<s>[INST] 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.
 [/INST] $$\left(\frac{48}{10}\right)^x=1000 \Longleftrightarrow $$
$$\left(\frac{24}{5}\right)^x=1000 \Longleftrightarrow x=\frac{\log_{10}(1000)}{\log_{10}\left(\frac{24}{5}\right)}$$

$$\left(\frac{8}{10}\right)^x=1000 \Longleftrightarrow $$
$$\left(\frac{4}{5}\right)^x=1000 \Longleftrightarrow y=\frac{\log_{10}(1000)}{\log_{10}\left(\frac{4}{5}\right)}$$
So:
$$\frac{1}{x}-\frac{1}{y}\Longrightarrow$$
$$\frac{1}{\frac{\log_{10}(1000)}{\log_{10}\left(\frac{24}{5}\right)}}-\frac{1}{\frac{\log_{10}(1000)}{\log_{10}\left(\frac{4}{5}\right)}}=$$
$$\frac{\log_{10}\left(\frac{24}{5}\right)}{\log_{10}(1000)}-\frac{\log_{10}\left(\frac{4}{5}\right)}{\log_{10}(1000)}=$$
$$\frac{\log_{10}\left(\frac{24}{5}\right)}{3}-\frac{\log_{10}\left(\frac{4}{5}\right)}{3}=$$
$$\frac{\log_{10}\left(\frac{24}{5}\right)-\log_{10}\left(\frac{4}{5}\right)}{3}=$$
$$\frac{\log_{10}(6)}{3}$$
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