Question

20 points for anyone that can solve the question ​

Answer 1

`{3}^(x) = {9}^(y) - - - eqn(i) \\ 4 ^(xy) = {2}^(x - 2) - - - eqn(ii) \\ from \: (i) \: introduce \: log_(10)\: \: \\ log( {3}^(x) ) = log( {9}^(y) ) \\ x log(3) = y log(9) \\ x log(3) = 2y log(3) \\ x = 2y - - - eqn(iii) \\ substitute \: for \: x \: in \: eqn(ii) \\ {4}^{ {2y}^(2) } = {2}^(2(y - 1)) \\ {4}^{ {2y}^(2) } = {4}^((y - 1)) \\ from \: indice \: laws \\ {2}^{ {y}^(2) } = y - 1 \\ introduce \: log_(10) \\ {y}^(2) log(2) = log(y) - log(1) \\ log(2) = {y}^( - 2) log (y \: - \: 1) \\ {y}^( - 2) (2 - y - 1) = 10 \\ \frac{2}{ {y}^(2) } - (1)/(y) - \frac{1}{ {y}^(2) } = 10 \\ 1 - y = 10 {y}^(2) \\ 10 {y}^(2) + y - 1 = 0 \\`

hope that step is enough to give you the two values of y, coz I gat no calc here with me.

hint: use the quadratic equation