If 4^x2^x+y=8, 9^x+y3^5y=243, x and y are real numbers, then xy equals:

1966 AMC 12 Problem 16

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16.

\begin{gathered} \dfrac{4^x}{2^{x+y}}=8,\\ \dfrac{9^{x+y}}{3^{5y}}=243, \end{gathered}

x

and

y

are real numbers, then

xy

equals:

1\dfrac34

4

6

12

-4

Rewrite both equations with bases

2

and

3

Equate exponents to obtain two linear equations in

x

and

y

The first equation becomes

2^{x-y}=2^3,

x-y=3.

The second becomes

3^{2x-3y}=3^5,

2x-3y=5.

Solving gives

y=1,

x=4,

and hence

xy=4.

Therefore, the correct answer is B.