Question 244644
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3\log_3(x)\ +\ 4\log_3(y)\ -\ 4\log_3(z)]


First use:


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


to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_3\left(x^3\right)\ +\ \log_3\left(y^4\right)\ -\ \log_3\left(z^4\right)]


Then use the fact that the sum of the logs is the log of the product, that is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(x) + \log_b(y) = \log_b(xy)]


to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_3\left(x^3y^4\right)\ -\ \log_3\left(z^4\right)]


Then use the fact that the difference of the logs is the log of the quotient, that is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(x) - \log_b(y) = log_b\left(\frac{x}{y}\right)]


To write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_3\left(\frac{x^3y^4}{z^4}\right)]


and you are done.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\left(3\ln\left(x\right)\ -\ \ln\left(x-1\right)\right)]


Distribute the 2:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6\ln\left(x\right)\ -\ 2\ln\left(x-1\right)]


Then continue in the same fashion as illustrated in the first problem to finish the second one.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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