+128475 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
