Question 242511
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I sincerely hope you meant:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_7\left(\frac{2x}{7}\right)]


rather than:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_{10}^7\left(\frac{2x}{7}\right)]


which is what you actually wrote.


Going with my initial assumption:


The difference of the logs is the log of the quotient:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(x) - \log_b(y) = \log_b\left(\frac{x}{y}\right)]


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_7\left(\frac{2x}{7}\right)\ =\ \log_7\left(2x\right)\ -\ \log_7\left(7\right)]


The sum of the logs is the log of the product:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(x) + \log_b(y) = \log_b(xy)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_7\left(2x\right)\ -\ \log_7\left(7\right)\ =\ \log_7\left(x\right)\ +\ \log_7\left(2\right)\ -\ \log_7\left(7\right)]


The log to the base b of b is 1:


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_7\left(x\right)\ +\ \log_7\left(2\right)\ -\ \log_7\left(7\right)\ =\ \log_7\left(x\right)\ +\ \log_7\left(2\right)\ -\ 1]



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