I think what you've stated is
if so it can be adjusted as,
remove the logs and proceed
I don't see where "log2(0)" has anything to do with that. What you have is equivalent to
and, while it is impossible to take the logarithm of 0, the logarithm, to any base, of 1 is equal to 0.
which, of course, is the same as x= 2x- 1.
Well, yes! Since ( and are inverse functions) that tells you immediately that 2x- 1= x.I know 2^(log2(2x-1)) = 2^log2(x) but does that help me?
Or, you could argue that since is a one to one function, gives, immediately, a= b. That is, because , you must have x= 2x- 1.
Bit stuck on this one, please help.
Angus