Originally posted by: blinky8225
for instance... f(bx) = log((b^2)x) = log(b) + log(bx) = log(b) + f(x)
So... f(bx) = log(b) + f(x)
How so?Originally posted by: StevenYoo
Originally posted by: blinky8225
for instance... f(bx) = log((b^2)x) = log(b) + log(bx) = log(b) + f(x)
So... f(bx) = log(b) + f(x)
OK now you're just fucking with us.
Originally posted by: Unitary
Assuming logb(x) means log base b we get,
f(bx)=logb(bx)
=logb(b)+logb(x) (Product to Sum for Logs)
=1+logb(x)
=1+f(x)
Originally posted by: Unitary
Assuming logb(x) means log base b we get,
f(bx)=logb(bx)
=logb(b)+logb(x) (Product to Sum for Logs)
=1+logb(x)
=1+f(x)
Eh his notation just confused me. I didn't realize b was meant to be the base. I thought he meant log(bx).Originally posted by: eLiu
Originally posted by: Unitary
Assuming logb(x) means log base b we get,
f(bx)=logb(bx)
=logb(b)+logb(x) (Product to Sum for Logs)
=1+logb(x)
=1+f(x)
ding ding ding i dunno wtf the rest of you were talking about.
Originally posted by: blinky8225
Eh his notation just confused me. I didn't realize b was meant to be the base. I thought he meant log(bx).Originally posted by: eLiu
Originally posted by: Unitary
Assuming logb(x) means log base b we get,
f(bx)=logb(bx)
=logb(b)+logb(x) (Product to Sum for Logs)
=1+logb(x)
=1+f(x)
ding ding ding i dunno wtf the rest of you were talking about.