Written by Colin+ in ninja maths.

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.

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.

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 $a$.

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.

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.

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.

## MathbloggingAll

How would you work out $3^{0.7}$? http://t.co/Ma0oertXoY

## MarkkuOpe

Matikkaninja laskee, nyt päässään 3^0,7. Hyvää kamaa, #UrMaa13

http://t.co/bvEtYPBvYj

## Josh Jordan

lg[3] is approximately 0.477, so we can write 3^0.7 = (10^0.477)^0.7 = 10^(0.477 · 0.7) = 10^0.3339. Now we just have to figure out what that is.

One way to convert 10^0.3339 to a number in decimal notation is to first convert the exponent to a number in the range [0.57, 0.72] (the so-called “log canoe” for fans of the major system). This helps because if x is a number in this range, then 10^x is just 10x – 2, an approximation which is accurate to two significant figures. One way to get the exponent number into this range is to add lg[2] (approximately 0.301). Note that doing this doubles the number, so we’ll have to divide our result by two at the end to correct for that.

Writing it out, we have 10^0.3339 = 1 · 10^0.3339 = (1/2 · 2) · 10^0.3339 ~ 1/2 · 10^0.301 · 10^0.3339 = 1/2 · 10^(0.301+0.3339) = 1/2 · 10^0.6349 ~ 1/2 · 10^(10 · 0.6349 – 2) = 1/2 * (6.349 – 2) = 1/2 * (4.34) ~ 2.17. Pretty close.