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.