# How do you solve log_8 (1) + log_9 (9) + log_5 (25) + 3x= 6?

Jan 27, 2016

I found $x = 1$

#### Explanation:

Here we can take advantage of the definition of log:
${\log}_{a} x = y \to x = {a}^{y}$
so that we get:
$0 + 1 + 2 + 3 x = 6$
$3 x = 3$
and
$x = 1$

Remember that:
${8}^{0} = 1$
${9}^{1} = 9$
${5}^{2} = 25$

Jan 27, 2016

$x = 1$

#### Explanation:

To solve this problem, we need to remember severals logarithmic properties.

${\log}_{a} a = 1$ , given $a$ is any positive number, $a > 0$
${\log}_{a} 1 = 0$
${\log}_{a} {a}^{n} = n$

We have

${\log}_{8} \left(1\right) + {\log}_{9} \left(9\right) + \log 5 \left(25\right) + 3 x = 6$

$0 + 1 + {\log}_{5} \left({5}^{2}\right) + 3 x = 6$
$0 + 1 + 2 + 3 x = 6$
Combine like terms

$3 + 3 x = 6$

$3 x = 3$

$x = 1$