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.

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    $\begingroup$ Related: math.stackexchange.com/questions/493109/… $\endgroup$ – Zev Chonoles Aug 22 '16 at 6:32
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    $\begingroup$ What do you mean the 4 main calculus? Do you mean the four basic arithmetic operations? $\endgroup$ – Joel Reyes Noche Aug 22 '16 at 6:33
  • $\begingroup$ If the calculator can calculate any radical just use $\sqrt[b]{a}=a^{1/b}$. $\endgroup$ – Henrik supports the community Aug 22 '16 at 6:33
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    $\begingroup$ @Joel Reyes Noche Yes. $\endgroup$ – Taha Akbari Aug 22 '16 at 6:33
  • $\begingroup$ Do you mean that the only root it can calculate is the square root? $\endgroup$ – Joel Reyes Noche Aug 22 '16 at 6:34

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.


$0.414214 \approx 0.011010100000100111101_2$.

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        3
1.316074 3.948222
1.987013 5.961039
2.441524 7.324572
2.706395 8.119185
1.299239 3.897717
1.021484 3.064452
1.323087 3.969261
1.411489 4.234467
2.057782 6.173346

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

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.

  • $\begingroup$ truly wonderful answer. +1 perfect example of a great answer which goes in-depth and describes each step in very clear manner which can be understood with bare minimum mathematical knowledge. $\endgroup$ – Paramanand Singh Aug 23 '16 at 8:43
  • $\begingroup$ @ParamanandSingh: Thank you for your kind encouragement! I always try to make my answers as simple and self-contained as possible, but the more advanced answers, like many of my answers on logic, require basic prerequisite knowledge, and I can only direct readers to references. =) $\endgroup$ – user21820 Aug 23 '16 at 8:59
  • $\begingroup$ Yeah I agree that it is not always possible to explain everything down to lowest level, but if there is an effort in that direction, it definitely helps. Answers which use sophisticated tools are also good in the sense that they can make a reader excited about such powerful tools (who doesn't want access to powerful tools?), but still my preference is for simpler tools because they can cater to wider audience. $\endgroup$ – Paramanand Singh Aug 23 '16 at 9:04
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    $\begingroup$ @ParamanandSingh: Yup. And I like this kind of question too. Students often just press the calculator buttons and not care about how one could do the calculation efficiently by hand. But I believe learning how to do it ourselves (even if only once) is far more instructive than people think. $\endgroup$ – user21820 Aug 23 '16 at 11:23

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.


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.

  • $\begingroup$ You beat me to the answer. The only worry is that, if the calculator is itself limited to $7$ digits of precision, does the final calculation get the $6$-th decimal place correct? The binary expansion is exact, but I don't know if the errors accumulated in the final step can make the last digit incorrect. $\endgroup$ – user21820 Aug 22 '16 at 8:43
  • $\begingroup$ @user21820 I have the same concern but most calculator doesn't display all the digits it has. The question only state the calculator can display up to $7$ digits. It didn't say anything about how many digits the calculator is using for internal calculation. $\endgroup$ – achille hui Aug 22 '16 at 8:51
  • $\begingroup$ I just used a program to generate the results under assumption of always rounding at each step. It gets the correct answer in this case 'by luck'! I had that concern because most simple calculators I know really don't store any extra digit. I figure this out by subtracting what is shown. All scientific calculators I've tried this on will reveal 2 to 3 extra digits, but simple calculators just give zero. $\endgroup$ – user21820 Aug 22 '16 at 8:58

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

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 $3^{\sqrt{2}}$

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    $\begingroup$ This is correct with arbitrary precision, but if I remember right it is not practical since it takes too long to get just a few decimal places. Also, are you sure the error accumulation is small? To solve the first issue you can use identities, such as $e^x = (e^{x/2^m})^{2^m}$ for suitable $m$, but I recall that it makes the second issue worse. If you can provide the numerical computation, it will answer these concerns. $\endgroup$ – user21820 Aug 22 '16 at 9:12
  • $\begingroup$ $$\lfloor{3^{\sqrt{3}}\rfloor}=?$$ $\endgroup$ – studentsmath Dec 1 '20 at 17:18

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


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