Question

How do you solve (1-p)^n <= 0.4? By taking the log with base 1-p, it would give n <= something, but intuitively this is wrong, as there should be a minimum value for n, not a maximum. Also, 0<=p<=1.

Answer 1

take the log using some other base

Answer 2

like base 10

Answer 3

it doesn't actually matter which base you use

Answer 4

Why does it not matter? I need a bound for n (i.e. n >= something)

Answer 5

well take the log of both sides, you get log((1-p)^n) <= log(0.4)

Answer 6

which is n * log(1-p) <= log(0.4)

==> log(1-p) <= log(0.4) / n

Answer 7

you can do the transformation no matter the base

Answer 8

or if you're solving for n, then it's n <= log(0.4) / log(1-p)

Answer 9

and that is true no matter if it's base 10 or base 100 or base 16 or base whatever

Answer 10

"or if you're solving for n, then it's n <= log(0.4) / log(1-p)" -> right, that's what I get

Answer 11

but that doesn't seem right

Answer 12

because there should be a minimum not a maximum for n

Answer 13

explain?

Answer 14

well the larger the n, the smaller (1-p)^n will be

Answer 15

well that expression, log(0.4)/log(1-p), may be negative, in which case you can just multiply the whole thing through by -1 which switches the <= to >=

Answer 16

wait actually, it will never be negative. log(0.4) is negative, and so is log(1-p) since p is between 0 and 1. Therefore it is positive as a whole...?

Answer 17

you haven't given me any context to answer this, why should there be any constraint on n at all? what is n?

Answer 18

n is just an integer, p is a probability (between 0 and 1).
I want to find the smallest n such that (1-p)^n <= 0.4

Answer 19

so it's log(0.4) / log(1-p)

Answer 20

the <= vs >= is because the math doesn't "know" that you have a constraint on the values of P. for example if I'd said P was between 1 and infinity it might look right.

Answer 21

or rather, between -1 and negative infinity, i guess