How do you solve $log_2 2x + log_2(x-1)=2$ ?

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Answer 1 · 2016-03-11T05:21:42

2

log here is a binary (base 2) log.
log a + log b = log ab.

$log (2x(x-1)) = 2$
Inversely,
$2x(x-1) = 2^2$
$x^2-x-2 = (x+1)(x-2)=0$

The roots are $-1 and 2$ .
x > 1. The only real solution is x = 2.

How do you solve log_2 2x + log_2(x-1)=2log_2 2x + log_2(x-1)=2?