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- October 20th 2007, 11:23 AM #1

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## 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?

- October 20th 2007, 12:19 PM #2

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- October 20th 2007, 02:40 PM #3

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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

- October 20th 2007, 04:23 PM #4
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?

- October 21st 2007, 12:52 AM #5

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- October 21st 2007, 08:18 AM #6
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 defineas the number such that .**e**

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

- October 22nd 2007, 06:34 AM #7

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- October 22nd 2007, 10:05 AM #8

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- October 22nd 2007, 10:48 AM #9

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