# How to Find This without Calculator?

1. Aug 27, 2015

### basty

How do I find $2^{1.2}$ without using calculator?

2. Aug 27, 2015

### andrewkirk

Since it's equal to $2\sqrt[5]{2}$ and $\sqrt[5]{2}$ is irrational you won't be able to.

3. Aug 27, 2015

### Ssnow

You can use Taylor expansion in $1$ for a good approximation...

$2^{x} \approx 2+2\ln{2}(x-1)+(\ln{2})^{2}(x-1)^{2}+\cdots$

putting $x=\frac{12}{10}$...

4. Aug 27, 2015

### HallsofIvy

Or: if $y= 2^{1.2}$ then $log(y)= 1.2 log(2)$ and you can look up the logarithms in a logarithm table.

Or use a slide rule!

5. Aug 27, 2015

### jasonRF

Newton's method is a great way to find $\sqrt[5]{2}$.
https://en.wikipedia.org/wiki/Newton's_method
This converges very quickly, using 1 as my first guess, the 3rd iteration gave a fractional error of $3\times 10^{-5}$, the 4th gave fractional error of $1\times 10^{-9}$.

EDIT: brain freeze! This is not so calculator friendly after all - you need to compute 4th powers of decimals to make this work. You would want a calculator for this. Ooops!

jason

6. Aug 27, 2015

### jasonRF

This was nagging me on my drive home - so here is a hand calculation using Newtons method for $\sqrt[5]{2}$
. If $s_n$ is the $n^{th}$ estimate of the root, then I get the following for the $p^{th}$ root of $x$:
$$s_{n+1} = \frac{1}{p}\left[(p-1) s_n + \frac{x}{s_n^{p-1}} \right]$$
For us p=5, x=2, so,
$$s_{n+1} = \frac{1}{5}\left[4 s_n + \frac{2}{s_n^{4}} \right]$$

if I use $s_0 = 1$, then $s_1 = 1.2$. For $s_2$ I get
$$s_{2} = \frac{1}{5}\left[ 4.8 + \frac{2}{2.0736} \right] = \frac{1}{5}\left[ 4.8 + \frac{1}{1.0368} \right] \approx \frac{1}{5}\left[ 4.8 + 1 - 0.0368 \right] = \frac{5.7632}{5} \approx 1.15$$

So I get $2^{1.2} \approx 2.30$. Better than 1% accuracy in this case....

jason

7. Aug 27, 2015

### Staff: Mentor

I would use this approach:

21.2 = 2*20.2

$2^{0.2} = e^{0.2 \ln 2} \approx e^{0.2 \cdot 0.7} = e^{0.14} \approx 1+0.14$
2*1.14 = 2.28, better than 1% accuracy and the above calculation works without pen and paper if you know ln(2) ≈ 0.7.

Better: ln(2)=0.693 or 1% smaller than 0.7, and we can add the second order for the exponential (using 0.14 ≈ 1/7) => 20.2≈1+0.14-0.0014 + 1/2*1/72 = 1.1486 because 1/72 is about 1/50 and 1/2*1/50=0.01.
2*1.1486 = 2.2972 - an error of just 0.01%.