If (2^x) (2^y) = 8 and (9^x)(3^y) = 81, then (x,y) equals:

A. (1,2)
B. (2,1)
C. (1,1)
D. (2,2)
E. (1,3)


I understand how to cancel out things for the first half of the problem. For the second half, GMATPrep v2 says that (3^2x)(3^y) = 3^4 becomes 32x + y = 34. Can somebody explain how that happens?

If (2 to the power of X)(2 to the power of Y) and (9 to the power of X) (3 to the power of Y) = 81 then (X,Y) =

(1,2)

(2,1)

(1,1)

(2,2)

(1,3)

you could simply plug in answer choices into the equations and see which one is equal...
answer A.

If (2^x) (2^y) = 8 and (9^x)(3^y) = 81, then (x,y) equals:

A. (1,2)
B. (2,1)
C. (1,1)
D. (2,2)
E. (1,3)

I think the plug-in method works best in the actual exam when you're fighting against the clock, or when you've exhausted all possible angles of attack on a notoriously difficult problem. But during preparation, I believe it is essential to understand how and why the concept work to solve these questions.

To solve this question you should be familiar with the following rules :

- \(a^x * a^y = a^(x+y)\) (1) (notice that for this rule to work, we must have the same base, as in "a")
- \((a^x)^y = a^(x*y)\) (2)
- \(a^x = a^y\) therefore \(x = y\) (3) (same as with rule n°1, we need to have the same base)

Considering our problem, we have :

\(2^x*2^y = 8\) which, according to (1) will yield \(2^(x+y) = 8 = 2^3\) which, according to (3) will yield : \(x + y = 3\)
\(9^x*3^y = 81\)
Since \(9 = 3^2\), then the above equation becomes : \((3^2)^x*3^y = 81\). According to (2), we can write : \(3^(2*x)*3^y = 81\) which, according to (1) will yield \(3^(2x+y) = 81 = 3^4\), which according to (3), will yield : \(2x + y = 4\)

We end up with a linear system of two equations :
\(x + y = 3\)
\(2x + y = 4\)

(The solution of which is x = 1 and y = 2. Which is answer choice A.), Please solve how X=1 and Y=2

Hope that helped :-D