Author 
Topic 

ckitout
New Member
USA
2 Posts 
Posted  05/20/2008 : 06:14:26

I need to be able to show the steps and justifications for proving e^ln x = x 


skeeter
Advanced Member
USA
5634 Posts 
Posted  05/20/2008 : 08:36:24

let y = e^{ln(x)}
take the log of both sides ...
ln(y) = ln[e^{ln(x)}]
using the power property of logs ...
ln(y) = ln(x)*ln(e)
since ln(e) = 1 ...
ln(y) = ln(x)
so ... y = x.



newton
Junior Member
USA
4 Posts 
Posted  05/25/2008 : 16:23:10

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. Similarly 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 etc. 


Ultraglide
Advanced Member
Canada
299 Posts 
Posted  05/29/2008 : 17:41:39

Just a note  the domain of lnx is x>0, so you only get the portion of y=x where x>0. 


pitoten
Average Member
USA
8 Posts 
Posted  03/31/2010 : 02:08:29

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 


Hawkster93
New Member
Lebanon
1 Posts 
Posted  07/19/2013 : 16:35:10

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. :] 


someguy
Advanced Member
Canada
143 Posts 
Posted  07/28/2013 : 14:06:25

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.) 


nikkor180
New Member
USA
1 Posts 
Posted  12/18/2013 : 19:37:37

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].
Rich 
Edited by  nikkor180 on 12/18/2013 19:40:31 



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