Question

20 points for anyone that can solve the question ​

Answer 1

Answer:

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

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