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

1 Answer
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*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#