Is there any simple method to calculate $\sqrt x$ without using logarithm Suppose that we don't know logarithm, then how we would able to calculate $\sqrt x$, where $x$ is a real number? More generally, is there any algorithm to calculate $\sqrt [ n ]{ x } $ without using logarithm? More simple techniques would be nice.
Here is a simple technique used to approximate square roots by Persian author Hassan be al-Hossein:
For example: $\sqrt {78}\approx 8\frac { 14 }{ 17 } $ , where $8$ is the nearest integer root of $78$, $14 = 78 - 8^2$, $17 = 2 \times 8 + 1$.
if $n=2^k$ we can use the method above. 
For example, for $k=2$ Lets calculate $\sqrt [ 4 ]{ 136 } $: $$\sqrt [ 4 ]{ 136 } =\sqrt { \sqrt { 136 }  } \approx \sqrt { 11\frac { 136-{ 11 }^{ 2 } }{ 11\times 2+1 }  } =\sqrt { 11\frac { 15 }{ 23 }  } \\ \sqrt { 11\frac { 15 }{ 23 }  } \approx 3\frac { 11\frac { 15 }{ 23 } -{ 3 }^{ 2 } }{ 3\times 2+1 } =\frac { 544 }{ 161 } =3.38\\$$ The exact result is$$ \sqrt [ 4 ]{ 136 } =3.4149\cdots$$ The method approximates well, but it is working for only $n=2^k$ as I know. 

The continued fraction method works like this:  Suppose $x = a^2 + b$, where $a = \lfloor \sqrt x \rfloor$. Then 
$$
\begin{align}
x &= \sqrt{a^2 + b}\\
x-a &=  \sqrt{a^2 + b} - a\\
\frac{1}{x-a} &= \frac{1}{\sqrt{a^2 + b} - a}\\
              &= \frac{1}{\sqrt{a^2 + b} - a}\frac{\sqrt{a^2 + b} + a}{\sqrt{a^2 + b} + a}\\
              &= \frac{\sqrt{a^2 + b} + a}{b}\\
              &= \frac{x + a}{b}
\end{align}
$$
Substitute, and get:
$$
\begin{align}
x &= a + (x-a)\\
  &= a + \frac{b}{a+x}\\
  %= a + \frac{b}{2a+\frac{b}{a+x}}\\
x &= a+\cfrac{b}{2a+\cfrac{b}{2a+\cfrac{b}{2a + \dots}}}
\end{align}
$$
Now, this is not a simple continued fraction.  However, if one divides the numerator and denominator of $\frac{b}{2a+x}$ by $b$, then one can eventually get a periodic simple continued fraction, and one approximates by the convergents.  The above expression turns out to be faster.
