(x)]


the other way to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\left(\frac{1}{3}\right)^{-8}]


Use


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a^{-n}\ =\ \frac{1}{a^n}]


to write


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\left(3^8\right)]


And finally, use your calculator to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln(6561)]


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