Question 374910
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3\ +\ \ln(\sqrt{x\,-\,2})\ =\ 4]


Add -3 to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln(\sqrt{x\,-\,2})\ =\ 1]


Use *[tex \Large a^{\frac{1}{n}} = \sqrt[n]{a}] to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln(x\,-\,2)^{\frac{1}{2}}\ =\ 1]


Use *[tex \Large \log_b(x^n)\ =\ n\log_b(x)] to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{2}\ln(x\,-\,2)\ =\ 1]


Which is to say:


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


Then use *[tex \Large y = \log_b(x) \ \ \Rightarrow\ \ b^y = x] to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^2\ =\ x\ -\ 2]


Which is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ e^2\ +\ 2\ \approx\ 9.389]


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