What is the possible value of $2·((1+\tfrac{1}{100})^{100})$?
Google will give $2·((1+\tfrac{1}{100})^{100}) = 5.40962765884$.
How can I find the possible value without Google or a calculator?
How can I solve an equation like $(x^{100})$ or $(x^{20000})$?
Or like $(1.01^{100})$?
On
To compute $(1+\frac1{100})^{100}$ with an error $<10^{-4}$, expand: $$\left(1+\frac1{100}\right)^{100} =1+100\frac1{100}+{100\choose 2}\frac1{10000}+\ldots$$ until you notice that the summands fall below a suitable theshold (which should happen at ${100\choose 8}\frac1{10^{16}}$).
On
When you face situations such as $(x+1)^n$ where $x$ is small, you can use the binomial theorem for a few terms. In your case where $n=100$, the first terms are $$1+100 x+4950 x^2+161700 x^3+3921225 x^4+75287520 x^5$$ If you replace $x$ by $\frac{1}{100}$, you see that the successive terms contribute less and less. Uisng these terms, the value of the above expression is $2.70344$ and twice this number leads to $5.40688$ which is very close to the exact value $5.40963$.
But, going to very large values, remember what Trafalgar Law wrote.
On
I would go as follows:
Let $y=2(1+\frac{1}{100})^{100}$ so $\mathrm{ln}(y)=\mathrm{ln}(2)+100\mathrm{ln}(1.01)$. A first-order Taylor expansion around $\mathrm{ln}(1.01)$ is quite accurate and equal to $\mathrm{ln}(1)+1*(1.01-1)=0.01$. Therefore, $\mathrm{ln}(y)=\mathrm{ln}(2)+100*0.01=\mathrm{ln}(2)+1=\mathrm{ln}(2)+\mathrm{ln}(e)=\mathrm{ln}(2e)$. Finally, undo the transformation in both sides so $y=2e$.
Hint : Do you know about the limit that $$\lim_{x\to \infty}\left(1+\frac{1}{x}\right)^x = e$$