What is wrong with this solution of $ (2x)^{\ln 2} = (3y)^{\ln 3}$

Question

What is wrong with this solution of

Solve for $ (2x)^{\ln2 }= (3y)^{\ln3 }$

$3^{\ln x} = 2^{\ln y} $

Here is my solution-

$ (2x)^{\ln2 }= (3y)^{\ln3 }$

$ 2^{\ln2x }= 3^{\ln3y }$

$ 2^{(\ln x+\ln 2)} = 3^{(\ln3 + \ln y)}$

$ 2^{\ln x } 2^{\ln2 }= 3^{\ln3} 3^{\ln y}$

$ 2^{\ln x} = 3^{\ln y}$

Thus the answer is $x = y = 1$

But this is not the answer, it is $x = \frac{1}{2}$

Answer 1

Problem is in last second line $2^{ln2}$ is not equal to $3^{ln3}$

Answer 2

Taking the logarithm on both sides we get $$\ln(2)\ln(2x)=\ln(3)\ln(3y)$$ so $$\ln(2)(\ln(2)+\ln(x))=\ln(3)(\ln(3)+\ln(y))$$ Can you finish?

Answer 3

$(2\,x)^{\ln(2)}=(3\,y)^{\ln(3)}$ is obviously true for $x=\frac{1}{2}$ and $y=\frac{1}{3}$.

The mistake in the solution above is $(2\,x)^{\ln(2)}\ne 2^{\ln(2\,x)}$ and likewise $(3\,y)^{\ln(3)}\ne 3^{\ln(3\,y)}$, rather $(2\,x)^{\ln(2)}=e^{\ln(2\,x) \ln(2)}$ and likewise $(3\,y)^{\ln(3)}=e^{\ln(3\,y) \ln(3)}$. For $x=\frac{1}{2}$ and $y=\frac{1}{3}$ $e^{\ln(2\,x) \ln(2)}=e^{\ln(3\,y) \ln(3)}$ is equivalent to $1=1$ since since $\ln(1)=0$ and $e^0=1$.