# How do you solve 9^x=3^(x-2)?

Nov 11, 2015

$x = - 2$

#### Explanation:

First of all, we need the rule which states that

${\left({a}^{b}\right)}^{c} = {a}^{b c}$.

Then, we observe that $9 = {3}^{2}$, and so the expression becomes

${\left({3}^{2}\right)}^{x} = {3}^{x - 2}$

Applying the forementioned rule, we get

${3}^{2 x} = {3}^{x - 2}$

Now, if two powers of three are equal, then the exponents must be equal. In fact, it's not possible to give two different exponents to three and obtain the same result. This means that, in technical terms, the exponential is an injective function. So, if ${3}^{2 x} = {3}^{x - 2}$, then it must be

$2 x = x - 2$

which we can easily solve: subtractin $x$ from both sides, we get

$2 x - x = x - 2 - x$ and thus $x = - 2$

Nov 11, 2015

I found $x = - 2$

#### Explanation:

You could write $9 = {3}^{2}$ and so:
${3}^{2 x} = {3}^{x - 2}$
to be equal the exponents must be equal:
$2 x = x - 2$
$x = - 2$