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Why is x^(1/x) = e^((1/x)lnx)? 
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#1
Dec2612, 07:57 PM

P: 9

or why is ?
thx in advance! 


#2
Dec2612, 08:00 PM

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Do you understand why
[tex]x=e^{ln(x)}[/tex] Think about the definition of the logarithm. 


#3
Dec2612, 08:08 PM

P: 9

I don't understand why.



#4
Dec2612, 08:16 PM

P: 454

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



#6
Dec2612, 08:24 PM

P: 9




#7
Dec2612, 08:26 PM

P: 9

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



#8
Dec2612, 08:26 PM

HW Helper
P: 2,264

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
Dec2612, 08:32 PM

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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
Dec2612, 08:41 PM

P: 9




#11
Dec2612, 08:45 PM

P: 9




#12
Dec2612, 08:48 PM

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[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
Dec2612, 08:51 PM

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


#14
Dec2612, 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]



#16
Dec2612, 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] 


#17
Dec2612, 09:24 PM

P: 9




#18
Dec2712, 02:47 PM

HW Helper
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P: 2,841

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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