Quadratic Equation and Inequalities

We have
(2x)ln2 = (3y)ln3 (1)
3lnx = 2lny (2)
⇒(log x)(log3) = (log y) log2
⇒ log y =

(logx)(log3)

log2

(3)
Taking log both sides of Eq. (1), we get
(log 2){log2 + log x} = log3{log3 + log y}
(log2)2+(log2)(logx) = (log3)2+

(log3)2(logx)

log2

from Eq. (3)
⇒ (log2)2−(log3)2 =

(log3)2−(log2)2

log2

(logx)
⇒ - log 2 = log x
⇒ x =

1

2

⇒ x0 =

1

2

.