Let `(x_0,y_0)` be the solution of the following equations: `(2x)^ln2 = (3y)^ln3 and 3^(lnx) = 2^(lny)`Then value of `x_0` is:
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`(2x)^(ln2) = (3y)^ln3`
Taking `ln` both sides,
`=>(ln2) (ln2x) = (ln3)(ln3y)`
`=>(ln2) (ln2 +lnx) = (ln3)(ln3+lny)->(1)`
Now, we will take the second equation,
`3^(ln x) = 2^(ln y)`
Taking `ln` both sides,
`=>(ln x)(ln 3) = (ln y)(ln 2)`
`=> (ln y) = ((ln x)(ln 3))/(ln 2)`
Putting value of `ln y` in (1),
`(ln2) (ln2 +lnx) = (ln3)(ln3+((ln x)(ln 3))/(ln 2))`
`=>(ln x)(ln 2- ((ln 3)^2/ln2)) = (ln3)^2-(ln2)^2`
`=>(ln x)/(ln 2)((ln 2)^2- (ln 3)^2) = (ln3)^2-(ln2)^2`
`=>(ln x)/(ln 2) = -1`
`=>(ln x) = ln (2)^-1`
`=> x = 2^-1 => x = 1/2`, which is the required value of `x_0`.