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**{7/3}****Member**- Registered: 2013-02-11
- Posts: 210

Help me prove this:

for some constant a[i cannot use the fact this is ln(x)]*Last edited by {7/3} (2013-05-22 18:43:33)*

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**bob bundy****Moderator**- Registered: 2010-06-20
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hi {7/3}

Are you wanting a proof from first principles? I usually start with the derivative of a^x, then e^x, then reverse these for the log.

If you may assume d(e^x)/dx = e^x then it will only take a few lines.

Bob

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**{7/3}****Member**- Registered: 2013-02-11
- Posts: 210

Proof from first principles will be better

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**bob bundy****Moderator**- Registered: 2010-06-20
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This is how I do powers and logs:

For the function

at (0,1), the derivative is:

Even though I don't know what that is, it will have a value; let's say k.

Now the derivative at other points

So all graphs in the family have the property that the gradient function at x is a^x times the gradient at (1,0)

In the family there will be one value of a for which k = 1

Call that one a = e

then

so

Now suppose

Taking logs base e for the first expression:

Differentiating wrt x

which means we now know the value of k ... and

so [still working on this last bit but I think I'll post before I lose it all]

No good. I seem to be stuck here because if k - ln a this becomes ln x and I was trying hard to avoid that. I seem to have gone too far and proved the log is base e. I'll come back to it later after a think.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bob bundy****Moderator**- Registered: 2010-06-20
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OR

I suspect there's a circular argument lurking here as power series probably depend on natural logs somewhere, but maybe it's ok.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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Hi Bob

The Taylor series uses the derivatives, so you are still using the derivative of log there. Unless we get the Taylor series of log in a different manner.

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**bob bundy****Moderator**- Registered: 2010-06-20
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OR

This was posted by yeyui on

http://forums.xkcd.com/viewtopic.php?f=17&t=36281

I have kept the post but edited the variables to suit your problem.

First some change of variable magic:

Now apply this magic to the function of interest

So this function has the property f(ab)=f(a)+f(b) which means that it is some logarithm.

Now "just" evaluate it at any particular point to show that it is the right one.

Bob

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**anonimnystefy****Real Member**- From: The Foundation
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I really like that proof. Really elegant.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

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**{7/3}****Member**- Registered: 2013-02-11
- Posts: 210

That was an awesome proof,but i need one more favor,if f'(x)=f(x) and f(0)=1 than f(x)=a^x for some constant a,how do i prove this?

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**bob bundy****Moderator**- Registered: 2010-06-20
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There are a whole set of functions that differentiate to give themselves. But they are all multiples of each other.

That is, if f'(x) = f(x) and g'(x) = g(x) then f = kg for some constant k.

Proof:

Consider

Show this is equal to zero which means that h(x) = constant.

I think that should enable you to do what you want.

Bob

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**{7/3}****Member**- Registered: 2013-02-11
- Posts: 210

Thanks

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