Question 193996
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_4{(2x+1)} = \log_4{(x-3)} + \log_4{(x+5)}]


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_4{(2x+1)} = \log_4{\left\{(x-3)(x+5)\right\}}]


If two logs to the same base are equal, then their arguments must be equal, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x+1 = (x-3)(x+5)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x+1 = x^2 + 2x - 15]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2 - 16 = 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x - 4)(x + 4) = 0]


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


or 


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


But if *[tex \LARGE x = -4] then *[tex \LARGE 2x + 1 = -7], and the domain of the log function is *[tex \LARGE \left(0,\infty\right)].  Hence, you must exclude *[tex \LARGE x = -4], and the solution set for the given equation is *[tex \LARGE \{4\}]


Check:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_4{(2(4)+1)} =^? \log_4{((4)-3)} + \log_4{((4)+5)}]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_4{(9)} =^? \log_4{(1)} + \log_4{(9)}]


But *[tex \LARGE log_b{(1)} = 0\ \ \forall\ b \geq 0], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_4{(9)} = \log_4{(9)}] Checks.


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