Why is x^(1/x) = e^((1/x)lnx)?

  • Thread starter lucas7
  • Start date
In summary, the conversation discusses the definition and properties of logarithms and exponentiation, how they are related, and how logarithms can be used to define exponentiation. The conversation also includes a proof using logarithms and exponents, and the concept of inverse functions.
  • #1
lucas7
9
0
or why is
2gxjj15.jpg
?
thx in advance!
 
Mathematics news on Phys.org
  • #2
Do you understand why

[tex]x=e^{ln(x)}[/tex]

Think about the definition of the logarithm.
 
  • #3
I don't understand why.
 
  • #4
lucas7 said:
or why is
2gxjj15.jpg
?
thx in advance!

Try to make logarithm of the both sides. Regards.
 
  • #5
lucas7 said:
I don't understand why.

So what is your definition of the logarithm?
 
  • #6
logaritmo1.jpg
 
  • #7
I know it is a very basic definition but that is it for me.
 
  • #8
Logarithms can be used to define exponentiation as

[itex]u(x)^{v(X)}=e^{v(x) \log(u(x))}[/itex]

Otherwise it can be proved as a theorem.

Note also that e^x is continuous is used in your example to justify moving the limit past e.
 
  • #9
lucas7 said:
logaritmo1.jpg

OK, so you're saying that [itex]x=ln(b)[/itex] iff [itex]e^x=b[/itex].

So take an arbitrary b. Then we can of course write [itex]ln(b)=ln(b)[/itex]. Define [itex]x=ln(b)[/itex]. The "iff" above yields directly that [itex]b=e^x = e^{ln(b)}[/itex].
 
  • #10
lurflurf said:
Logarithms can be used to define exponentiation as

[itex]u(x)^{v(X)}=e^{v(x) \log(u(x))}[/itex]

Otherwise it can be proved as a theorem.

Note also that e^x is continuous is used in your example to justify moving the limit past e.

Do you know where can I find this theorem? I do not understand why one is equal to the otherI am doing some limits exercises and I came up with [tex]({\frac{n+3}{n+1}})^{n}[/tex] limit when n->infinity. And to calculate this limit wolphram alpha show this "formula" but I just can't understand. I have some basic knowledge of logarithm.
 
  • #11
micromass said:
OK, so you're saying that [itex]x=ln(b)[/itex] iff [itex]e^x=b[/itex].

So take an arbitrary b. Then we can of course write [itex]ln(b)=ln(b)[/itex]. Define [itex]x=ln(b)[/itex]. The "iff" above yields directly that [itex]b=e^x = e^{ln(b)}[/itex].

I got it. But I fail to apply it for my case, when x has an exponential. Like [tex]{x}^{1/x}={e}^{(1/x)lnx}[/tex]
 
  • #12
lucas7 said:
I got it. But I fail to apply it for my case, when x has an exponential. Like [tex]{x}^{1/x}={e}^{(1/x)lnx}[/tex]

So you understand why [itex]b=e^{ln(b)}[/itex]?? Good. Now apply it with [itex]b=x^{1/x}[/itex]. Then you get

[tex]x^{1/x} = e^{ln\left(x^{1/x}\right)}[/tex]

Do you agree with this? Now apply the rules of logarithms: what is [itex]ln(a^b)=...[/itex]. Can you apply this identity with a=x and b=1/x ?
 
  • #13
micromass said:
So you understand why [itex]b=e^{ln(b)}[/itex]?? Good. Now apply it with [itex]b=x^{1/x}[/itex]. Then you get

[tex]x^{1/x} = e^{ln\left(x^{1/x}\right)}[/tex]

Do you agree with this? Now apply the rules of logarithms: what is [itex]ln(a^b)=...[/itex]. Can you apply this identity with a=x and b=1/x ?

Alternative proof: apply [itex]e^{ln(b)}=b[/itex] on x=b. Then

[tex]x=e^{ln(x)}[/tex]

and thus

[tex]x^{1/x} = \left(e^{ln(x)}\right)^{1/x}[/tex]
 
  • #14
[tex]x=ln({x}^{1/x})[/tex] so [tex]{e}^{x}={x}^{1/x}[/tex] and [tex]{e}^{ln({x}^{1/x})}={x}^{1/x}[/tex]
 
  • #15
lucas7 said:
[tex]x=ln({x}^{1/x})[/tex]

Why is this true? This isn't correct for all x.
 
  • #16
Look:

[tex]x = ln b[/tex]

[tex]{e}^{x}=b[/tex]

[tex]{e}^{ln b}=b[/tex]

[tex]b={x}^{1/x}[/tex]

[tex]y=ln{x}^{1/x}[/tex]

[tex]{e}^{y} = {x}^{1/x}[/tex]

[tex]{e}^{ln{x}^{1/x}}={x}^{1/x}[/tex]


now
a=x, b=1/x
[tex]ln({a}^{b})=c[/tex]

[tex]ln({x}^{1/x}) = c[/tex]

[tex]{e}^{c}={x}^{1/x}[/tex]

[tex]{x}^{1/x}={e}^{ln{x}^{1/x}}[/tex]
 
  • #17
micromass said:
Why is this true? This isn't correct for all x.

because x=ln(x1/x)... that was my first statement, should be correct for all x I think
 
  • #18
lucas7,

The Natural logarithm function is the INVERSE of the exponential function (for base, e). That is why the method works.

How do you start with a number, x, put into a function, and then put this function into another function, and the outcome be x? One function undoes the effect of the other function.

y=e^x
y is the value along the vertical number line, x is the value along the horizontal number line. What if you switch x and y?
x=e^y
What function is this? What is "y"? We call this the natural logarithm function,
y=ln(x), and this is the inverse of y=e^x.

Pick an x, any x. e^(ln(x))=x, and ln(e^x))=x
 
  • #19
[tex]e^{ln(x)}=x[/tex] because they are inverse functions. [tex]f(f^{-1}(x))=x[/tex] by definition
 

What is x^(1/x)?

x^(1/x) is an exponential expression where the base (x) is raised to the power of the reciprocal of itself (1/x). It can also be written as x^(1/x) = e^((1/x)lnx).

What is e?

e is a mathematical constant that is approximately equal to 2.71828. It is the base of the natural logarithm and is often used in exponential and logarithmic functions.

Why is x^(1/x) equal to e^((1/x)lnx)?

This is a property of logarithmic and exponential functions. When a logarithmic function and an exponential function have the same base, they cancel each other out and result in the original base. In this case, the base is x and the logarithm and exponential functions have the same base of e.

What is the significance of e^((1/x)lnx)?

e^((1/x)lnx) is a special form of the exponential function that is used to solve equations involving exponents and logarithms. It is also commonly used in calculus and other advanced mathematical concepts.

Can x^(1/x) be simplified further?

Yes, x^(1/x) can be simplified further to e. This is because as x approaches infinity, the expression approaches the value of e. However, for any finite value of x, x^(1/x) cannot be simplified any further.

Similar threads

Replies
1
Views
1K
  • General Math
Replies
7
Views
389
Replies
2
Views
987
Replies
2
Views
928
  • General Math
Replies
19
Views
850
  • General Math
2
Replies
66
Views
4K
  • General Math
Replies
2
Views
959
Replies
4
Views
801
Replies
1
Views
892
Replies
3
Views
575
Back
Top