Calculating 6 decimal digits of $3^{\sqrt2}$ using a calculator. 
How can we calculate $3^{\sqrt2}$ to 6 decimal digits, using only a
  semi-basic calculator (Which has the square root too) and a pen and
  paper?

I asked this question from my teacher and he gave me a hint: "Compute the binary digits of $\sqrt2$." So what i did was this: Suppose $A=\sqrt2$ can be written in binary as: $A=a_0/a_1a_2a_3...$ and $a_i\in 0,1$. We have: $2^kA=a_0a_1a_2...a_k/a_{k+1}a_{k+2}...$ so $a_k$ is the first digit of the binary expansion of $[2^kA]$ and thus we can calculate it using the calculator. But i don't know how to go further from here.

Basically you're approximating $\sqrt{2}$ by a sum $1/2^{n_1} + 1/2^{n_2} + 
\ldots + 1/2^{n_k}$.  So 
$$ 3^\sqrt{2} \approx 3^{1/2^{n_1}} 3^{1/2^{n_2}} \ldots 3^{1/2^{n_k}}$$ and each
$3^{1/2^{m}}$ can be found by starting with $3$ and hitting the square root button $m$ times.
