Question

If ${log}_2(x+y)$=${log}_3(x-y)=\frac{log25}{log0.2},$ find the value of x and y.

Answer 1

We have,

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

$\Rightarrow {\log }_2(x+y)={\log }_3(x-y)=\frac{\log 25}{\log \frac{2}{10}}$

$\Rightarrow {\log }_2(x+y)={\log }_3(x-y)=\frac{\log 25}{\log \frac{1}{5}}$

$\Rightarrow {\log }_2(x+y)={\log }_3(x-y)=\frac{\log 5^2}{\log 5^{-1}}$

$\Rightarrow {\log }_2(x+y)={\log }_3(x-y)=\frac{2\log 5}{-\log 5}$

$\Rightarrow {\log }_2(x+y)={\log }_3(x-y)=-2$

Now,

${\log }_2(x+y)=-2,{\log }_3(x-y)=-2$

$(x+y)=2^{-2}......\left(1\right),(x-y)=3^{-2}......\left(2\right)$

Solving equation (1) and (2) to, and we get,

$x=\frac{13}{72}$ and $y=\frac{5}{72}$

Hence, this is the answer.