Explain why $\exp(-7 \log_{10} n)$ approximates $1/n^3$ so well

Question

I was graphing a few functions, and discovered that the graphs of $\exp(-7 \log_{10} n)$ approximates $1/n^3$ are almost the same. Can anyone explain why this is so? Is there a general result for this phenomenon?

Answer 1

If you mean $\log_{10}$, then

$$\exp (-7 \log_{10} n) = \exp \left(-7\frac{\ln n}{\ln 10}\right) = n^{-7/\ln 10}$$

Now $7/\ln 10 \approx 3.040$, so this should be fairly similar to $n^{-3}$, at least for $n$ not too far from $1$.