SOLUTION: {{{9^(x+2)=240+9^x}}}

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Question 513557: 9%5E%28x%2B2%29=240%2B9%5Ex

Answer by kingme18(98) About Me  (Show Source):
You can put this solution on YOUR website!
The first thing to recognize is that 9%5E%28x%2B2%29=9%5Ex+%2A+9%5E2 by definition of properties of exponents. That is equivalent to saying 81%2A9%5Ex on the left hand side of your equation.

You now have 81%2A9%5Ex=240%2B9%5Ex. If you subtract 9%5Ex from both sides, you have 81%2A9%5Ex-9%5Ex=240. The left hand side simplifies to 80%2A9%5Ex (you could factor out 9%5Ex and be left with 9%5Ex%2A%2881-1%29=9%5Ex%2A80 or you could recognize that 81*something-same something=80*something).

Now, you have 80%2A9%5Ex=240. Divide by 80 on both sides and get 9%5Ex=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%283%2C9%29=?
4) Get like bases. 9 is the same as 3%5E2, so this could be written as %283%5E2%29%5Ex=3%5E1, or 3%5E%282x%29=3%5E1. Since the bases are equal, the exponents must be equal, and 2x=1, so x=1%2F2.

No matter how you do it, you should get x=1%2F2.