Question 306917
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Your teacher is right.  Don't you just hate it when that happens?



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log(x\ +\ 9)\ -\ \log(x)\ =\ 3]


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log(\frac{x\ +\ 9}{x})\ =\ 3]


Use the definition of logarithm


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10^3\ =\ \frac{x\ +\ 9}{x}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1000x\ =\ x\ +\ 9]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 999x\ =\ 9]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{9}{999}\ =\ \frac{1}{111}]


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