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:
$x+1=2x-2$
$x=3$

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