Using equation to find value of $1/x - 1/y$ $$\left(\frac{48}{10}\right)^x=\left(\frac{8}{10}\right)^y=1000$$
What is the value of $\frac{1}{x}-\frac{1}{y}$?
I have already used that when $48$ divided by $10$ then it becomes $4.8$ and when $8$ divided by $10$ then it becomes $0.8$ by getting $10^x$ and $10^y$ to make it simpler but I cannot proceed further.

$$\left(\frac{48}{10}\right)^x=\left(\frac{8}{10}\right)^y=1000$$   
$$48^x = 10^{x+3} \; \; \; \& \; \; \; 8^y = 10^{y+3}$$   
$$x+3  =  x \log {48} \Rightarrow \; \; \; 1 + \frac{3}{x} = \log {48}$$   
$$y+3 = y \log {8} \Rightarrow \; \; \; 1 + \frac{3}{y} = \log {8}$$   
$$3 \times (\frac{1}{x} - \frac{1}{y}) = \log {48} - \log {8} = \frac{\log_2 {48}}{3}$$  
$$\frac{1}{x} - \frac{1}{y} = \frac{\log_2 {48}}{9} \approx 0.62055$$
