Question

Solve the system of the equations $(ax{)}^{\log a}=(by{)}^{\log b};b^{\log x}=a^{\log y}$ where $a>0,b>0$ and $a\ne b,ab\ne 1$

Answer 1

$(ax{)}^{\log a}=(by{)}^{\log b}$

$\Rightarrow a^{\log a}{x}^{\log a}=b^{\log b}{y}^{\log b}$

$\Rightarrow a^{\log a}{a}^{\log x}=b^{\log b}{b}^{\log y}$

$\Rightarrow a^{\log a+\log x}=b^{\log b+\log y}$

$\Rightarrow a^{\log ax}=b^{\log by}$

Taking log both sides,

$\Rightarrow {\log }_{a}ax{\log }_{a}a={\log }_{a}by{\log }_{a}b$

$\Rightarrow {\log }_{a}ax={\log }_{a}by{\log }_{a}b$

$\Rightarrow \frac{{\log }_{a}ax}{{\log }_{a}b}={\log }_{a}by$

$\Rightarrow {\log }_b\left(ax\right)={\log }_a\left(by\right)$

$\Rightarrow {\log }_b^a+{\log }_{b}x={\log }_{a}b+{\log }_{a}y$

$\Rightarrow {\log }_{b}a-{\log }_{a}b={\log }_{a}y-{\log }_{b}x$

$\Rightarrow {\log }_{b}a-\frac{1}{{\log }_{b}a}={\log }_{a}y-{\log }_{b}x$

$\Rightarrow ({\log }_{b}a{)}^2-1={\log }_{b}a{\log }_{a}y-{\log }_{b}a{\log }_{b}x$

$\Rightarrow ({\log }_{b}a{)}^2-1={\log }_{b}y-{\log }_{b}a{\log }_{b}x$ __(1)

we have $b^{\log x}=a^{\log y}$

$\Rightarrow {\log }_{b}x{\log }_{b}b={\log }_{b}y{\log }_{b}a$

$\Rightarrow {\log }_{b}x={\log }_{b}y{\log }_{b}a$ __(2)

Substituting (2) in (1)

$\Rightarrow ({\log }_{b}a{)}^2-1={\log }_{b}y-{\log }_{b}a({\log }_{b}y{\log }_{b}a)$

$\Rightarrow ({\log }_{b}a{)}^2-1={\log }_{b}y(1-{\log }_{b}a{)}^2$

$\Rightarrow {\log }_{b}y=-1\therefore y=b^{-1}\therefore y=\frac{1}{b}$

$\Rightarrow {\log }_{b}x={\log }_b{a}^{-1}\Rightarrow x=a^{-1}$

$\therefore x=\frac{1}{a},y=\frac{1}{b}$