Let $(x_0,y_0)$ be the solution of the following equations $(2x)^{\ln 2}=(3y)^{\ln 3}, 3^{\ln x}=2^{\ln y}$. Then $x_0$ is:-

Question

Let $(x_0,y_0)$ be the solution of the following equations $(2x)^{\ln 2}=(3y)^{\ln 3}, 3^{\ln x}=2^{\ln y}$. Then $x_0$ is:-

My attempt is as follows:-

$$x^{\ln3}=y^{\ln2}$$ $$y=x^{\frac{\ln3}{\ln2}}$$

$$(2x)^{\ln 2}=\left(3x^{\frac{\ln 3}{\ln 2}}\right)^{\ln 3}$$ $$\dfrac{2^{\ln 2}}{3^{\ln 3}}=x^{\frac{\ln^23}{\ln2}-\ln2}$$

$$x=\left(\dfrac{2^{\ln 2}}{3^{\ln 3}}\right)^{\frac{\ln2}{\ln^23-\ln^22}}$$

But actual answer is $\dfrac{1}{2}$. I think I did the right procedure but unfortunately not getting the answer at end. Please help me in figuring out what is wrong with my solution.

Answer 1

Take natural log both sides, $$\ln{x} = \left(\dfrac{\ln{2}}{\ln^23 - \ln^22}\right)(\ln^22 - \ln^2 3)$$ $\implies \ln x = -\ln 2$

$x = \dfrac{1}{2}$

Answer 2

If you plug \begin{eqnarray*} x=\left(\dfrac{2^{\ln 2}}{3^{\ln 3}}\right)^{\frac{\ln2}{\ln^23-\ln^22}} \end{eqnarray*} into your casio (other brands of calculator are avaiable) ... you do indeed get $0.5$.

Alternatively take logarithms of both equations (& using an obvious notation) we have the following simultaneous equations \begin{eqnarray*} l_2^2+l_2 l_x = l_3^2+l_3 l_y \\ l_3 l_x =l_2 l_y. \end{eqnarray*} These are easily solved (multiply the first equation by $l_3$ and the second by $l_2$ then subtract) \begin{eqnarray*} l_x=-l_2 \end{eqnarray*} which gives $x=1/2$.

Answer 3

Let's multiply the first equation for the second one by getting:

$e^{ln(2)^{2}}.x^{ln(6)}= e^{ln(3)^{2}}.y^{ln(6)}$

that is

$y=\frac{2x}{3}$

So the first equation becomes:

$e^{ln(2)^{2}}.x^{ln(2)}= 3^{ln(2)}.x^{ln(3)}$,

$3^{-ln(2)}.e^{ln(2)^{2}}.x^{-ln(\frac{3}{2}}=1$

Applying the logarithm to the first and second members, we get:

$ln(x).ln(\frac{3}{2})=ln(2)^{2}-ln(2).ln(3)$

$ln(x)=\frac{ ln(2)^{2}-ln(2).ln(3)}{ ln(\frac{3}{2})}=-ln(2)$,

$x=\frac{1}{2}$,

$y=\frac{1}{3}$.