Thread: 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?

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 , then , so by the definition:



(this is taking of )

Similarly, if , then , so taking of this last expression we have:



RonL

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

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 is invertible if there exists a function such that:



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

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:





No calculus needed, just the definition of natural log.

RonL

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: .
Further, they define e as the number such that .

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

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

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 with initial value

RonL

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?

How about

e^(ln(x)) = N
Take the natural logs of both sides,
ln(x) *ln(e) = ln(N)
ln(x) = ln(N)
N = x