- #1
azabak
- 32
- 0
Playing around with logarithms I found an interesting property that "log b^n(a^n) = log b(a)". Then I tried to find some kind of proof that this is right and not only a coincidence. Ι made a gereral formula for any value of both n's (α and β) so that "log b^β(a^α) = x". Therefore "a^α = b^(β*x)" ; "a = b^(β*x/α)" ; "log b(a) = β*x/α" ; "x = (α/β)*log b(a)". And therefore "log b^β(a^α) = (α/β)*log b(a)".