from observation 2 answers of this are and but is the one which not sure how got.

if I take the log of each side but don't see what you can do with this? thnx ahead for help(Cool)

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- Nov 23rd 2011, 01:32 PMbigwavex^2=2^x

from observation 2 answers of this are and but is the one which not sure how got.

if I take the log of each side but don't see what you can do with this? thnx ahead for help(Cool) - Nov 23rd 2011, 01:35 PMSironRe: x^2=2^x
It's very hard to solve, 2 is indeed a solution by observation but I think the best way to determine the other solutions is by using the graph of the 2 functions.

- Nov 23rd 2011, 01:37 PMalexmahoneRe: x^2=2^x
- Nov 23rd 2011, 01:44 PMbigwaveRe: x^2=2^x
I did use Wolfram|Alpha but they used the

*product log function*? which I have never used. apparently one can only graph and estimate. or close in like Newton. - Nov 25th 2011, 05:38 AMymarRe: x^2=2^x
I remember trying to prove as a first-year student that the equation where is a positive real number, has exactly two positive real solutions whenever . I don't remember solving it so I must have failed.

- Nov 25th 2011, 08:02 AMJJacquelinRe: x^2=2^x
In general, this kind of equation is solved thanks to numerical computation.

In case of x²=2^x the roots are :

x = 2 (obvious)

x = -0.766664695962123..

In the general case, the roots cannot be expressed in form of a combination of a finit number of elementary functions. Analytically, the solutions are expressed thanks to a special function, the Lambert function W(X)

x = -2 W(X)/ln(2) where X= -ln(2)/2

Since the Lambert function is multi-valuated two values are obtained for x.

More generally the roots of equation a(x^b)=c^x, with a,b,c constant parameters, are :

x = -b W(X)/ln(c) where X = - ln(c) / (b (a^(1/b)))

Depending on the values of the parameters, there are two, or one, or no solution(s). - Nov 25th 2011, 09:58 AMbigwaveRe: x^2=2^x
well for whatever it is worth here is a graph of

- Nov 25th 2011, 11:06 AMTKHunnyRe: x^2=2^x
Deciding where to start looking can be a bit of a trick. In this case, reasonable results can be obtained by a quick linearization or maybe a quadratic, since x^2 is quadratic already.

Produces x = 1.405 and x = -0.712

If we move up to the quadratic...

We get x = 1.691 and x = -0.778

Moving up to the cubic is a WHOLE LOT MORE WORK, but does manage an approximation of all three solutions,

x = 1.871, x = -0.765, and x = 12.582

That third one isn't pretty, but it could lead to something.