# How do you find the exact value of 8^(log_8 6-log_8 9)?

Jul 26, 2017

$\frac{2}{3.}$

#### Explanation:

Recall that, ${a}^{x} = b \ldots . \left(1\right) . \iff x = {\log}_{a} b \ldots \ldots \ldots . \left(2\right) .$

Hence, if we subst. $x$ in $\left(1\right) ,$ we have, ${a}^{{\log}_{a} b} = b .$

Accordingly, ${8}^{{\log}_{8} 6} = 6 , \mathmr{and} , {8}^{{\log}_{8} 9} = 9.$

Therefore, The Reqd. Value=${8}^{\left({\log}_{8} 6\right) - \left({\log}_{8} 9\right)} ,$

$= \frac{{8}^{{\log}_{8} 6}}{{8}^{{\log}_{8} 9}} ,$

$= \frac{6}{9} ,$

$= \frac{2}{3.}$