How do you solve #log_8(x+1) = log_8 (2x-2)#?

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Answer 1 · 2015-05-10T03:28:43

Remember thar #a^(log_a(x))=x#
So you can write:
#8^(log_8(x+1))=8^(log_8(2x-2))#
and so:
#x+1=2x-2#
#x=3#