By some properties of logarithms, rewrite the equation as
[tex]2\log(a-2b) = \log(a) + \log(b) \\\\ \log(a-2b)^2 = \log(ab)[/tex]
so that
(a - 2b)² = ab
Expand the left side:
a ² - 4ab + 4b ² = ab
Rearrange terms to get a quadratic equation in a/b :
a ² - 5ab + 4b ² = 0
b must be greater than 0, otherwise log(b) doesn't exist, and the same goes for a. So we can divide by b ² to get
a ²/b ² - 5a/b + 4 = 0
Factorize and solve for a/b :
(a/b - 4) (a/b - 1) = 0
==> a/b = 4 or a/b = 1
However, if a/b = 1, then a = b makes a - 2b = -b. But we must have b > 0, so we omit the second solution and end up with
a/b = 4
