Dear Uncle Colin

I have to work out the solution to $x^7 = 14$ to the nearest tenth without a calculator. Can you help?

- Getting One-Seventh Power, Evaluating Root

Hi, GOSPER, and thanks for your message!

In case the Ninja is listening – and the Ninja is always listening – I wouldn’t dream of using a calculator to find that the answer is about 1.458, so “1.5” is the answer.

Here’s how I’d do it.

Starting with easy things to work out, $1^7 = 1$ and $2^7 =128$, so we’re clearly between those. $(\sqrt{2})^7 = 8 \sqrt{2}$, which is between 11 and 11.5; $\left(\frac{1}{2}\right)^7 = \frac{2187}{128}$, which is a shade over 17.

That’s good, but not brilliant – we’ve narrowed the possible answers down to 1.4 and 1.5. The way to determine which? Try $1.45^7$.

One could do that longhand. I am not going to do that. I have better things to do with my life.

Instead, I am going to work out $\left( \frac{3}{2} - \frac{1}{20}\right)^7$ using the binomial expansion.

I’ll do a thing I swore I’d never do after discovering another way, and factor out $\left(\frac{3}{2}\right)^7$ to get $\left(\frac{3}{2}\right)^7 \left( 1 - \frac{1}{30}\right)^7$.

At this point, we stop and think for a moment. We can work out that 14 is the same as $\frac{1792}{128}$, so we want to compare the expansion part to $\frac{1792}{2187}$ – if it’s smaller than that, then 1.5 wins (because the solution is between 1.45 and 1.5); if it’s larger, then 1.4 is the answer. And, as an estimate, $\frac{1792}{2187}$ is about $\frac{18}{22}$, or 0.82.

So what’s the expansion? $\left(1 - t^7\right) = 1 - 7t + 21t^2 - 35t^3 + 35t^4 - 21t^5 + 7t^6 - t^7$. We’re not going to need all of those terms (or at least, I hope not.)

If we stick $t=\frac{1}{30}$ in there, the first two terms give $1 - \frac{7}{30}$, or $\frac{23}{30}$, which is a bit less than 0.8. Adding the third gives $\frac{23}{30} + \frac{21}{900}$, so we’re at $\frac{237}{300}$, which is 0.79$ – and the remaining terms are not going to be large enough to get past 0.8, let alone 0.82.

Where are we? That means that $1.45^7 < 14$, but $1.5 > 14$, so $14^{\frac{1}{7}} = 1.5$ to one decimal place.

Alternative approaches

Some things I considered:

  • Taking logs is a good strategy, if you know your logs. However, the root is quite close to 1.45 so you’d need to be pretty accurate.
  • Using the generalised binomial expansion of $\left(\frac{2187}{128} - x\right)^{\frac{1}{7}}$ with $x = \frac{395}{128}$ – this works, but the error analysis is non-trivial.
  • Working out $\left(\frac{16}{11}\right)^7$ – again, this will work, but requires a lot of slightly speculative effort (and you may as well just do $\left(\frac{29}{20}\right)^7$).
  • Using the binomial expansions of $(\sqrt{2} + t)^7$ and $(1.5 - u)^7$, setting both to be 14, solving for $t$ and $u$ (using the first two or three terms). There’s still a bit of error analysis here, but this is the most workable alternative for me.

If you, dear reader, have an approach I’ve missed, please do let me know!

In the meantime, GOSPER – I hope that helps!

- Uncle Colin