Question 975676
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\ln(x\ -\ 6)\ =\ -4]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln(x\ -\ 6)\ =\ -2]


Use the definition of the logarithm function:


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


Where the base is *[tex \LARGE e]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -2 = \ln(x\ -\ 6) \ \ \Rightarrow\ \ e^{-2} = x\ -\ 6]


Which is to say


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ e^{-2}\ +\ 6]


for the exact answer.  Use your calculator for a decimal approximation to the precision of your choice.


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \