Solve the system of equations Logx 3y 2 logx 4y 1 Solve th

Solve the system of equations.

Log_x (3y) = 2 log_x (4y) = 1 Solve the system of equations.

logx(3y) =2 ---(1)

logx(4y) =1 ---(2)

equation (1): 3y = x^2 ( Use log property: loga(x) = y ---> x = a^y )

equation (2) : logx(4y) =1

4y =x

So, substitute x = 4y in equation (1) : 3y = (4y)^2

3y -16y^2 =0

y(3 -16y)=0

y = 3/16

and x= 4y =3/4

Solution :(x,y) = ( 3/4, 3/16)