Question

Let\[L(x)=\int_{1}^{x}\frac{dt}{t}.\]How can I show that there is a unique number \(e\), such that\[L(e)=\int_{1}^{e}\frac{dt}{t}=1?\]

Answer 1

using integral properties again only?

Answer 2

That's correct.

Answer 3

hi again. i hope you got a chance to take a look at the last problem like this that you posted

Answer 4

and if you have any questions about it, let me know. this one is easier

Answer 5

I did, and also derived other rules such as \(L(a^r)=rL(a)\), etc. I got stuck on this one, however.

Answer 6

again the idea is that you know the derivative is
\(\frac{1}{x}\) and so your function is strictly increasing on \((1,\infty)\) therefore it is one to one, and therefore there exists SOME number \(e\) such that \(f(e)=1\)
oh and i guess you have to show that it is continuous as well,

Answer 7

L is a continuous function.
L(1) =0
L is increasing
and L goes to infity with x.
By the intermedaite value theorm, there is e such that L(e)=1
It is unique since L is striclty increasing

Answer 8

in your notation \(L(e)=1\) at this point no one is saying that \(e\) is the \(e\) your are familiar with, just some number \(e\)
what eliassaab said, although you might need to say something about why you know the function is continuous

Answer 9

That's it. :) Thank you, guys!