$$\left(\frac{48}{10}\right)^x=\left(\frac{8}{10}\right)^y=1000$$
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.
On
From $4.8^x = 1000$ we get $4.8 = 1000^{1/x}$ and similarly for the $y$ term we get $0.8 = 1000^{1/y}$ by taking the $x$ and $y$-root of both sides respectively. Hence we get $$\frac{4.8}{0.8} = \frac{1000^{1/x}}{1000^{1/y}}$$
This yields $$6 = 1000^{1/x - 1/y}$$ so $$\bbox[10px, border: solid red 2px]{\frac{1}{x} - \frac{1}{y} = \log_{1000} 6}$$ by applying the $1000$-base logarithm of both sides of the equation.
On
$$\left(\frac{48}{10}\right)^x=1000 \Longleftrightarrow $$ $$\left(\frac{24}{5}\right)^x=1000 \Longleftrightarrow x=\frac{\log_{10}(1000)}{\log_{10}\left(\frac{24}{5}\right)}$$
$$\left(\frac{8}{10}\right)^x=1000 \Longleftrightarrow $$ $$\left(\frac{4}{5}\right)^x=1000 \Longleftrightarrow y=\frac{\log_{10}(1000)}{\log_{10}\left(\frac{4}{5}\right)}$$
So:
$$\frac{1}{x}-\frac{1}{y}\Longrightarrow$$ $$\frac{1}{\frac{\log_{10}(1000)}{\log_{10}\left(\frac{24}{5}\right)}}-\frac{1}{\frac{\log_{10}(1000)}{\log_{10}\left(\frac{4}{5}\right)}}=$$
$$\frac{\log_{10}\left(\frac{24}{5}\right)}{\log_{10}(1000)}-\frac{\log_{10}\left(\frac{4}{5}\right)}{\log_{10}(1000)}=$$
$$\frac{\log_{10}\left(\frac{24}{5}\right)}{3}-\frac{\log_{10}\left(\frac{4}{5}\right)}{3}=$$
$$\frac{\log_{10}\left(\frac{24}{5}\right)-\log_{10}\left(\frac{4}{5}\right)}{3}=$$
$$\frac{\log_{10}(6)}{3}$$
On
Let us make it more general considering $$a^x=b^y=c$$ Take logarithms (using any base) $$x \log(a)=y\log(b)=\log(c)$$ So $$x \log(a)=\log(c)\Longleftrightarrow x=\frac{\log(c)}{\log(a)}\Longleftrightarrow \frac 1x=\frac{\log(a)}{\log(c)}$$ $$y \log(b)=\log(c)\Longleftrightarrow y=\frac{\log(c)}{\log(b)}\Longleftrightarrow \frac 1y=\frac{\log(b)}{\log(c)}$$ $$\frac 1x-\frac 1y=\frac{\log(a)}{\log(c)}-\frac{\log(b)}{\log(c)}=\frac{\log(a)-\log(b)}{\log(c)}=\frac{\log(\frac ab)}{\log(c)}$$ which simplifies if you use $\log_c(.)$ as Zain Patel proposed.
On
$$\left(\frac{48}{10}\right)^x=\left(\frac{8}{10}\right)^y=1000$$
$$48^x = 10^{x+3} \; \; \; \& \; \; \; 8^y = 10^{y+3}$$
$$x+3 = x \log {48} \Rightarrow \; \; \; 1 + \frac{3}{x} = \log {48}$$
$$y+3 = y \log {8} \Rightarrow \; \; \; 1 + \frac{3}{y} = \log {8}$$
$$3 \times (\frac{1}{x} - \frac{1}{y}) = \log {48} - \log {8} = \frac{\log_2 {48}}{3}$$
$$\frac{1}{x} - \frac{1}{y} = \frac{\log_2 {48}}{9} \approx 0.62055$$
HINT :
$$4.8=1000^{\frac 1x},\ \ \ 0.8=1000^{\frac 1y}$$ So, $$1000^{\frac 1x-\frac 1y}=\frac{1000^{1/x}}{1000^{1/y}}=\frac{4.8}{0.8}$$