Why can we integrate ln?

Why can we integrate ln?

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

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

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".

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

 

what is e^ln3 ?

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+.

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

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?

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

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.

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]

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

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.

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!