How do you solve for $log_9 27 = x$ ?

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Answer 1 · 2015-11-25T15:30:32

It is

$x=log_9 27=log_9 3^3=3*log3/log9=3*(log3/log3^2)=3/2$

Finally $x=3/2$

Answer 2 · 2015-11-25T15:32:20

I found: $x=3/2$

We can use the definition of log:
$log_bx=a->x=b^a$
and write:
$27=9^x$ that we can write as:

$3^3=3^(2x)$
for the two terms to be equal also the exponents must be equal, so:
$3=2x$
and:
$x=3/2$

How do you solve for log_9 27 = xlog_9 27 = x?