Simplify the logarithim expression

[vorcil]

not actually homework, but things I need to know:

1: simplify the expression e^{a ln b} (write it in a way that doesn'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

thank you in advance

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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]

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 e^a+b=e^ae^b, but you're not adding a and ln b.

 

[vorcil]

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?

 

[ianprime0509]

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.

 

[vorcil]

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]

For #3, let y = ln a.

y = ln a <==> a = e^y
Then a^b = (e^y)^b = e^by

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