Question 513557
The first thing to recognize is that {{{9^(x+2)=9^x * 9^2}}} by definition of properties of exponents.  That is equivalent to saying {{{81*9^x}}} on the left hand side of your equation.


You now have {{{81*9^x=240+9^x}}}.  If you subtract {{{9^x}}} from both sides, you have {{{81*9^x-9^x=240}}}.  The left hand side simplifies to {{{80*9^x}}} (you could factor out {{{9^x}}} and be left with {{{9^x*(81-1)=9^x*80}}} or you could recognize that 81*something-same something=80*something).


Now, you have {{{80*9^x=240}}}.  Divide by 80 on both sides and get {{{9^x=3}}}.  A couple of ways to think about this:
1) 9 to what power gives me 3?
2) The square root of 9 is 3; what's the exponent for square root?
3) {{{log(3,9)}}}=?
4) Get like bases. 9 is the same as {{{3^2}}}, so this could be written as {{{(3^2)^x=3^1}}}, or {{{3^(2x)=3^1}}}.  Since the bases are equal, the exponents must be equal, and {{{2x=1}}}, so {{{x=1/2}}}.


No matter how you do it, you should get {{{x=1/2}}}.