\(2^x*2^y=8\) --> \(2^{x+y}=2^3\) --> \(x+y=3\);
\(9^x*3^y=81\) --> \(3^{2x}*3^y=3^4\) --> \(3^{2x+y}=3^4\) --> \(2x+y=4\);

Solving: \(x+y=3\) and \(2x+y=4\) we get \(x=1\) and \(y=2\).

Answer: A.

Hope it's clear.
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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)

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.

Hope that helped

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

If base is same, we add powers.
So,
x+y= 3 (3 because 2^3=8 and we are comparing/equating powers)
2x+y= 4

Let's subtract the two equations:
2x+y=4
-x-y=-3

x= 1
and thus y=2

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