I was just fiddling around with a calculator and calculating powers of numbers really close to $1$ like $1.01,1.001\dots$ trying to find at what value they exceed $2$.

This got me thinking if I could find that value of the exponent which causes it to be greater than $2$, or at least approximate it to some extent. The $2$ is really just a random choice, please feel free to comment if a different number would be better.

Formally, I define $$f(n)=1+10^{-n} \space\space\space, n\in\mathbb{N}$$ I want to find the smallest integer $k$ such that for a given $n$, such that $$(f(n))^k>2$$

Running a computer program yielded the first few values easily.
1. For $n=1,f(n)=1.1,k=8$.
2. For $n=2,f(n)=1.01,k=70$
3. For $n=3,f(n)=1.001,k=694$
4. For $n=4, f(n)=1.0001,k=6932$
5. For $n=5, f(n)=1.00001, k=69316$

There seems something fishy about these values. The values are quite close to the multiples of 10 of the previous $k$. This leads me to believe there might exist a solution something like $6.93...\times 10^{n-1}$

Can a solution be found mathematically? I tried to open $(1+10^{-n})^k$ using binomial expansion, but I couldn't get anywhere from there.

Also, I couldn't decide on much relevant tags so if anyone could add any, that would be great.


Well, you can solve it using logarithms. If $$(1+10^{-n})^k>2$$ then, as logarithm is increasing, $$k \log(1+10^{-n})>\log(2)$$ and thus $$k >\frac{\log(2)}{\log(1+10^{-n})}.$$ Note that $\log(1+10^{-n})>0.$ As you want integer answer, we have $$k = \left\lceil\left(\frac{\log(2)}{\log(1+10^{-n})}\right)\right\rceil,$$ where $\left\lceil x \right\rceil$denotes the ceiling function.

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  • 2
    $\begingroup$ And, if $n$ is large, by Taylor $k \approx \left\lceil\left(10^n {\log(2)}\right)\right\rceil$ and $\log(2)=0.693147\cdots$. $\endgroup$ – Claude Leibovici Apr 6 '15 at 9:07
  • $\begingroup$ Perfect. $\log(2)=0.693$ so that's where that was coming from. $\endgroup$ – Arpan Apr 6 '15 at 9:09

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