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algebra intermediate

Problem

If

\frac{4 ^{x}}{2 ^{x + y}} = 8

and

\frac{9 ^{x + y}}{3 ^{5 y}} = 243

,

x

and

y

real numbers, then

x y

equals:

(A)

\frac{12}{5}

(B)

4

(C)

6

(D)

12

Solution

\frac{4 ^{x}}{2 ^{x + y}} = 8 ⟹ 4^{x} = 2^{x + y + 3} ⟹ 2 x = x + y + 3 ⟹ x = y + 3

.

\frac{9 ^{x + y}}{3 ^{5 y}} = 243 ⟹ 9^{x + y} = 3^{5 y + 5} ⟹ 2 x + 2 y = 5 y + 5 ⟹ 2 x = 3 y + 5

. So,

2 y + 6 = 3 y + 5 ⟹ y = 1 ⟹ x = 4

. Therefore,

x y = 4

or

4

Final answer

B