Let the principal/initial sum of money be P
Let the number of years be n
Interest rate is 7% → 7/100 or 0.07 of principal sum

Total sum including interest after n years

= P(1 + 7/100)ⁿ
= P(1.07)ⁿ

Total interest after n years
= Total sum including interest - principal sum
= P(1.07)ⁿ - P
= P( (1.07)ⁿ - 1)


What we need to solve for :

P( (1.07)ⁿ - 1) > 2P

(1.07)ⁿ - 1 > 2

(1.07)ⁿ > 3

n > log3(1.07)

n > ln3 / ln1.07

n > 16.238 (approximately)

Minimum number of years = 17

Ironically enough this is tagged as an E Maths question, so guess and check is the likely method to proceed.

Non-log way: (iterative)


Start from some value of n.

(You can test with calculator first to see where is a closer value of n to start.

So we don't have to guess too many times on the answer paper. Just show 1-2 iterations before and after the desired one will do )



When n = 15

1.07ⁿ = (1.07)^15 ≈ 2.759 (4s.f) < 3

when n = 16 ,

1.07ⁿ = (1.07)^16 ≈ 2.952 (4s.f) < 3

when n = 17,

1.07ⁿ = (1.07)^17 ≈ 3.159 (4s.f) > 3

when n = 18,

1.07ⁿ = (1.07)^18 ≈ 3.380 (4s.f) > 3


Minimum n we need = 17

Minimum number of years = 17

Nothing ironic though. Iteration way or log way will still be awarded full credit.

Students who learn both A and E Math can use a mix of techniques and should, unless specifically told to use a certain method in the question.

Yes this is an e math question. I haven't really started learning a math yet, cuz I'm sec 2, and I'm just trying to learn a bit of both a math and e math, b4 next year when I will be sec 3

Wow you're starting so early.

Very hardworking.

Hope the above working will help you

Haha thanks, I'm just thinking that maybe it will be benefit me and maybe give me a headstart.

Yes it will. Keep up the good work. It's a good use of the long holiday period

Okay, thank you