- #1

guifb99

- 10

- 1

_{b}(x

^{2}-1)≈log

_{b}(x)

This is an easy and useful way to calculate the log of any natural number, including primes, it won't ever give a precise result, obviously (because of the -1), but as "x

^{2}-1" will always have divisors smaller than "x", you can easily calculate the approximation by using the property that log

_{c}(a*b)=log

_{c}(a) + log

_{c}(b).

It could actually just be n

^{-1}log

_{b}(x

^{n}-k)≈log

_{b}(x) but it would be harder to calculate depending on which "k" or which "n" you use and won't be as useful for school purposes.

Obviously, since ½log

_{b}(x

^{2})=log

_{b}(x), the closer "k" is to 0, the more precise the result will be. And the closer "n" is to infinity, also the more precise the result will be, since the value of "k" will become less and less significant as "x

^{n}" gets bigger.

*It's good to notice that for a "k" bigger than one, it will fail miserably.