ABC26GN2057 · Logarithms

$\log _{10} x+\log _{10} y=3$ $$\log _{10} x-\log _{10} y=1$$ Adding (i) and (ii), we get: $2 \log _{10} x=4$ or $\log _{10} x=2 \Rightarrow \quad \therefore x=10^{2}=100$. Also, $\log _{10} y=3-\log _{10} x=3-2=1 \Rightarrow y=10^{1}=10$. Hence, $x=100, y=10$. Another method: $$\begin{aligned} & \log _{10} x+\log _{10} y=3 \Rightarrow \log _{10}(x y)=3 \Rightarrow x y=10^{3}=1000 . \\ & \log _{10} x-\log _{10} y=1 \\ & \Rightarrow \log _{10}\left(\frac{x}{y}\right)=1 \Rightarrow \frac{x}{y}=10^{1}=10 \Rightarrow x=10 y \Rightarrow 10 y \cdot y=1000 \\ & \quad \Rightarrow 10 y^{2}=1000 \Rightarrow y^{2}=100 \Rightarrow y=10 . \\ & \therefore x=10 y=10 \times 10=100 . \end{aligned}$$