x^2=2^x

**bigwave** · November 23rd 2011, 12:32 PM

$x^2=2^x$

from observation 2 answers of this are $2$ and $4$ but $x=-0.0767$ is the one which not sure how got.

if I take the log of each side $2log(x)=xlog(2)$ but don't see what you can do with this? thnx ahead for help

**Siron** · November 23rd 2011, 12:35 PM

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.

**alexmahone** · November 23rd 2011, 12:37 PM

Originally Posted by bigwave

$x^2=2^x$

from observation 2 answers of this are $2$ and $4$ but $x=-0.0767$ is the one which not sure how got.

if I take the log of each side $2log(x)=xlog(2)$ but don't see what you can do with this? thnx ahead for help

You need to use Newton's method or a computer program (like MATLAB).

**bigwave** · November 23rd 2011, 12:44 PM

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.

**ymar** · November 25th 2011, 04:38 AM

I remember trying to prove as a first-year student that the equation $a^x=x^a,$ where $a$ is a positive real number, has exactly two positive real solutions whenever $a\neq e$ . I don't remember solving it so I must have failed.

**JJacquelin** · November 25th 2011, 07:02 AM

Originally Posted by bigwave

$x^2=2^x$

from observation 2 answers of this are $2$ and $4$ but $x=-0.0767$ is the one which not sure how got.

if I take the log of each side $2log(x)=xlog(2)$ but don't see what you can do with this? thnx ahead for help

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).

**bigwave** · November 25th 2011, 08:58 AM

well for whatever it is worth here is a graph of $x^2=2^x$

**TKHunny** · November 25th 2011, 10:06 AM

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.

$x^{2} = 1 + x\cdot log(2)$ Produces x = 1.405 and x = -0.712

If we move up to the quadratic...

$x^{2} = 1 + x\cdot log(2) + \frac{1}{2}\cdot\left(x\cdot log(2)\right)^{2}$

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.

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