Question 216568
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We know that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log(1) = 0]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0 = -0]


so  solve


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x - 1 = 1]


for *[tex \LARGE x]


to get the value(s) of *[tex \LARGE x] for which


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log(x-1)=-\log(x-1)]



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