Question

I’m sure ive had this in school before, but i cant remember what is this thing called as.

I have arbitrary number and i need to know how many times i can multiply it by 0.9 (or any other value 0-1) until theres less than x left from the original number.

in a loop format it would look like:

num = 4654;
mult = 0.9;
limit = 140;
count = 0;
while(num >= limit){
    num *= mult;
    count++;
}

But is this even possible to be done without a loop? something with logarithms?

Answer 1

Note that

num * (0.9)^k <= limit

is the inequality you wish to satisfy for some integer k, and you seek the smallest such k. Then

(0.9)^k <= limit / num

and

k * log(0.9) <= log(limit / num)

so that

k >= log(limit / num) / log(0.9)

where the inequality reverses because log(0.9) < 0. Thus, take the smallest integer k larger than log(limit / num) / log(0.9).

So, take the ceiling of log(limit / num) / log(0.9).

Of course, this generalizes by replacing 0.9 by r where r is your multiplier from (0, 1).