While, I was solving a problem of Chemistry [Solid State] when I encountered an equation like : $$a^3 = 3.612 \times 10^{-23} $$

Where, a is just a quantity [Actually, it is the length of a cubic unit cell.]

I was required to find the cube-root of $a^3$ as I had to use $a$ to find another quantity which was actually my answer.

I know that the cube-root of $a^3$ here can be found by using logarithm.

$$\begin{align} & \log{a^3} = \log{(3.612 \times 10^{-23})} \\ & 3\log{a} = \log{(36.12 \times 10^{-24})} \\ & 3\log{a} = \log{(36.12)} + \log{(10^{-24})} \\ & 3\log{a} = \log{(36.12)} - 24 \\ & \log{a} = \cfrac{\log{(36.12)} - 24}{3} \end{align} $$

By using logarithm table, I found $\log{(36.12)} = 1.557$

Now, just putting this : $$\begin{align} & \log{a} = \cfrac{1.557 - 24}{3} \\ & \\ & \log{a} = \cfrac{-22.443}{3} \end{align} $$

Now, taking antilog both sides, this becomes : $$a = antilog(-7.481) $$

Now, by using anti-log table, I got this as $$\boxed{a = 3.303 \times 10^{-8} } $$

Now, as you all might have noticed that this is a lot longer method and this has larger probability of a student doing a mistake in the calculations etc.

Is there a shorter or any proper method to find cube-root of a number like stated above?

  • $\begingroup$ To be honest, I really don't know that which tag will be the best for this question. If you have any ideas for proper tag for this, please feel free to add the tag by editing the post. Thanks! $\endgroup$ – Kushashwa Ravi Shrimali Jul 1 '14 at 14:18
  • $\begingroup$ From $a^3 = 36.12\cdot 10^{-24}$, you can go directly to $a = \sqrt[3]{36.12}\cdot 10^{-8}$, and getting the cube root of $36.12$ gives you a little less opportunity of screwing up. $\endgroup$ – Daniel Fischer Jul 1 '14 at 14:22
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    $\begingroup$ See p. 413 of Gilbert A. Christian and George Collar's 1899 book A New Arithmetic, Theoretical and Practical, for example. $\endgroup$ – Dave L. Renfro Jul 1 '14 at 14:36
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    $\begingroup$ Out of curiosity, I looked at the reference (books.google.com/books?id=z_gSAQAAMAAJ&pg=PA413) in the prior comment, and I think the logarithm method seems faster and less error-prone; certainly it is more direct. $\endgroup$ – David K Jul 1 '14 at 15:55
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    $\begingroup$ And a little humor on the same topic, from A Space Child's Mother Goose: "Little Jack Horner / Sits in a corner / Extracting cube roots to infinity ..." improbable.com/airchives/paperair/volume3/v3i5/mg35.htm $\endgroup$ – David K Jul 1 '14 at 15:59

Given a log/antilog table and no calculator, your approach is probably optimal. Without such a table, I think you are better off proceeding as:

$$\left ( 3.612 \cdot 10^{-23} \right )^{1/3} \\ = \left ( 36.12 \cdot 10^{-24} \right )^{1/3} \\ = 36.12^{1/3} \cdot 10^{-8}$$

Then do bisection to find $36.12^{1/3}$. Note that $3^3=27<36.12$ and $4^3=64>36.12$. (If you can't find a small interval quickly, just use $[1,10]$.) So now you keep checking midpoints until you get your desired level of precision: $3.5^3 > 36.12$, $3.25^3<36.12$, $3.375^3>36.12$, etc. If at some point the arithmetic gets to be too messy, you can round, at the expense of some convergence speed. For example, you could check $3.4$ instead of $3.375$, then $3.4^3>36.12$, so you could check $3.3$. Then $3.3^3<36.12$, so you have $a=(3.35 \pm 0.05) \cdot 10^{-8}$.

You can also do Newton's method, which in this case is $x_{k+1} = \frac{2 x_k}{3} + \frac{c}{3 x_k^2}$ where you want $c^{1/3}$, but this is hard to do by hand, because the fractions quickly become very awkward, even with rounding.

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    $\begingroup$ With bisection, the final result is an interval $[a,b]$ where you know the root is located. Given this interval, the best single guess for the root is $(a+b)/2$, which is within $(b-a)/2$ of every point in $[a,b]$. So in my example, my final interval was $[3.3,3.4]$. That is, I know, from my procedure, that $36.12^{1/3}$ is between $3.3$ and $3.4$. Therefore the root is within $0.05$ of $3.35$. $\endgroup$ – Ian Jul 2 '14 at 0:28

I think your method is probably the best way to solve for the cube root of a number.

I found some sites that offer quick solution towards finding a cube root, but they assume that your answer will be expressed as a whole number.

Link 1: https://www.math.hmc.edu/~benjamin/papers/Cubing.pdf

Link 2: http://www.careeranna.com/calculate-cube-roots-in-less-than-5-seconds/

  • $\begingroup$ @KushashwaRaviShrimali I'm actually very disappointed as well that there is no algorithm to find cube roots. Some sites only offer methods that only yield integer answers. I'll try and search online; I'm very intrigued to see how the algorithm will work. $\endgroup$ – Varun Iyer Jul 1 '14 at 14:37
  • $\begingroup$ @Varun Iyer: For an algorithm to find cube roots numerically, see my comments to the OP's question. Such algorithms are fairly common in really old textbooks, and I gave a link for one such textbook. $\endgroup$ – Dave L. Renfro Jul 1 '14 at 14:44

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