Question

If ${log}_2(x+y)={log}_3(x-y)=\frac{\log 25}{\log 0.2},$ find the values of $x$ and $y$.

Answer 1

${\log }_2(x+y)={\log }_{0.2}25,{\log }_3(x-y)={\log }_{0.2}25$

$\Rightarrow x+y=2^{{\log }_{0.2}25},x-y=3^{{\log }_{0.2}25}$

$\Rightarrow x+y=2^{{\log }_{\frac{2}{10}}25},x-y=3^{{\log }_{\frac{2}{10}}25}$

$\Rightarrow x+y=2^{{\log }_{\frac{1}{5}}25},x-y=3^{{\log }_{\frac{1}{5}}25}$

$\Rightarrow x+y=2^{{\log }_{5^{-1}}25},x-y=3^{{\log }_{5^{-1}}25}$

$\Rightarrow x+y=2^{-2{\log }_{5}5},x-y=3^{-2{\log }_{5}5}$

$\Rightarrow x+y=2^{-2},x-y=3^{-2}$

$\Rightarrow x+y=\frac{1}{4},x-y=\frac{1}{9}$

$\Rightarrow x+y+x-y=\frac{1}{4}+\frac{1}{9}$

$\Rightarrow 2x=\frac{9+4}{36}=\frac{13}{36}$

$\Rightarrow x=\frac{13}{72}$

Substitute $x=\frac{13}{72}$ in $x+y=\frac{1}{4}$ we get

$\Rightarrow y=\frac{1}{4}-x=\frac{1}{4}-\frac{13}{72}=\frac{18-13}{72}=\frac{5}{72}$

$\therefore x=\frac{13}{72},y=\frac{5}{72}$