Lawn

[gene]

Why can we integrate ln?

[Firedragon52]

Why can we integrate ln?

The natural log of what? Do you mean, How can we integrate the ln of anything?

[matt grime]

Because, in its domain (strictly positive Reals) it is continuous, and hence Riemann integrable.

[gene]

Because, in its domain (strictly positive Reals) it is continuous, and hence Riemann integrable.

 

i understand the domain being continous but i'm no sure with "Riemann integrable".

[gene]

Ok. here's an additional sub-on question.

 

what is e^ln3 ?

[matt grime]

The domain is not the continuous thing, the function is. Riemann integration is the integration you are doing. Every continuous function on an interval [a,b] can be integrated. Sometimes that integral can be expressed as in general terms as a nice function like the thing you're integrating: eg integral of 2x is x^2 + k for some k. We say it is an elementary integral if we can find a function that differentiates to give the thing we are integrating (the integrand). Mostly that isn't true though.

 

exp and log are inverse functions between R and R+.

[123rock]

e^ln3=3

 

Proof:

x=b^y; y=logbx by definition of a logarithm. substitute b with e, and y with ln3, what do you get?

 

ln3=logex, logex=lnx, thus ln3=lnx, and thus x=3, and e^ln3=3

[gene]

rock. i sort of get the gist of it. but is if i were to apply your method to this eqn.

 

y=e^(0.5ln3) + 3e^(-0.5ln3)

 

would it work?

[123rock]

I can only simplify it to [e^(ln3)+3]/e^(0.5ln3)

 

I guess you have to find out what ln3 is and 1/2ln3.

 

Edit: oh nvm 4/e^(0.5ln3) since e^ln3 is 3.

 

e^0.5ln3+3(e^-0.5ln3)=[e^ln3/e^(0.5ln3)]+[e^0/e^(0.5ln3)]=[3/e^(0.5ln3)]+[1/e^(0.5ln3)]=4/e^(0.5ln3)

 

substitute e^0.5ln3 for e^(ln3-0.5ln3)=e^ln3/e^0.5ln3 and e^-0.5ln3=e^(0-0.5ln3)=e^0/e^0.5ln3

[gene]

I can only simplify it to [e^(ln3)+3]/e^(0.5ln3)

 

I guess you have to find out what ln3 is and 1/2ln3.

 

Edit: oh nvm 4/e^(0.5ln3) since e^ln3 is 3.

 

e^0.5ln3+3(e^-0.5ln3)=[e^ln3/e^(0.5ln3)]+[e^0/e^(0.5ln3)]=[3/e^(0.5ln3)]+[1/e^(0.5ln3)]=4/e^(0.5ln3)

 

substitute e^0.5ln3 for e^(ln3-0.5ln3)=e^ln3/e^0.5ln3 and e^-0.5ln3=e^(0-0.5ln3)=e^0/e^0.5ln3

 

Apparently, the answer is suppose to prove that it gets to 2 square root 3.

 

But i don't see how i can get such a perfect answer without trying to evaluate 2 square root 3.

[matt grime]

you do know that rlogx = log(x^r)? so that your thing to simplify is

 

[math]\sqrt{3} + \frac{3}{\sqrt{3}} = 2\sqrt{3}[/math]

[gene]

yup. i don't see what you mean... simplify where?

[matt grime]

post 8, you asked how to manipulate

 

e^(0.5log3)...

 

and got some long post off 123 rock, or rather a post containing a long string of numbers.

 

well, i'm not sure what they were getting at, but the answer, which you don't see how to get, is what I just wrote if you apply the log laws to the original expression.

[gene]

OH... !!! i see now.

Chey, i thought that e^(0.5ln3) = e^0.5 X e^(ln3)

 

got my laws mixed up.

So, it is actually, e^(ln3) to the power of 0.5. So that's how you get square root of 3!

 

ThANKS!