How to calculate $3^{\sqrt{2}}$ with a simple calculator? How to calculate $3^{\sqrt{2}}$ with a simple calculator ?. What is a simple calculator here ?:


*

* It is a calculator which can only do the $4$ main calculus and radicals $\left(\,\sqrt{}\,\right)$.

*And it can only show up to seven digits.

*We want to calculate $3^{\sqrt{2}}$ with this calculator up to $6$ decimals.


In the question is written that the question has a nice solution don't find the answer just by using the calculator.
What to do here ?. I tried to divid it to a number, multiply, etc$\ldots$ But I can find a good way to calculate it.
 A: You may compute $\sqrt{2}$ and write it in binary $\sqrt{2}=(1.0110..)_2=(b_0.b_1b_2b_3...)_2$ This should be feasible on your calculator (just multiplying by 2 picking up and taking away the 1's that appears).  
Then $$3^{\sqrt{2}} = \prod_{k\geq 0} 3^{b_k/2^k} = 3 \times 3^{1/4} \times 3^{1/8} \times ...$$ 
And the factors $3^{1/2^{k+1}}=\sqrt{3^{1/2^{k}}}$, $k\geq 0$ may be computed recursively taking square-roots.
A: Notice $$2^{23}\sqrt{2} \approx 11863283.20303145\ldots
\quad\implies\quad \sqrt{2} \approx \frac{11863283}{2^{23}}
$$
On my casio calculator (fx3900Pv), I can compute
$\displaystyle\;3^{\frac{11863283}{2^{23}}}$ using following $57$ key strokes.
$$\begin{align}
3\;
& \sqrt{} \times 3 =\\
& \sqrt{} \sqrt{} \sqrt{} \times 3 = \sqrt{} \times 3 = \sqrt{} \times 3 = \sqrt{} \times 3 =\\
& \sqrt{} \sqrt{} \sqrt{} \times 3 =\\
& \sqrt{} \sqrt{} \sqrt{} \sqrt{} \sqrt{} \sqrt{} \times 3 =\\
& \sqrt{} \sqrt{} \times 3 =\\
& \sqrt{} \sqrt{} \times 3 = \sqrt{} \times 3 =\\
& \sqrt{} \sqrt{} \times 3 =\\
\end{align}
$$
My calculator gives me $4.728804262$. Compare this with the exact value $$3^{\sqrt{2}} \approx 4.7288043878374149478942833404160053668397164242548\ldots$$
this is accurate to $6$ decimal places.
A: $$y=3^\sqrt{2}$$
$$\log y=\sqrt{2}\log 3$$
the formula of $\log 3$ is
$$\log 3=\log 2+\frac{1}{1*3}+\frac{1}{2*3^2}+\frac{1}{3*3^3}+...$$
and the formula of $\log 2$ is
$$\log 2=\frac{1}{1*2}+\frac{1}{2*2^2}+\frac{1}{3*2^3}+...$$
then we can calculate the $\sqrt{2}\log 3$.
after that use the series of $e^x$ and plug $x=\sqrt{2}\log 3$ to find the value of $3^{\sqrt{2}}$
A: if your calculator can calculate integer roots and exponentiate by integers just plug N=10000 or more
$$
\lim_{N\to\infty}\left(\sqrt{2} \left(\sqrt[N]{3}-1\right)+1\right)^N=3^{\sqrt 2}
$$
this works in my 10+ years old calculator.
to obtain that limit:
$$
3^{\sqrt 2}=\left(3^{\frac{\sqrt 2}{N}}\right)^N\approx\left(1+\frac{\sqrt 2}{N}\log 3\right)^N=\left(1+\frac{\sqrt 2}{N}N\log 3^{\frac{1}{N}}\right)^N=\left(1+\sqrt 2\log{ (1+(3^{\frac{1}{N}}-1))}\right)^N\approx\big[\sqrt{2} (\sqrt[N]{3}-1)+1\big]^N
$$
