<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] 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. 
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