Question 471922
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Presuming the normal convention of an unspecified base denoting base 10 logarithms, and using:


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


We can deduce that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y = \log_{10}(1000) \ \ \Rightarrow\ \ 10^y = 1000]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ 3]


Substituting:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{3}\log(x)\ -\ 3\ =\ 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{3}\log(x)\ =\ 6]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log(x)\ =\ 18]


Then using


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


again we can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 18 = \log_{10}(x) \ \ \Rightarrow\ \ 10^{18} = x]


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