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

[lucas7]

or why is

?
thx in advance!

 

[micromass]

Do you understand why

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

Think about the definition of the logarithm.

 

[lucas7]

I don't understand why.

 

[sweet springs]

or why is

?
thx in advance!

Try to make logarithm of the both sides. Regards.

 

[micromass]

I don't understand why.

So what is your definition of the logarithm?

 

[lucas7]

 

[lucas7]

I know it is a very basic definition but that is it for me.

 

[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.

 

[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 $b=e^x = e^{ln(b)}$.

 

[lucas7]

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.

 

[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 $b=e^x = e^{ln(b)}$.

I got it. But I fail to apply it for my case, when x has an exponential. Like $${x}^{1/x}={e}^{(1/x)lnx}$$

 

[micromass]

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 ?

 

[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

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

 

[lucas7]

$$x=ln({x}^{1/x})$$ so $${e}^{x}={x}^{1/x}$$ and $${e}^{ln({x}^{1/x})}={x}^{1/x}$$

 

[micromass]

$$x=ln({x}^{1/x})$$

Why is this true? This isn't correct for all x.

 

[lucas7]

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
$$ln({a}^{b})=c$$

$$ln({x}^{1/x}) = c$$

$${e}^{c}={x}^{1/x}$$

$${x}^{1/x}={e}^{ln{x}^{1/x}}$$

 

[lucas7]

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]

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

 

[pierce15]

$$e^{ln(x)}=x$$ because they are inverse functions. $$f(f^{-1}(x))=x$$ by definition