If (2x)ln2 = (3y)ln3, 3lnx = 2lny and (x0, y0) is the solution of these equations, then x0 is

Let (2x)ln2 = (3y)ln3 — (i)
3lnx = 2lny — (ii)
(i) ⇒ (ln2)(ln2x) = (ln3)(ln3y) ⇒ (ln2)(lnx + ln2) = (ln3)(ln3 + lny)
Let lnx = a, lny = b
⇒ (ln2)(a + ln2) = (ln3)(ln3 + b) — (iii)
(ii) ⇒ (lnx)(ln3) = (ln2)(lny)
Substituting a, b ⇒ a(ln3) = b(ln2) — (iv)
Substituting the value of b from (iv) in (iii), we get
a(ln2) + (ln2)² = (ln3)² + a(ln3)²/(ln2)
On simplifying we get,
a = ln2((ln3)² - (ln2)²)/(ln2)² - (ln3)² ⇒ a = -ln2
⇒ lnx = -ln2 ⇒ x = 1/2