# How do you solve 3^(x+1) + 3^x = 36?

Apr 21, 2016

$x = 2$

#### Explanation:

First we to need to know a property of exponents with more than 1 term:
${a}^{b + c} = {a}^{b} \cdot {a}^{c}$
Applying this, you can see that:
${3}^{x + 1} + {3}^{x} = 36$
${3}^{x} \cdot {3}^{1} + {3}^{x} = 36$
${3}^{x} \cdot 3 + {3}^{x} = 36$
As you can see, we can factor out ${3}^{x}$:
$\left({3}^{x}\right) \left(3 + 1\right) = 36$
And now we rearrange so any term with x is on one side:
$\left({3}^{x}\right) \left(4\right) = 36$
$\left({3}^{x}\right) = 9$

It's should be easy to see what $x$ should be now, but for the sake of knowledge (and the fact that there are much harder questions out there), I'll show you how to do it using $\log$

In logarithms, there is a root which states: $\log \left({a}^{b}\right) = b \log \left(a\right)$, saying that you can move exponents out and down from the brackets. Applying this to where we left off:
$\log \left({3}^{x}\right) = \log \left(9\right)$
$x \log \left(3\right) = \log \left(9\right)$
$x = \log \frac{9}{\log} \left(3\right)$
And if you type it into your calculator you'll get $x = 2$