Question 977515
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The difference of the logs is the log of the quotient, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_5(x)\ -\log_5(x\ -\ 10)\ =\ \log_5(k) \Right\ \log_5\left(\frac{x}{x\,-\,10}\right)\ =\ \log_5(k)]


Now, if you consider the fact that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(\alpha)\ =\ \log_b(\beta)] if and only if *[tex \LARGE \alpha\ =\ \beta]


You should be able to finish this yourself.


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \