How do you solve for log_9 27 = x?

It is

$x = {\log}_{9} 27 = {\log}_{9} {3}^{3} = 3 \cdot \log \frac{3}{\log} 9 = 3 \cdot \left(\log \frac{3}{\log} {3}^{2}\right) = \frac{3}{2}$

Finally $x = \frac{3}{2}$

Nov 25, 2015

I found: $x = \frac{3}{2}$

Explanation:

We can use the definition of log:
${\log}_{b} x = a \to x = {b}^{a}$
and write:
$27 = {9}^{x}$ that we can write as:

${3}^{3} = {3}^{2 x}$
for the two terms to be equal also the exponents must be equal, so:
$3 = 2 x$
and:
$x = \frac{3}{2}$