Question

Solve the equations for x and y : ${\left(3x\right)}^{10g3}={\left(4y\right)}^{\log 4}and{4}^{\log x}=3^{\log y}$ .

Answer 1

${\left(3x\right)}^{\log 3}={\left(4y\right)}^{\log 4},4^{\log x}=3^{\log y}$
$\Rightarrow \left(\log 3\right)\left(\log 3x\right)=\left(\log 4\right)\left(\log 4y\right)$ and $\left(\log x\right)\left(\log 4\right)=\left(\log y\right)\left(\log 3\right)$
$\Rightarrow \left(\log 3\right)(\log 3+\log x)=\left(\log 4\right)(\log 4+\log y)$ and $\left(\log x\right)\left(\log 4\right)=\left(\log y\right)\left(\log 3\right)$
$\Rightarrow \left(\log 3\right)(\log 3+p)=\left(\log 4\right)(\log 4+q)$ and $p\left(\log 4\right)=q\left(\log 3\right)$ (where $p=\log x$ and $q=\log y$)
$\Rightarrow \left(\log 3\right)(\log 3+\frac{q\log 3}{\log 4})=\left(\log 4\right)(\log 4+q)$ (eliminating p)
$\Rightarrow {\left(\log 3\right)}^2-{\left(\log 4\right)}^2=\frac{{\left(\log 4\right)}^2-{\left(\log 3\right)}^2}{\log 4}q$
$\Rightarrow q=-\log 4$
$\Rightarrow \log y=\log 4^{-1}$
$\Rightarrow y=\frac{1}{4}$
Now $p\left(\log 4\right)=q\left(\log 3\right)$
$\Rightarrow p\left(\log 4\right)=-\left(\log 4\right)\left(\log 3\right)$
$\Rightarrow p=-\left(\log 3\right)$
$\Rightarrow \log x=\log 3^{-1}$
$\Rightarrow x=\frac{1}{3}$