Question

Please help me out.....

Answer 1

here is the answer
I think u got a perfect method

Answer 2

Hope u like my process
=====================
$= > \bf {a}^{3 - x} {b}^{5x} = {a}^{x + 5} {b}^{3x} \\ \\ = > \frac{ {b}^{5x} }{ {b}^{3x} } = \frac{ {a}^{x + 5} }{ {a}^{3 - x} } \\ \\ = > {b}^{5x - 3x} = {a}^{(x + 5) - (3 - x)} \\ \\ = > {b}^{2x} = {a}^{2x + 2} \\ \\ = > {a}^{2} \times {a}^{2x} = {b}^{2x} \\ \\ = > {a} = ( \frac{b}{a} ) ^{x} \\ \\ \: \: \: \: \: \bf \underline{taking \: \: log \: \: on \: \: both \: \: sides} \\ \\ = > log( {( \frac{b}{a} )}^{x} ) = log(a) \\ \\ = > \bf \: x log( \frac{b}{a} ) = \underline{ \: log(a) \: }$
____________________________
Hope this is ur required answer

Proud to help you