calculus help

[roboskier]

If f(x)=logbX, then f(bx)=


We have really only used natural log this year, last time I used logs was a couple years ago so I'm struggling with this question haha.

thanks.

 

[Howard]

10^f(x) = bx

I'm not that good at math, so this approach might be wrong.

Do you mean log base e when you say natural log? If so, change 10 to e.

EDIT: Damn, apparently I suck at math.

 

[StevenYoo]

this isn't a calc problem.

if f(x)=logbX, then f(bx) = log b(bx)

or

f(bx) = log (b^2)x

 

[blinky8225]

Unless an upper-case X is different than a lower-case x, StevenYoo's answer should be correct. Although if you use the properties of logs, you can get some interesting results.

for instance... f(bx) = log((b^2)x) = log(b) + log(bx) = log(b) + f(x)
So... f(bx) = log(b) + f(x)

 

[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.

 

[blinky8225]

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.

How so?

Are you saying that equation is wrong or are you trying to say something else?

That equation is correct. Try plugging numbers in a calculator.
I just though that it would be interesting to show that if you should choose to multiply the original parameter by a factor "b", the new result would be the original result added to log(b).

 

[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)

 

[roboskier]

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)

fuck that was simple... Thanks for the help!

 

[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.

 

[blinky8225]

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.

Eh his notation just confused me. I didn't realize b was meant to be the base. I thought he meant log(bx).

 

[roboskier]

Originally posted by: blinky8225

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.

Eh his notation just confused me. I didn't realize b was meant to be the base. I thought he meant log(bx).

haha no problem, wasn't sure how to type b as the base, I guess I should have just said it...