How to calculate $3^{\sqrt{2}}$ with a simple calculator?

Question

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.

Answer 1

I'll give you the general procedure for calculating $x^y$ for arbitrary reals $x,y$ such that $x > 0$ on a calculator with limited precision.

First note that we only need to care about the case when $0 < y < 1$.

$3^\sqrt{2} = 3 \times 3^{\sqrt{2}-1} \approx 3 \times 3^{0.414214}$.

Next express $y$ in binary. To do so repeat ( ×2 then ( -1 if the value is at least $1$ ) ), and the first digits form the binary expansion.

0.414214
0.828428
1.656856
1.313712
0.627424
1.254848
0.509696
1.019392
0.038784
0.077568
0.155136
0.310272
0.620544
1.241088
0.482176
0.964352
1.928704
1.857408
1.714816
1.429632
0.859264
1.718528

$0.414214 \approx 0.011010100000100111101_2$.

Finally compute $\prod_{k=1} x^{a_k/2^k}$ where $a_k$ is the $k$-th digit in the binary expansion of $y$. To do so efficiently, start with 1, and go in reverse order from the last to the first digit, at each step doing ( √ then ( × $x$ = if the current digit is a $1$ ) ).

[It is very helpful if your calculator also has a memory function, which you should use to store $x$ first so that you can just recall it at each multiplication.]

Here are all the intermediate results assuming your calculator rounds to $7$ significant digits on every operation.

1
1        3
1.732051
1.316074 3.948222
1.987013 5.961039
2.441524 7.324572
2.706395 8.119185
2.849418
1.688022
1.299239 3.897717
1.974264
1.405085
1.185363
1.088744
1.043429
1.021484 3.064452
1.750558
1.323087 3.969261
1.9923
1.411489 4.234467
2.057782 6.173346
2.484622
1.576268

$3^\sqrt{2} \approx 1.576268 \times 3 \approx 4.728804$.

As you can see it turns out the answer you get is correct to $7$ significant digits. It is accidental in this case, because even the final multiplication alone will force the result to be an exact multiple of $3$, and it so happens that the answer correct to $7$ digits is also a multiple of $3$. In general you expect at least the last digit to be inaccurate.

Answer 2

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.

Answer 3

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.

Answer 4

$$y=3^\sqrt{2}$$ $$\log y=\sqrt{2}\log 3$$

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 $3^{\sqrt{2}}$

Answer 5

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: $$ 3^{\sqrt 2}=\left(3^{\frac{\sqrt 2}{N}}\right)^N\approx\left(1+\frac{\sqrt 2}{N}\log 3\right)^N=\left(1+\frac{\sqrt 2}{N}N\log 3^{\frac{1}{N}}\right)^N=\left(1+\sqrt 2\log{ (1+(3^{\frac{1}{N}}-1))}\right)^N\approx\big[\sqrt{2} (\sqrt[N]{3}-1)+1\big]^N $$