Question 328860
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Your answer sheet is correct.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \log_3(12)\ -\ 2\log_3(2)] 


First use:


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


to write


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_3(12)\ -\ 2\log_3(2)\ =\ \log_3(12)\ -\ \log_3(2^2)\ =\ \log_3(12)\ -\ \log_3(4)]


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


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


to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_3(12)\ -\ \log_3(4)\ =\ \log_3(\frac{12}{4})\ =\ \log_3(3)].


Then use the definition of the logarithm function


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y = \log_b(x) \ \ \Rightarrow\ \ b^y = x]


To write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \log_3(3)\ \ \Rightarrow\ \ 3^y\ =\ 3]


Then use the fact that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^1\ =\ x\ \forall\ x\ \in\ \mathbb{R}]


to write


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ 1]


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