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

by lucas7
Tags: e1 or xlnx, x1 or x
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micromass
#2
Dec26-12, 08:00 PM
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Do you understand why

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

Think about the definition of the logarithm.
lucas7
#3
Dec26-12, 08:08 PM
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I don't understand why.

sweet springs
#4
Dec26-12, 08:16 PM
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Why is x^(1/x) = e^((1/x)lnx)?

Quote Quote by lucas7 View Post
or why is ?
thx in advance!
Try to make logarithm of the both sides. Regards.
micromass
#5
Dec26-12, 08:20 PM
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Quote Quote by lucas7 View Post
I don't understand why.
So what is your definition of the logarithm?
lucas7
#6
Dec26-12, 08:24 PM
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lucas7
#7
Dec26-12, 08:26 PM
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I know it is a very basic definition but that is it for me.
lurflurf
#8
Dec26-12, 08:26 PM
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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.
micromass
#9
Dec26-12, 08:32 PM
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Quote Quote by lucas7 View Post
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].
lucas7
#10
Dec26-12, 08:41 PM
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Quote Quote by lurflurf View Post
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.
lucas7
#11
Dec26-12, 08:45 PM
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Quote Quote by micromass View Post
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]
micromass
#12
Dec26-12, 08:48 PM
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Quote Quote by lucas7 View Post
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 ?
micromass
#13
Dec26-12, 08:51 PM
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Quote Quote by micromass View Post
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]
lucas7
#14
Dec26-12, 09:01 PM
P: 9
[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]
micromass
#15
Dec26-12, 09:09 PM
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Quote Quote by lucas7 View Post
[tex]x=ln({x}^{1/x})[/tex]
Why is this true? This isn't correct for all x.
lucas7
#16
Dec26-12, 09:22 PM
P: 9
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]
lucas7
#17
Dec26-12, 09:24 PM
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Quote Quote by micromass View Post
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
symbolipoint
#18
Dec27-12, 02:47 PM
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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


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