Question 201509
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If two logs to the same base are equal, then their arguments must be equal, that is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(a) = \log_b(c) \ \ \Leftrightarrow\ \ a = c,\ \forall\ a, c\ \in\ R,\ a > 0,\ c > 0]


And since *[tex \Large \ln(a)] is the same thing as *[tex \Large \log_e(a)], given *[tex \Large \ln(4x)=\ln(x-5)], you can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4x = x - 5]


Solve for *[tex \Large x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = -\frac{5}{3}]


But that value of *[tex \Large x] means that *[tex \Large 4x < 0].  Hence *[tex \Large x = -\frac{5}{3}\ \ \Rightarrow\ \4x = -\frac{20}{3}] is not in the domain of *[tex \Large \ln].  Therefore there is no solution.






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