Question

If ${\log }_2(x+y)={\log }_3(x-y)=\frac{\log 25}{\log 0.2}$, then $x=\frac{13}{k}$ and $y=\frac{5}{k}$.

Fink value of $\frac{k}{9}$.

Answer 1

$Let{\log }_2(x+y)={\log }_3(x-y)=\frac{\log 25}{\log 0.2}=P{\log }_2(x+y)=P;x+y=2^P{\log }_3(x-y)=P;x-y=3^P\frac{\log 25}{\log 0.2}=P;\frac{\log 5^2}{\log 5^{-1}}=P\log 25=P\log 5^{-1}\log 5^2=\log 5^{-P}P=-2\text{Substituting value of P}x+y=2^{-2}=\frac{1}{4}...\left(1\right)x-y=3^{-2}=\frac{1}{9}...\left(2\right)\text{Adding eq.(1) and eq.(2), we get}2x=\frac{1}{4}+\frac{1}{9}=\frac{9+4}{36}=\frac{13}{36}So,x=\frac{13}{72}\text{Substituting value of x in eq.(1)}y=\frac{1}{4}-\frac{13}{72}=\frac{18-13}{72}=\frac{5}{72}$