0

I’m trying to find a good approximation for $a-\log_{10}\left(10^a-1\right)$, with $a$ being a large positive real number.

I think the result should be in the order of magnitude of $10^{-a}$, but I can’t find a way to prove whether or not this is the case.

NoChance
  • 6,427
LSD_Sumus
  • 1
  • 2
    You could use $\ln(1-x)\approx -x$ for small $x$ – Andreas Lenz Apr 30 '23 at 12:16
  • 1
    See https://math.stackexchange.com/questions/977586/is-there-an-approximation-to-the-natural-log-function-at-large-values – NoChance Apr 30 '23 at 12:20
  • $f(x)-f(x-1)=f'(c)$, for $x-1<c<x$, and $f(t)=log_{10} x$, $f'(t)={1\over t .\ln(10)}$. For $x=10^a$, we get your result is approximately ${1\over 10^a .\ln(10)}$ – Thomas Apr 30 '23 at 15:05
  • @Thomas, thank you, that should do, do you happen to have any source for it? – LSD_Sumus Apr 30 '23 at 17:40

0 Answers0