Discussion Forum

 Let ($x_{0},y_{0})$ be the solution of the following equations  $(2x)^{\log 2}=(3y)^{\log 3}$  $3^{\log x}=2 ^{\log y}$, then $x_{0}$ is equal to 

Answer:

Explanation:

Taking log on both  sides

 $\log 2.\log(2x)= \log 3(\log 3y)$

$\Rightarrow$   $\log 2(\log 2+\log x)$

= $\log 3(\log 3+\log y)$....(i)

and $\log x.\log 3= \log y \log 2$

$\log y= \frac{ \log x. \log 3}{ \log 2}$...(ii)

 from Eq.(i) and (ii) we get

 $\log 2( \log 2+\log x)=$

                      $\log 3.\left\{ \log 3+\frac{\log x.\log 3}{\log 2}\right\}$

 $\Rightarrow$  $(\log 2)^{2}+\log 2.\log x=$

 $(\log 3)^{2}+\frac{ (\log 3)^{2}}{ (\log 2)}. \log x$

$\Rightarrow$  $\log x.\left\{\frac{(\log 3)^{2}}{\log 2}-\log 2\right\}=  (\log 2)^{2}-(\log 3)^{2}$

$\Rightarrow$   $\log x.\left\{\frac{(\log 3)^{2}-(\log 2)^{2}}{\log 2}\right\}=  (\log 2)^{2}-(\log 3)^{2}$

$\Rightarrow$   $\log x=-\log 2= \log 2^{-1}$

$\therefore$  $x= \frac{1}{2}$