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Find \(\frac{a}{b}\)when \(2\log{(a -2b)} = \log(a) + \log(b).\)

 Apr 2, 2019
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\(2\log(a-2b) = \log((a-2b)^2)=\log(a)+\log(b) = \log(ab)\\ (a-2b)^2 = ab\\ a^2-4ab+4b^2 = ab\\ a^2 - 5ab + 4b^2 = 0\\ a = \dfrac{5b\pm\sqrt{25b^2-16b^2}}{2}\\ a = \dfrac{5b\pm 3b}{2} = b,4b\\ \text{however, }a\neq b \text{ as that would result in a negative number}\\ \text{as the argument to the log function in the original equation. So }\\ a=4b\\ \dfrac a b = \dfrac 1 4\)

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 Apr 2, 2019
edited by Rom  Apr 2, 2019

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