Question

solve for x and y : 3^x * 3^-y= 9 and 2^y*4^-x=1/8

Answer 1

Answer:

${3}^{x} \times {3}^{ - y} = 9....(given) \\ \implies \: {3}^{x - y} = {3}^{2} \\ \implies \: x - y = 2......(1) \\ next \\ {2}^{y} \times {4}^{ - x} = \frac{1}{8} \\ \implies \: {2}^{y} \times {2}^{ - 2x} = (\frac{1}{2})^{3} \\ \implies \: {2}^{y - 2x} = {2}^{ - 3} \\ \implies \: y - 2x = - 3.......(2) \\ adding \: equations \: (1) \: and \: (2) \\ x - y = 2 \\ y - 2x = - 3 \\ - - - - - - \\ - x = - 1\implies \: \huge \fbox{x = 1 }\\ substituting \: x = 1 \: in \: equation \: (1) \\ 1 - y = 2 \\ \implies \: \huge \fbox{y = - 1}$