/ Maths Query
Trying to help my daughter with her GCSE revision...
Is there an elegant way of calculating the annual percentage increase if a business increases in value from £250k to £325k over 6 years? I can only do it iteratively
Thanks in advance!
(325/250) to the power of (1/6)
edit: to explain
If r is the annual rate, 325 = 250 * r^6.
You solve for r by first dividing both sides by 250, then taking the 6th root of both sides, equivalent to raising to the power of 1/6. That gives the rate as an annual fractional increase, which is about 1.044, so that’s 4.4%.
Sixth root of 325/250, minus 1, times 100?
I'm intrigued. The question is simple but the maths seems A-level. It's like the questions used to explain the use of logarithms.
The opposite, what is the sales value of a 250 company that grows 4.4% annually for 6 years, is easier. And part of explaining geometric series.
I think you need to find out what element of maths she is trying to learn with this exercise. Simply showing how to solve won't help her, but this is the maths you need to reverse compound interest - which I think is an important skill.
I intuitively do this with logs, but your easy is faster and more elegant.
I’ve spent nearly 16 years thinking pretty much my only use to her would be to help her with maths, but I’m useless. GCSEs seem to have gone from piss-easy to nearly A-level this year, combined with me having done my O-levels over 30 years ago...
>The question is simple but the maths seems A-level.
This is something which has always been on the GCSE syllabus. The new spec can ask questions in more complex scenarios, but this is something which would have appeared on the old one too.
I don't know why anyone would try to solve this with logarithms, it's just a rearranged equation which would appear on a calculator paper. You'd only need logarithms to solve for the power (which is an A-Level topic).
As Philip says, just showing her the solution isn't in itself helpful. The key thing here is that she understands how compound interest works (i.e. where the power comes from) I train my students to set up the equation as if they were calculating the interest (as wintertree did) and then to rearrange. Setting up an equation is a really important skill on the new specification.
> I don't know why anyone would try to solve this with logarithms
If you had log and anti-log tabled to hand but no calculator it would make sense and would be the easiest way for sure. I’ve worked with someone who still had rote memorised log tables in their head from 40 years previously.
If I didn’t have a calculator, and didn’t have the necessary tables, I would partially solve it in my head to get r^6 = 1.3. Then as this isn’t much more than 1, I would take an estimate of r = 1 + (1.3 - 1) / 6 = 1.05 as my upper bound and half that amount above 1, r = 1.025, as my lower bound and do a few rounds of bisection to get the answer, with a lot of pen and paper multiplication.
Or perhaps I’d calculate the r^6 for 4 values between and including my bounds, plot them on a paper graph and draw a smooth curve, before reading off the answer.
Log tables definitely win in the absence of a calculator!
There’s presumably a pen and paper algorithm for computing fractional powers more directly, but I don’t know it. Edit: okay, I googled it - computers either use log/anti-log, bisection or Newton’s method. Two of which are embodied in my all time favourite WTF piece of code - https://en.m.wikipedia.org/wiki/Fast_inverse_square_root
> I’ve worked with someone who still had rote memorised log tables in their head from 40 years previously...
This makes sense. It also explains why it wouldn't have occurred to me; I'm not old enough ;-)
> I don't know why anyone would try to solve this with logarithms,
Because, without access to a calculator, but with access to log/antilog tables, you can compute a numerical result.
And, when we were being taught maths, those were the exam conditions.
That sixth root computation isn't something I can do in my head. Or on paper...
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