is there Any way to calculate nth Root without calculator?

Question

is there anyway to solve complicated roots without Calc e.g

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$S=108^{1/5}$

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i always get stuck on these type of calculations. Calculator restricted here in school level competitive exams.

Answer 1

In HS-level competitive exams, you need to know how to find roots which can be simplified to an exact expression.

Thus, you would be expected to be able to simplify expressions such as $$ 8^{-\frac{2}{3}}, \;\;\;343^{\frac{1}{3}}, \;\;\;1024^{\frac{1}{10}}, \;\;\;\left({\small{\frac{81}{16}}}\right)^{\frac{3}{4}} $$ but not $108^{\frac{1}{5}}$.

As far as approximations, for HS-level competitions, you would be expected to find simple bounds for roots that don't simplify exactly. Thus, for the expression $108^{\frac{1}{5}}$, you would be expected to be able bound it between $2$ and $3$, but you wouldn't be required to compute it by hand to the nearest thousandth.

If the competition assumes Calculus, then Newton's method could be used to get more accurate approximations, but that's more of a HW-type problem than a competition-type problem.

Edit:

In your later comments, you indicate that while calculators are not allowed, log tables _are_ allowed.

Based on that information, Michael Grant's comment outlines the standard log-based approach. Still, it's tedious work, and not a typical requirement for the solution of a HS-level competition-type problem.

Here's a link

$\qquad$ How to find a cube root of numbers?

which discusses the relative merits of a log-based approach.

Answer 2

If you have log tables, you use $\log \left(108^{(1/5)}\right)=\frac 15 \log (108)$ With linear interpolation, you should get full table accuracy. If you have a simple calculator, you can just compute better and better approximations. Maybe you know $2^5=32,3^5=243$, so $2.5$ seems a natural place to start. You can find $2.5^5=97.65625$ so you need the fifth power to be $\frac {108}{97.65625}\approx 1.106$ times larger. Now we can use $(1+x)^n \approx 1+nx$ for $x\ll 1$ so we want the root to be about $1.02$ times larger, giving $2.5 \cdot 1.02 \approx 2.55$. You can then find $2.55^5\approx 107.82$ and call that good enough. You can do all of this last by hand, but it will be a fair amount of work.

Answer 3

The result is irrational anyway, so here is an iterative scheme:

Note that $$ S = \frac45 S + \frac{108}{5 S^4} $$ is solved by $S = 108^{1/5}$. This can be used to build an iterative formula which can be handled by a simple calculator, using that $S^4$ can be calculated by 4 times multiplying: $$ S_{new} = \frac45 S_{old} + \frac{108}{5 S_{old}^4} $$ One can start even with a very rough guess, e.g. $S_{old} = 3$. Then $$ S_{new} = \frac{12}{5} + \frac{108}{5\cdot 81} = \frac83 \simeq 2.67 $$ which is already closer to $S = 108^{1/5} \simeq 2.55085$. We use $S_{new}$ as an $S_{old}$ in the iterative scheme again: $$ S_{new} = \frac{32}{15} + \frac{108}{5 \cdot (\frac{8}{3} )^4} = \frac{39329}{15360} \simeq 2.56048 $$ One more step gives $$ S_{new} = \frac45 \cdot \frac{39329}{15360} + \frac{108}{5 \cdot (\frac{39329}{15360})^4} \simeq 2.55092 $$ The relative deviation from the true value is $0.0027$ percent !

This can be iterated further to give better results.

The formal reason why this works is Banach's fixed point theorem.