- #1
vorcil
- 398
- 0
not actually homework, but things I need to know:
1: simplify the expression [tex] e^{a ln b} [/tex] (write it in a way that dosen't involve logarithims)
i want to prove the identities:
lnab = lna + lnb
lna^b = blna
[tex] \frac{d}{dx}lnx = \frac{1}{x} [/tex]
i also want to derive the useful assumption
[tex] ln(1+x) ~ aprrox = x [/tex]
thank you in advance
---------------------------------------------
my working
1: [tex] e^{a ln b} = e^a * e^{ln b} = e^a * b [/tex]
that's as far as i got, i think that's correct but I'm not sure, because i used a way that INVOLVES logarithims
2:proving lnab = lna + lnb
not quite sure how to prove this..
3:proving lna^b = blna
again not sure how to prove this, I've just remembered the rule and assumed it was right
4:proving d/dx lnx = 1/x
should i be using the taylor series to prove these?
i haven't officially been taught the taylor series yet, but i understand how it works
1: simplify the expression [tex] e^{a ln b} [/tex] (write it in a way that dosen't involve logarithims)
i want to prove the identities:
lnab = lna + lnb
lna^b = blna
[tex] \frac{d}{dx}lnx = \frac{1}{x} [/tex]
i also want to derive the useful assumption
[tex] ln(1+x) ~ aprrox = x [/tex]
thank you in advance
---------------------------------------------
my working
1: [tex] e^{a ln b} = e^a * e^{ln b} = e^a * b [/tex]
that's as far as i got, i think that's correct but I'm not sure, because i used a way that INVOLVES logarithims
2:proving lnab = lna + lnb
not quite sure how to prove this..
3:proving lna^b = blna
again not sure how to prove this, I've just remembered the rule and assumed it was right
4:proving d/dx lnx = 1/x
should i be using the taylor series to prove these?
i haven't officially been taught the taylor series yet, but i understand how it works