Question: Approximate this number to the nearest whole number without a calculator:

$2^{\sqrt{5}}$

I don't know how to do this problem. Can you help me? My answer key says $5$.

Also, how do I approximate $2^\pi$ and $3^e$? The answers are $9$ and $20$, respectively.

Answer: 
First note that 

$$2.2 = 11/5 <\sqrt{5} < 9/4 = 2.25.$$

You can prove this by showing that $(11/5)^2$ is less than $5$, and that $(9/4)^2$ is more than $5$.

Now we estimate $2^{2.25}$, by showing that it is less than $5$. To do this, we raise both sides of the inequality $2^{9/4} < 5$ to the power of $4$ to obtain an equivalent inequality, namely
$$2^9 < 5^4.$$
Since this inequality is easily checked, we've proved that $2^{2.25} < 5$.

Finally we estimate $2^{2.2}$, by showing that it is more than $4.5$. The inequality
$$2^{11/5} > 9/2$$
is equivalent to the inequality obtained by raising both sides to the power of $5$:
$$2^{11} > (9/2)^5, \quad \text{ equivalent in turn to } 2^{16}>9^5.$$
The last inequality can be checked by hand.

Together, these calculations show that $2^{\sqrt{5}}$ is more than $4.5$, but less than $5$.
From an algebraic point of view, let us consider $$A=2^{\sqrt x}=e^{\sqrt x \log(2)}$$ and use the first term of the Taylor expansion built at $x=4$ (which is the closest perfect square). So $$A=4+(x-4) \log (2)+O\left((x-4)^2\right)$$ So, for $x=5$, $$2^{\sqrt 5}\approx 4+\log(2)=4.69315$$ the "correct " value being $4.71111$; thus an error of $0.38$%. Now, take the closest integer.