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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 $78$, $14=78-8^2$, $17=2\times8+1$.

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$$ \sqrt [ 4 ]{ 136 } =3.4149...$$ The method approximates well, but it is working for only $n=2^k$ as I know.

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.

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 $78$, $14=78-8^2$, $17=2\times8+1$.

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 $78$, $14=78-8^2$, $17=2\times8+1$.

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$$ \sqrt [ 4 ]{ 136 } =3.4149...$$ The method approximates well, but it is working for only $n=2^k$ as I know.