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

by lucas7
Tags: e1 or xlnx, x1 or x
 P: 9 or why is ? thx in advance!
 PF Patron Sci Advisor Thanks Emeritus P: 15,673 Do you understand why $$x=e^{ln(x)}$$ Think about the definition of the logarithm.
 P: 9 I don't understand why.
P: 449

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

 Quote by lucas7 or why is ? thx in advance!
Try to make logarithm of the both sides. Regards.
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 Quote by lucas7 I don't understand why.
So what is your definition of the logarithm?
 P: 9
 P: 9 I know it is a very basic definition but that is it for me.
 HW Helper P: 2,079 Logarithms can be used to define exponentiation as $u(x)^{v(X)}=e^{v(x) \log(u(x))}$ 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.
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 Quote by lucas7
OK, so you're saying that $x=ln(b)$ iff $e^x=b$.

So take an arbitrary b. Then we can of course write $ln(b)=ln(b)$. Define $x=ln(b)$. The "iff" above yields directly that $b=e^x = e^{ln(b)}$.
P: 9
 Quote by lurflurf Logarithms can be used to define exponentiation as $u(x)^{v(X)}=e^{v(x) \log(u(x))}$ 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 $$({\frac{n+3}{n+1}})^{n}$$ 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.
P: 9
 Quote by micromass OK, so you're saying that $x=ln(b)$ iff $e^x=b$. So take an arbitrary b. Then we can of course write $ln(b)=ln(b)$. Define $x=ln(b)$. The "iff" above yields directly that $b=e^x = e^{ln(b)}$.
I got it. But I fail to apply it for my case, when x has an exponential. Like $${x}^{1/x}={e}^{(1/x)lnx}$$
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 Quote by lucas7 I got it. But I fail to apply it for my case, when x has an exponential. Like $${x}^{1/x}={e}^{(1/x)lnx}$$
So you understand why $b=e^{ln(b)}$?? Good. Now apply it with $b=x^{1/x}$. Then you get

$$x^{1/x} = e^{ln\left(x^{1/x}\right)}$$

Do you agree with this? Now apply the rules of logarithms: what is $ln(a^b)=...$. Can you apply this identity with a=x and b=1/x ?
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P: 15,673
 Quote by micromass So you understand why $b=e^{ln(b)}$?? Good. Now apply it with $b=x^{1/x}$. Then you get $$x^{1/x} = e^{ln\left(x^{1/x}\right)}$$ Do you agree with this? Now apply the rules of logarithms: what is $ln(a^b)=...$. Can you apply this identity with a=x and b=1/x ?
Alternative proof: apply $e^{ln(b)}=b$ on x=b. Then

$$x=e^{ln(x)}$$

and thus

$$x^{1/x} = \left(e^{ln(x)}\right)^{1/x}$$
 P: 9 $$x=ln({x}^{1/x})$$ so $${e}^{x}={x}^{1/x}$$ and $${e}^{ln({x}^{1/x})}={x}^{1/x}$$
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 Quote by lucas7 $$x=ln({x}^{1/x})$$
 P: 9 Look: $$x = ln b$$ $${e}^{x}=b$$ $${e}^{ln b}=b$$ $$b={x}^{1/x}$$ $$y=ln{x}^{1/x}$$ $${e}^{y} = {x}^{1/x}$$ $${e}^{ln{x}^{1/x}}={x}^{1/x}$$ now a=x, b=1/x $$ln({a}^{b})=c$$ $$ln({x}^{1/x}) = c$$ $${e}^{c}={x}^{1/x}$$ $${x}^{1/x}={e}^{ln{x}^{1/x}}$$