# Back in 1940...

That @solvemymaths is an excellent source of puzzles and whathaveyou:

Meanwhile, back in 1940 when everything was basically shit... pic.twitter.com/A5eKXOunFC

— Ed Southall (@solvemymaths) October 7, 2017

How *would* you find $\sqrt[3]{\frac{1-x^2}{x}}$ when $x=0.962$, using log tables or otherwise?

I would start by trying to make the numbers nicer: I note that $x=(1-0.038)$, which means we can rewrite the expression (using difference of two squares) as:

$\sqrt[3]{\frac{(1.962)(0.038)}{1-0.038}}$

Does that help? Well, maybe. It’s easy enough to estimate $\ln(1.962)$ - 1.962 is a small amount - 1.9% - short of 2, so $\ln(1.962)\approx \ln(2) - 0.019$.

Similarly, on the bottom, $\ln(0.962) \approx - 0.038$.

But how about $\ln(0.038)$? We could work out $\ln(38) - \ln(1000)$, which is simple enough. Thirty-eight is a little more than 36, and $\ln(38) \approx 2\ln(6) + \frac{1}{18}$. Meanwhile, $\ln(1000) = 3\ln(10)$.

So, the logarithm of everything under the square root is $\ln(1.962)+\ln(0.038)-\ln(0.962)$, which we estimate as $\ln(2) - 0.19 + 2\ln(6) + \frac{1}{18} - 3\ln(10) + 0.38$.

Ugly as decimals are, we can write $\frac{1}{18}$ as $0.0\dot{5}$ and combine everything: we get $\ln(72) - \ln(1000) + 0.13$ or $+0.14$, give or take.

Now, we need the cube root of that, so we’re going to divide everything by 3. The final two terms are easy enough: $\frac{1}{3} \ln(1000) = \ln(10)$ and we’ll call the decimal bit 0.045. How about $\ln(72)$? Well, $\ln(72) = \ln(8) + \ln(9)$, and a third of that is $\ln(2)$ plus a third of $2.196$, from memory, which is $0.732$.

OK: so we now have $\ln(2) + 0.732 - \ln(10) + 0.045$, which is $0.779 - \ln(5)$. $\ln(5) \approx 1.608$, so we end up with -0.828 as the logarithm of the answer.

What about $e^{-0.828}$? It’s $e^{-0.693} \times e^{-0.135}$, so a fair guess would be 13.5% smaller than a half, which is 0.43 or so.

A brief play with the calculator says that the answer is 0.426 - which, for a by-hand estimate, isn’t bad at all.