# How do you solve 27^(x-3)=1/9?

Nov 29, 2015

Express as powers of $3$ and hence find $x = \frac{7}{3}$

#### Explanation:

If $a > 0$ then ${a}^{b c} = {\left({a}^{b}\right)}^{c}$

So:

${3}^{3 \left(x - 3\right)} = {\left({3}^{3}\right)}^{x - 3} = {27}^{x - 3} = \frac{1}{9} = {3}^{-} 2$

Since $f \left(x\right) = {3}^{x}$ is a one-one function from $\mathbb{R} \to \left(0 , \infty\right)$, the Real solutions of this must have equal exponents:

$3 \left(x - 3\right) = - 2$

Hence:

$3 x = - 2 + 9 = 7$

So $x = \frac{7}{3}$