- #1

- 393

- 0

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