Question

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

Answer 1

`x=2`

Explanation:

First we to need to know a property of exponents with more than 1 term:
`a^(b+c)=a^b*a^c`
Applying this, you can see that:
`3^(x+1)+3^x=36`
`3^x*3^1+3^x=36`
`3^x*3+3^x=36`
As you can see, we can factor out `3^x`:
`(3^x)(3+1)=36`
And now we rearrange so any term with x is on one side:
`(3^x)(4)=36`
`(3^x)=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(a^b)=blog(a)`, saying that you can move exponents out and down from the brackets. Applying this to where we left off:
`log(3^x)=log(9)`
`xlog(3)=log(9)`
`x=log(9)/log(3)`
And if you type it into your calculator you'll get `x=2`