Question

let x' and y' be solution of the following equations
(2x)^ln2=(3y)^ln3
3^lnx=2^lny
then x' is ?

Answer 1

see attachment....hope it will help.

Answer 2

Given:

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

To find:

x' =?

Answer:

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

Taking Log on both sides,

$\ln (\mathrm{x}) \ln (3)=\ln (\mathrm{y}) \ln (2)$

$\ln (\mathrm{y})=\frac{\ln (\mathrm{x}) \ln (3)}{\ln (2)}$

Now consider $2 x^{\ln 2}=3 y^{\ln 3}$,

$\ln (2)(\ln (2)+\ln (\mathrm{x}))=\ln (3) \quad(\ln (3)+\ln (\mathrm{y}))$

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

$(\ln { ( } 2))^{ 2 }+\ln { ( } 2)\cdot \ln { ( } { x })=(\ln { ( } 3))^{ 2 }+\ln { ( } 3)\cdot \frac { \ln { ( } { x })\ln { ( } 3) }{ \ln { ( } 2) }$

$-\ln (2)=\ln (\mathrm{x})$

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

Therefore, $x'=\frac { 1 }{ 2 }$ is the required answer.