So, there you are, stuck on a desert island, you’ve played your eight pieces of music, burnt the Bible and Shakespeare, and now you’re kicking yourself for not bringing a calculator as your luxury item. An emergency has broken out and it’s vital for your to work out $3^{0.7}$ as accurately as possible. “Between 2 and 3” isn’t going to cut it (although it might help).

What do you do? Well, I have a few approaches.

Using logs to estimate

You twirl your “WWTMND?” wristband for a moment and recognise that the Mathematical Ninja would take logs as a way to estimate things. $0.7 \ln(3) \simeq 0.7 \times 1.10 = 0.770$, and $e^{0.770} \simeq e^{\ln (2) + 0.077} = 2 \times e^{0.077} \simeq 2 \times 1.077 = 2.154$. (Spoiler alert: the correct answer is 2.157669 - so that’s not at all bad for an estimate drawn in the sand with a stick).

But we can get closer.

Binomial expansion

The binomial expansion is a good place to start - as long as you can turn it into the form $(a+b)^n$ where you know $a^{0.7}$ and $b < a$.

I don’t know about you, but I haven’t memorised my powers of 0.7. I know that $1^{0.7} = 1$, but that’s not much help. The next one I know is $1024^{0.7} = 128$, which you could use in $(1024 - 1021)^{0.7}$. You could do it that way, but you get the quickest convergence when $b$ is much smaller than $a$.

A better idea (still Binomial)

If $x = 3^{0.7},$ then $x^{10} = 3^7 = 2187$, so if we can work out the 10th root of $2187$, we’re there! Initially, though, that seems to have made the problem worse - the expansion isn’t valid for $(1024 + 1163)^{0.1}$.

Luckily, though, we can get a better number to start from than $2^{10}$, because we worked it out in the first paragraph - $2.154$ is a much closer guess, and you can use the regular C2 binomial expansion to work that out as 2150.1. It’ll take a little while, but you have plenty of time on your saltwater-blistered hands.

Now the fun bit

We’re going to solve $2187 = (2.154^{10} + x)^{0.1} = 2.154^{10} + 0.1 \times 2.154^{9} x - 0.09 \times 2.154^{8} x^2 + …$.

Using the first few terms, you can construct a horrible-but-workable quadratic and get an answer for $x$. It’s a lot of work.


You might prefer to use the Newton-Raphson method (I know I would): if you have a decent guess ($x_0$) for a zero of a function, you can usually find a better guess by working out $x_0 - \frac{f(x_0)}{f’(x_0)}$.

In this case, our function is $f(x) = x^{10} - 2187$, and $f’(x)= 10x^9$.

So, if we picked $x_0 = 2.15$ as our start point, we’d get a next guess of roughly $2 + \frac{2110.5}{9816.3} \simeq 2.15779$, which is a definite improvement.

And back to the old beginning…

A third option is to do what the Mathematical Ninja would do, given plenty of time and use the expansions in greater detail:

$3^{0.7} = e^{0.7 \ln (3) }$, so we need a good approximation for $\ln(3)$. The expansion $\ln(1+x) = x - \frac 12 x^2 + \frac 13 x^3 - …$ is only good if $-1 < x < 1$ - however, we can split $\ln(3)$ up as, for example, $\ln\left(\frac{7}{4}\right) + \ln\left(\frac{12}{7}\right)$ (the two fractions are either side of the square root of three). It’s more efficient to break the product down into smaller bits still (e.g. $\frac 75 \times \frac 75 \times \frac {75}{49}$ or whatever you like. Using as many terms as you like in the expansion will give you progressively more accurate estimates of $\ln(3)$, which you can easily multiply by 0.7.

Similarly, you can then apply the expansion for the exponential $e^x = 1 + x + \frac 1{2!} x^2 + \frac 1{3!}x^3 + …$ to your answer. Doing this entirely in fractions will take quite a lot of beach (the numbers get big, quickly), but you can get a fractional answer as accurate as you like, and then use long division to get a decimal answer. If you want it, obviously.