Question 194318
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I think you mean


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln{\frac{x}{5e}}]


in which case you are way off base.


First, the log of the quotient is the difference of the logs, so:


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


Second, the log of the product is the sum of the logs, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln{x} - \ln{5e} = \ln{x} - \left(\ln{5} + \ln{e}\right) = \ln{x} - \ln{5} - \ln{e}]


Finally, *[tex \LARGE \log_b{b} = 1] so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln{x} - \ln{5} - \ln{e} = \ln{x} - \ln{5} - 1]








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