Find \(\frac{a}{b}\) when \(2\log{(a -2b)} = \log{a} + \log{b}. \)
----------Thanks!
log (a-2b)2 = log(ab)
(a-2b)2 = ab
a^2 - 4ab+4b^2 = ab
a^2 - 5 ab + 4 b^2 = 0
(a-4b)(a-b) = 0 a = 4b or a = b a/b = 4 or 1 ( I think!)
1 will not work because you would wind up with a log of a negative in the first term ......