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 $x = \frac{1}{2}$
$\endgroup$3 Answers
$\begingroup$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?
$\endgroup$ 1 $\begingroup$$(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 $e^0=1$.
$\endgroup$ 0 $\begingroup$Problem is in last second line $2^{ln2}$ is not equal to $3^{ln3}$
$\endgroup$
