# AS & A-Level Maths: Where the Power and Log Laws Come From

You will need to know the power laws and log laws for AS and A-level maths. The power laws and log laws may be old friends, but do you know where they come from? Imagine you have seven £2 coins. If you wanted to write down how much money you had, you could write £2+ £2+ £2+ £2+ £2+ £2+ £2, but you’d never dream of doing that (for one thing, you’d probably lose count); you’d almost automatically write 7×£2 or £2×7.

In a similar way, if you multiply things together repeatedly, you could write £500× 1.01× 1.01× 1.01 more efficiently as £500×1.01³. In both cases, the notation clarifies what’s going on and, critically, how much of it needs to happen.

There’s also some kind of a pecking order emerging: adding turns into multiplication, which turns into . . . let’s call it powering. (This, it turns out, is where the horrors of BODMAS come from; the rule is simply that you do the most general operation first, unless brackets or groupings tell you otherwise.)

## So, where do the power laws come from?

Imagine you need to work out (3×3×3×3)× (3×3×3), which you’d happily turn into 34× 3³. You could rewrite it as a list of seven 3s multiplied together or as 37. If you had a 3s in the first list and b 3s in the second, you’d turn them into a list of (a+b) 3s. More generally still, this leads to the rule (xa)(xb)= xa+b.

Divide both sides by xb, and you get xa= (xa+b)/xb; if you divide the product of (a+b) xs by the product of b xs, b of them cancel out, leaving you with the product of a xs. You can rewrite this as xm/xn = xmn – so the first couple of power laws hold water.

Another consequence of the first power law is that if you multiply a power by itself several times – as in (xa)3 – you can write it as (xa)(xa)(xa)= x3a . In general, multiplying xa by itself b times gives you a list of b (xa)s, which you can turn into xab.

Roots aren’t quite as easy to see, but think about the example from the preceding paragraph: (xa)(xa)(xa)= x3a , which you can rewrite as (xb/3)(xb/3)(xb/3) = xb; it’s now clear that xb/3 is the cube root of xb. You can extend this idea to see that x1/k is the kth root of x.

## And the log laws?

The remaining log laws can be worked out in pretty much the same way: each of the log laws corresponds exactly to a power law!