Question 421859
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Presuming you are trying to represent base 6 logs, i.e., *[tex \Large \log_6(x)] as opposed to base 10 logs, *[tex \Large \log(6x)]:


First use "The sum of the logs is the log of the product":


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_6\left(x(x\ +\ 16)\right)\ =\ 2]


Then use the definition of the logarithm function:


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


to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 16x\ =\ 6^2\ =\ 36]


Standard form of the quadratic:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 16x\ -\ 36\ =\ 0]


The quadratic factors.  You can take it from here.  You will need to exclude one of the roots because it is not in the domain of the logarithm function.


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