Solve the following equation Watch

blackdragonthegreat
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#1
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e^x = 3(x^2)
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RichE
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(Original post by blackdragonthegreat)
e^x = 3(x^2)
There are three roots between -1,0 and 0,1 and 3,4 but I doubt any of them can be found exactly and you're going to have to use numerical methods.
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DFranklin
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You can do something with the Lambert-W function if you really want...
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doddy3
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e^x = 3(x^2) and the answer is,.....

{x = -2*LambertW(-(1/6)*sqrt(3)*ln(e))/ln(e)}, {x = -2*LambertW((1/6)*sqrt(3)*ln(e))/ln(e)}

Wow...!
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DFranklin
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Can simplify that a bit I think... (ln e???) Not entirely sure where the 3rd root is disappearing though, but the LambertW func is multivalued for suitably negative x.
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blackdragonthegreat
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(Original post by doddy3)
e^x = 3(x^2) and the answer is,.....

{x = -2*LambertW(-(1/6)*sqrt(3)*ln(e))/ln(e)}, {x = -2*LambertW((1/6)*sqrt(3)*ln(e))/ln(e)}

Wow...!
i dont see the pont in multiplying by lne then also dividing by it?
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rkd
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(Original post by blackdragonthegreat)
i dont see the pont in multiplying by lne then also dividing by it?
Looks like the ln e is an argument to the Lambert-W function.
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EvenStevens
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e^x = 3(x^2)

\ln {e}^{x} = \ln 3(x^2)

 x = \ln 3 \cdot \ln x^2

x = \ln 3 \cdot 2 \ln x

Then...uh... I dunno <_<
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blackdragonthegreat
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(Original post by EvenStevens)
e^x = 3(x^2)

\ln {e}^{x} = \ln 3(x^2)

 x = \ln 3 \cdot \ln x^2

x = \ln 3 \cdot 2 \ln x

Then...uh... I dunno <_<
your manipulation of logarithms is wrong
ln(3.x^2) is
ln3 +2lnx
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EvenStevens
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Whoops.. Thanks for that >_<
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blackdragonthegreat
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(Original post by EvenStevens)
Whoops.. Thanks for that >_<
thats alright, no probs.
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TSRreader
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Isn't ln3x^2=ln3+2lnx?

Edit:nvrm about above
I used a graphing calculator for ln(3)= x-2ln(x)
According to it, the answer was 0.91.
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doddy3
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By plotting e^x and 3x^2 in the same X-Y axis, you get to see the estimates of the would be right answers! otherwise you ll have to rely on Lamb£t-W function! how did I get it?! Well I used Maple !
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DFranklin
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e^x = 3x^2,

e^{x/2} = \pm\sqrt{3} x,

e^{x/2} = \pm\sqrt{12} (x/2),

e^{-x/2} =\pm \frac{1}{\sqrt{12}} 1/(x/2)

(x/2) e^{-x/2} = \pm \frac{1}{\sqrt{12}}. Write y = -x/2, then

ye^y =\pm \frac{1}{\sqrt{12}}.

Then y = \text{ LambertW}(\pm \frac{1}{\sqrt{12}}) (as LambertW is defined as the inverse function of f(y) = ye^y).

x = -2y = -2\text{ LambertW}(\pm \frac{1}{\sqrt{12}})
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blackdragonthegreat
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#15
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(Original post by DFranklin)
e^x = 3x^2,

e^{x/2} = \pm\sqrt{3} x,

e^{x/2} = \pm\sqrt{12} (x/2),

e^{-x/2} =\pm \frac{1}{\sqrt{12}} 1/(x/2)

(x/2) e^{-x/2} = \pm \frac{1}{\sqrt{12}}. Write y = -x/2, then

ye^y =\pm \frac{1}{\sqrt{12}}.

Then y = \text{ LambertW}(\pm \frac{1}{\sqrt{12}}) (as LambertW is defined as the inverse function of f(y) = ye^y).

x = -2y = -2\text{ LambertW}(\pm \frac{1}{\sqrt{12}})
how does rt3x go to rt12(x/2)

never mind
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