This one is killing me, I don't really know what to do, but I will tell you what I have tried. I thought that since the first 2 terms were being multiplied we can add the exponents, this just seemed wrong though. I thought maybe I needed to make the bases the same so I took 27^(3-x) and thought maybe 9^3^(3-x). This confused me once again though. I think i'm stuck.
9^2x * 27^(3-x) = 1/9
- Thread starter ochocki
- Start date
ochocki said:
9<SUP>2x</SUP> * 27<SUP>(3 - x)</SUP> = 1/9
You are on the right track....but the rule you're attempting to apply says that when you multiply powers of the same base, you add the exponents. And that's the difficulty here....we have two different bases.
HOWEVER....do you notice that both 9 and 27 are powers of 3?
(3<SUP>2</SUP>)<SUP>2x</SUP> * (3<SUP>3</SUP>)<SUP>(3 - x)</SUP> = 3<SUP>-2</SUP>
When you raise a power to a power, you MULTIPLY the exponents:
3<SUP>2*2x</SUP> * 3<SUP>3(3 - x)</SUP> = 3<SUP>-2</SUP>
3<SUP>4x</SUP> * 3<SUP>(9 - 3x)</SUP> = 3<SUP>-2</SUP>
You can add the exponents on the left-hand side since NOW we are multiplying powers of the same base:
3<SUP>9 + x</SUP> = 3<SUP>-2</SUP>
Can you take it from here?