1. ## exp and log

He,

I want to prove that exp(log(x))=log(exp(x))=x

Now the second one i diff. and get 1/exp(x) * exp(x) = 1
So log(exp(x))= x+ C, Calculation shows c = 0
But hopw do I get the first one

because when i diff. I get exp(log(x))/x , now I have to prove that exp(log(x))=x and so this doesn't help

What can I do now?

2. Originally Posted by kuntah
He,

I want to prove that exp(log(x))=log(exp(x))=x

There is nothing to prove, you just need the definition of the natural log.

If $y=\log(x)$, then $\exp(y)=x$, so by the definition:

$\exp(y)=\exp(\log(x))=x$

(this is taking $\exp$ of $y=\log(x)$)

Similarly, if $x=\log(y)$, then $\exp(x)=y$, so taking $\log$ of this last expression we have:

$\log(\exp(x))=\log(y)=x$

RonL

3. yes thanks but,

The problem I had to solve was to prove exp(log(x)=x
So I don't know actually yet, according to the exercise, that this is true
But I know the derivatives of exp and log, and I know there values in 0 and 1. So I use this:

Now if i take the derivative of exp(log(x)), I get exp(log(x))*1/x
This must be equal to 1 then. Then exp(log(x)) must be x, because then
exp(log(x))/x=1, but I can't use this, because this is what I have to prove
Can't I use something else?

Thanks

4. Originally Posted by kuntah
yes thanks but,

The problem I had to solve was to prove exp(log(x)=x
So I don't know actually yet, according to the exercise, that this is true
But I know the derivatives of exp and log, and I know there values in 0 and 1. So I use this:

Now if i take the derivative of exp(log(x)), I get exp(log(x))*1/x
This must be equal to 1 then. Then exp(log(x)) must be x, because then
exp(log(x))/x=1, but I can't use this, because this is what I have to prove
Can't I use something else?

Thanks
what does the relationship in question remind you of? it reminds me of the relationship between inverse functions. recall that, a function $f(x)$ is invertible if there exists a function $f^{-1}(x)$ such that:

$f \left( f^{-1}(x) \right) = f^{-1}(f(x)) = x$

so answering your question amounts to proving that the log is the inverse of the exponential function. can you do that?

5. Originally Posted by kuntah
yes thanks but,

The problem I had to solve was to prove exp(log(x)=x
So I don't know actually yet, according to the exercise, that this is true
But I know the derivatives of exp and log, and I know there values in 0 and 1. So I use this:

Now if i take the derivative of exp(log(x)), I get exp(log(x))*1/x
This must be equal to 1 then. Then exp(log(x)) must be x, because then
exp(log(x))/x=1, but I can't use this, because this is what I have to prove
Can't I use something else?

Thanks
But you do know:

"If , then ",

and:

"if , then "

which are all I used to show that:

$\exp(\log(x))=\log(\exp(x))=x$

No calculus needed, just the definition of natural log.

RonL

6. Originally Posted by kuntah
I want to prove that exp(log(x))=log(exp(x))=x
I would like to know more about the background of this question. There is a widely use calculus textbook by Salas and Hille that asks this very question. That text defines the logarithm function as: $Log(x) = \int\limits_1^x {\frac{{dt}}{t}} ,\quad x > 0$.
Further, they define e as the number such that $Log(e) = \int\limits_1^e {\frac{{dt}}{t}} = 1$.

From these, the authors develop the usual properties of both Log(x) and exp(x).

7. I defined exp as the solution to this differential equation

x'(t) = f(x(t)) with boundary values (0,1)

for the log I used the same definition

8. Originally Posted by kuntah
I defined exp as the solution to this differential equation

x'(t) = f(x(t)) with boundary values (0,1)

for the log I used the same definition
You mean $x'(t)=x(t)$ with initial value $x(0)=1$

RonL

9. Originally Posted by kuntah
He,

I want to prove that exp(log(x))=log(exp(x))=x

Now the second one i diff. and get 1/exp(x) * exp(x) = 1
So log(exp(x))= x+ C, Calculation shows c = 0
But hopw do I get the first one

because when i diff. I get exp(log(x))/x , now I have to prove that exp(log(x))=x and so this doesn't help

What can I do now?