#1**0 **

2 log ( a - 2b) = log (a) + log(b)

log ( a - 2b)^2 = log (ab) .....so....

(a - 2b)^2 = ab

a^2 - 4ab +4b^2 = ab

a^2 - 5ab + 4b^2 = 0 factor as

(a - 4b) ( a - b) = 0

Setting each factor to 0 and solving we have that

a - 4b = 0 or a - b = 0

a = 4b a = b

a/b = 4 a/b = 1

The second solution is no good because if a = b......then a - 2b = a - 2a = -a

Which means that log ( a - 2b) = log (-a) and is defined only if a is negative

But....on the right side of the original equation , log (a) produces a log of a negative which is not defined for real numbers

So

a / b = 4

CPhill Mar 17, 2019