Question

Please solve this for me with full steps please it’s urgent

Answer 1

Answer:

Step-by-step explanation:

logx$y^{2}$ <=> logx + 2logy = a (1)

log$x^{3}$y=b <=> 3logx + 6logy =b (2)

(1),(2)=> logx= $\frac{-3a+2b}{3}$

            logy= 3a-b

=> log$\frac{y}{x}$=logy-logx=3a-b -( $\frac{-3a+2b}{3}$) = $\frac{12a-5b}3}$

   log$\sqrt{xy}$ = $\frac{1}{2}$(logx+logy)=........ OK