METHOD 1
If you know that the number is a perfect 5th power, you can use deductive methods. To start, calculate the 5th power of integers from 1 to 9:
1, 32, 243, 1024, 3125, 7776, 16807, 32768, 59049. Note that the last digit is also 1 to 9. Then take the number and divide its digits into groups of 5 starting from the right.
For example: 6436343 becomes 64|36343. The first group is 64 and the nearest 5th power lower than this is 25=32. So the first digit of the answer is 2, and there are two groups, so the 5th root contains only two digits. The second group ends in 3 and we know that this last digit points to 3 as being the second digit of the 5th root, making the number 23.
Another example: 69343957 becomes 693|43957. 693 is between 35 and 45 so there are 2 digits in the 5th root, the first of which is 3, and the second is 7 because the last digit of the number is 7, so the 5th root is 37.
This works for perfect 5th powers in numbers up to 10 digits long. If the number is longer than this a deductive method can still be used to get the first and last digits.
METHOD 2
For numbers which are not perfect 5th powers, you could use an iterative method and calculus as shown below.
Suppose you wanted the 5th root of a number n. If the 5th root is represented by x, then x5=n.
Therefore x5-n=0. Let f(x)=x5-n, then the derivative of f(x)=f'(x)=5x4.
Use the iterative equation:
xn+1=xn-(xn5-n)/(5xn4) where x1, x2, x3, etc., are successive iterations of approximations of the fifth root of n. We just need a value x0 to get started. So we need an approximate estimate of the fifth root of n.
For n up to 32, x0=1; between 32 and 243, use 2; between 243 and 1024, use 3; etc.
EXAMPLE 5th root of 100, x0=2.
x1=x0-(x05-100)/(5x04)=57/20=2.85,
x2=2.85-(2.855-100)/(5(2.854))=2.583...,
x3=2.5157..., x4=2.5118..., x5=2.51188..., ...
Eventually we get a value which doesn't change (after an arbitrary number of decimal places).
The 5th root of 100 is irrational so there's no termination of the decimal.
Here's a method using the Intermediate Value Theorem IVT. We take an approximation to the 5th root of 100, say, 2.5=5/2. Raise to the 5th power (multiply by itself five times)=3125/32=97.66 approx, close to 100. So we know the 5th power of 100 is greater than 5/2 but less than 3 (35=243), therefore we need to find a value midway between (the average of) 2.5 and 3, which is 2.75; 2.755>100, so next we try (2.5+2.75)/2=2.625; 2.625>100. Next, try (2.5+2.625)/2=2.5625; 2.56255>100; next, 2.53125; 2.531255>100. However, with each successive attempt we are getting closer to 100. The process is tedious, time-consuming and involves increasingly more difficult computations.
METHOD 3
A variation on this method is to note that 3125/32=97.65625, which is only 2.34375 short of 100.
So (5/2+h)5=3125/32+2.34375=100.
But we can expand the binomial: (5/2)5+5(5/2)4h+10(5/2)3h2+10(5/2)2h3+5(5/2)h4+h5=100.
5(5/2)4h+10(5/2)3h2+10(5/2)2h3+5(5/2)h4+h5=100-3125/32=2.34375.
We know h must be small so higher powers of h must be even smaller. We can ignore them for the time being and just consider 5(5/2)4h as approximately equal to 2.34375.
Therefore h=2.34375/(5(5/2)4)=2.34375/(3125/16)=0.012 so the revised approximation is 2.5+0.012=2.512.
Now the process can be repeated: 2.5125=100.02260825945..., therefore (2.512+h)5≈100.02260825945...-0.02260825945..., so, dividing by 5(2.5124), we get h=0.0005678 approx. and so we arrive at a better approximation: 2.51143... and the process is repeated yet again until we get the accuracy we need.
And so on. But this is the same method (METHOD 1) as the iterative one we used earlier. And this is probably similar to the method that Newton devised to give us the iterative formula.
Whatever method is used, you can see how tedious it is to compute by hand. Calculators contain algorithms that can perform these calculations very quickly and that's why we use them!
METHOD 4
You can also use logarithms. Take the log of the number and divide it by 5 then look up the antilog. This is not really solving by hand.
