Question 383448
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I think you actually divided correctly, but you said it  backwards:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 17.2\ln(x\ +\ 2)\ =\ 24.5]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln(x\ +\ 2)\ =\ \frac{24.5}{17.2}\ \approx\ 1.4244]


Then use:


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


to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ 2\ =\ e^{\frac{24.5}{17.2}}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ e^{\frac{24.5}{17.2}}\ -\ 2]


is the exact answer.  A numerical approximation is a few calculator buttons away.


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