# Simplify the logarithim expression

not actually homework, but things I need to know:

1: simplify the expression $$e^{a ln b}$$ (write it in a way that dosen't involve logarithims)

i want to prove the identities:
lnab = lna + lnb
lna^b = blna
$$\frac{d}{dx}lnx = \frac{1}{x}$$

i also want to derive the useful assumption
$$ln(1+x) ~ aprrox = x$$

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my working
1: $$e^{a ln b} = e^a * e^{ln b} = e^a * b$$
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

vela
Staff Emeritus
Homework Helper

my working
1: $$e^{a ln b} = e^a * e^{ln b} = e^a * b$$
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
That's not correct. You're thinking of the rule ea+b=eaeb, but you're not adding a and ln b.

is it not,

b^a then?

i know that is one of the rules, $$e^{a ln b} = b^a$$
i can't prove that it's correct but i remember that,
so the first one is just b^a?

is it not,

b^a then?

i know that is one of the rules, $$e^{a ln b} = b^a$$
i can't prove that it's correct but i remember that,
so the first one is just b^a?

Yes, because you already know that $$x^{yz}$$ can also be expressed as $$(x^y)^z$$ so you simply apply that to your problem and simplify the logarithm. The solutions to problems 2 and 3 can also be derived from these laws of exponents.

Yes, because you already know that $$x^{yz}$$ can also be expressed as $$(x^y)^z$$ so you simply apply that to your problem and simplify the logarithm. The solutions to problems 2 and 3 can also be derived from these laws of exponents.

can someone show me the proof for 3 please?
only just managed to figure out a way for problem 2

x=ln(a)
y=ln(b)

a=e^x
b=e^y

a*b = e^x *e^y
a*b = e^(x+y)
ln(a*b) = ln(e^x + e^y)
ln(a*b) = x + y
ln(a*b) = ln(a) + ln(y)

Mark44
Mentor

For #3, let y = ln a.

y = ln a <==> a = ey
Then ab = (ey)b = eby

Now, rewrite this exponential equation as a log equation, and the result you're looking for will fall out.