Find a/b when 2log(a-2b) = log a + log b .

2log(a-2b) = loga + logb  so log(a-2b)² = logab

therefore  (a-2b)² = ab

expanding : a² -4ab + 4b² = ab

ie              a²  - 5ab + 4b² = 0

divide both sides by b² : a²/b² -5ab/b² + 4 =0

ie              (a/b)² - 5(a/b) + 4 = 0

solving the quadratic for a/b gives (a/b - 1)(a/b - 4) = 0

therefore  a/b =1 or a/b = 4

2 log (a - 2b  )  = loga + log b

log (a - 2b)^2  = log (a*b)

This implies that

(a - 2b)^2  =  ab

a^2 - 4ab + 4b^2  = ab

a^2 - 5ab + 4b^2  = 0

(a - 4b) (a - b)  = 0

This implies that

a = 4b   ⇒   a/b  =  4      or

a - b  = 0  ⇒   a  = b  ⇒   a/b  =  1

Wondering if  a/b =1 might be considered an extraneous (invalid) solution, as it would result in a

LOG (negative number)   ??

No I don't think so as squaring a negative gives a positive

ie 2log(a-2b) =log (a-2b)²

Perhaps the author of the question didn't believe it an issue otherwise they would have included that a-2b >= 0 in the question. I do see your dilemma but if you remember back to when you learnt all about logs and I guess it depends on how your teacher treated the topic. A logarithm as we know is the power of a base to obtain a given number. From this definition we get a whole new mathematics topic part of which is simplifications. The actual log law in question is that if we have a log of a power then :  log_bxⁿ  =  nlog_bx  not the other way round. Of course the other way has been adopted also to help with simplification. Maybe there are those who think its a case of which came first the chicken or the egg?

your very welcome...actually it's foghorn leghorn!!!....great to see a good sense of humour

Maybe it's a case of which fell off the log first....the chicken or the egg.....???