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Source:

March 22, 2006
logarithms

Problem Statement

If 4x2x+y=8\frac{4^x}{2^{x+y}}=8 and 9x+y35y=243\frac{9^{x+y}}{3^{5y}}=243, xx and yy are real numbers, then xyxy equals: (A) 125(B) 4(C) 6(D) 12(E) 4\text{(A)} \ \frac{12}{5} \qquad \text{(B)} \ 4 \qquad \text{(C)} \ 6 \qquad \text{(D)} \ 12 \qquad \text{(E)} \ -4