Question

Best answer gets the brainliest! Find $a/b$ when $2\log{(a -2b)} = \log{a} + \log{b}$.

Answer 1

By some properties of logarithms, rewrite the equation as

$2\log(a-2b) = \log(a) + \log(b) \\\\ \log(a-2b)^2 = \log(ab)$

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