# 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!
 Mentor P: 15,894 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.
Mentor
P: 15,894
 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: 2,111 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.
Mentor
P: 15,894
 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 $${x}^{1/x}={e}^{(1/x)lnx}$$
Mentor
P: 15,894
 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 ?
Mentor
P: 15,894
 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 $${e}^{ln({x}^{1/x})}={x}^{1/x}$$
Mentor
P: 15,894
 Quote by lucas7 $$x=ln({x}^{1/x})$$
Why is this true? This isn't correct for all 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 $$ln({a}^{b})=c$$ $$ln({x}^{1/x}) = c$$ $${e}^{c}={x}^{1/x}$$ $${x}^{1/x}={e}^{ln{x}^{1/x}}$$
P: 9
 Quote by micromass 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
 HW Helper P: 2,692 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