Problem: If 4x2x+y=8\dfrac{4^{x}}{2^{x+y}}=82x+y4x=8 and 9x+y35y=243,x\dfrac{9^{x+y}}{3^{5 y}}=243, x35y9x+y=243,x and yyy real numbers, then xyx yxy equals:
Answer Choices:
A. 125\dfrac{12}{5}512
B. 444
C. 666
D. 121212
E. −4-4−4 Solution:
4x2x+y=22x2x+y=2x−y=8=23∴x−y=3∴x=4,y=1\dfrac{4^{x}}{2^{x+y}}=\dfrac{2^{2 x}}{2^{x+y}}=2^{x-y}=8=2^{3} \quad \therefore \quad x-y=3 \quad \therefore x=4, y=12x+y4x=2x+y22x=2x−y=8=23∴x−y=3∴x=4,y=1
9x+y35y=32x+2y35y=32x−3y=243=35∴2x−3y=5∴xy=4\dfrac{9^{x+y}}{3^{5 y}}=\dfrac{3^{2 x+2 y}}{3^{5 y}}=3^{2 x-3 y}=243=3^{5} \quad \therefore 2 x-3 y=5 \quad \therefore x y=435y9x+y=35y32x+2y=32x−3y=243=35∴2x−3y=5∴xy=4