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.
On
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.
On
i can get $4 < 2^{\sqrt 5} < 5.$ use the fact that $\sqrt x$ is concave down so the tangent approximation $\sqrt 5 = \sqrt {4 + 1)} < \sqrt 4 + {1 \over 2 \sqrt 4} = {5 \over 4}.$ so we have $2 < \sqrt 5 < {5 \over 4}$ and $2 = {512 \over 256} < {625 \over 256}$ gives us $2^{1/4} < {5 \over 4}.$
so far we have, $2 < \sqrt 5 < {5 \over 4}$ and $2 ^{1/4} < {5 \over 4}$ exponentiating shows $4 = 2^2 < 2^\sqrt 5 < 2^{5 \over 4} = 2^2 2^{1 \over 4} < 4 {5 \over 4} = 5.$
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 $5$.