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 Pre-Calculus and Calculus
 show e ^ln x = x not graphically
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New Member

2 Posts

Posted - 05/20/2008 :  06:14:26  Show Profile  Reply with Quote
I need to be able to show the steps and justifications for proving
e^ln x = x
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Advanced Member

5634 Posts

Posted - 05/20/2008 :  08:36:24  Show Profile  Reply with Quote
let y = eln(x)

take the log of both sides ...

ln(y) = ln[eln(x)]

using the power property of logs ...

ln(y) = ln(x)*ln(e)

since ln(e) = 1 ...

ln(y) = ln(x)

so ... y = x.
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Junior Member

4 Posts

Posted - 05/25/2008 :  16:23:10  Show Profile  Reply with Quote
Another approach:

Let's look at e^ln 10
ln 10 means the power of 'e' required to give 10.
That power is 2.302585..
So e^ln 10 = e^2.302585.. = 10.
e^ln 5 = e^1.609437.. = 5

e^ln x
ln x is the power of 'e' that will give 'x'
Call that power 'k'
e^ln x = e^k = x
e^ln x = x.

Look also at common logs:
10^log 100 = 10^2 = 100
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Advanced Member

299 Posts

Posted - 05/29/2008 :  17:41:39  Show Profile  Reply with Quote
Just a note - the domain of lnx is x>0, so you only get the portion of y=x where x>0.
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Average Member

8 Posts

Posted - 03/31/2010 :  02:08:29  Show Profile  Reply with Quote
How about this?

let a real number (a) = ln (x) (x>0)

putting this into exponential form, we get

e^a = x, and subbing (a) back in from above, we get

e^(ln x) = x!!!

Edited by - pitoten on 03/31/2010 02:12:22
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New Member

1 Posts

Posted - 07/19/2013 :  16:35:10  Show Profile  Reply with Quote
Another way to prove that e^ln x = x is as follows :
for each x>0, if we derive e^ln x we get:
d/dx ( e^ln x)=(1/x)*(e^ln x)
if we divide by e^ln x and multiplying by dx on each side (noting that e^ln x is "always" different from zero) we get:
d(e^ln x)/(e^ln x) = dx/x
Now we integrate both sides to get :
ln (e^ln x) = ln x (1), but the natural log function is a bijection so for each x>0, if ln u = ln v ---> u=v, we conclude from (1) that e^ln x = x.
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Advanced Member

143 Posts

Posted - 07/28/2013 :  14:06:25  Show Profile  Reply with Quote
Hawkster, you need a constant of integration. Then show it is zero by evaluating at x=0.

Also, you have to worry about how you know the derivatives of the natural log and exponential functions. If you justified one derivative and then used this result to show the other, you shouldn't then use the derivatives to justify this equality.
If you can show how to find the derivatives of both ln(x) and exp(x) without using the fact that they are functional inverses of one another, then yes, you can use your knowledge of their derivatives to verify that they are functional inverses of each other.

(The last comment is due to your use of the word prove. Since you talk about bijections, you should be advanced enough to see this is not being unnecessarily picky.)
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New Member

1 Posts

Posted - 12/18/2013 :  19:37:37  Show Profile  Reply with Quote
Would a most simple proof not simply employ the definition of a logarithm? That is,

Definition: Let b, x > 0 with b not=1. logb[x] = y if and only if b^y = x.

Therefore, the proof is as follows:

By convention, loge[x] = ln x. Hence employing the preceding definition literally, gives e^(ln x) = [x].


Edited by - nikkor180 on 12/18/2013 19:40:31
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