How do you prove a^{x}=e^{xlna}?

  • Thread starter Lipi
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In summary, The conversation discusses proving the equivalence of a^{x} and e^{xlna}, with the help of logarithms. The conversation also touches upon the definition of logarithms and their properties, such as eln a= a and ln(ea)= a. The conversation ends with a reminder to understand the properties of logarithms before studying Euler's number.
  • #1
Lipi
3
0
It's not a homework, I just want to understand the e. Just don't know enough yet to grasp it.
Can someone help me prove that this is eaqual:

a[tex]^{x}=e^{xlna}[/tex]

for example

10[tex]^{x}=e^{xln10}[/tex]

How do you come from 10[tex]^{x}[/tex] to [tex] e^{xln10}[/tex]

Thanx.
 
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  • #2
Take logarithms of both sides of the equation, and see what you get.
 
  • #3
You get zero on both sides, that's all fine. I allready know it's the same.

But if you have a function y=a[tex]^{x}[/tex],

how do you get to y=e[tex]^{xlna}[/tex] being the same thing?
 
  • #4
Lipi said:
You get zero on both sides...
I don't know what you mean. How come, zero? What is the logarithm of a^x ? And of e^(x ln a) ?
 
  • #5
Oh i just got it.:shy:

Just found the rule e[tex]^{lna}=a[/tex].

Thanx for the kick in the right direction, i appreciate it:smile:
 
  • #6
Do you understand WHY eln a= a? Quite simply because "ln x" is DEFINED as the inverse function to ex! That should have been one of the first things you learned about logarithms.


(And, just to shortcut possible objections, yes, it is quite possible to define [itex]ln(x)= \int_1^x 1/t dt[/itex] and then define y= ex to be the inverse of THAT function. But still, because they are inverse function eln a= a and ln(ea)= a.
 
  • #7
HallsofIvy said:
(And, just to shortcut possible objections, yes, it is quite possible to define [itex]ln(x)= \int_1^x 1/t dt[/itex] and then define y= ex to be the inverse of THAT function. But still, because they are inverse function eln a= a and ln(ea)= a.

Well, to go even further, the only reason to call it a logarithm is because the integral behaves exactly like one, thus the properties of logarithms in general should have been studied before anything about Euler's number. :D
 

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