# How do you solve 1/81 - 9^(x-3) = 0?

Sep 24, 2015

color(blue)(x=1

#### Explanation:

$\frac{1}{81} - {9}^{x - 3} = 0$

We know that , $\frac{1}{81} = \frac{1}{9} ^ 2$

As per property
color(blue)(a^-m = 1/a^m

So,
$\frac{1}{81} = \frac{1}{9} ^ 2 = {9}^{-} 2$

The expression now becomes:

$\frac{1}{9} ^ 2 - {9}^{x - 3} = {9}^{-} 2 - {9}^{x - 3} = 0$

9^color(blue)(-2) =9^color(blue)((x-3)

As the bases are equal we can equate the powers:

$- 2 = x - 3$

$x = - 2 + 3$

color(blue)(x=1