Question 998190
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It really depends on the value of the logarithm base that you assume when the base is not specified.


*[tex \Large \log(x)] refers to *[tex \Large \log_2(x)] in computer science and information theory.


*[tex \Large \log(x)] refers to *[tex \Large \log_e(x)]  or the natural logarithm in mathematical analysis, physics, chemistry, statistics, economics, and some engineering fields.


*[tex \Large \log(x)] refers to *[tex \Large \log_{10}(x)] in various engineering fields, logarithm tables, and handheld calculators.


Either way, you use the same process to solve.  I'll assume base 10 to demonstrate.


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


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log(x\ -\ 2)\ =\ -2\ \Leftrightarrow 10^{-2}\ =\ (x\ -\ 2)]


Hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 2.01]


Check


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2.01\ -\ 2\ =\ 0.01\ =\ 10^{-2} \Rightarrow \log_{10}(x\ -\ 2)\ =\ -2]


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

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