So lately I came across this seemingly simple problem that I just can't get around. Solve this equation:
$x^2 = 2^x$
I cannot do this algebraically, while I refuse to believe it is impossible to solve this algebraically or at least without a graphic calculator (maybe use Newton's method or series approximations?).
I decided to simplify the problem:
$x = 2^x$
Again, I cannot find an exact way to solve this.
On
If you rearrange both sides and take logarithms:
$$e^{\ln(x)\cdot2} = 2^x$$
$$2\cdot\ln(x)=x\cdot\ln(2)$$
This type of equation, a relationship between a polynomial and a logarithm has been shown to have no solutions in standard analytic functions. As such, a new function - the Lambert $W$ function - was defined as the $W(z)=y$ such that $y=z\cdot e^z$.
In this case the problem can be solved by seeing that the equations have a symmetry in $x$ and $2$, hence $x=2$ is a solution.
On
There are two trivial solutions : $x=2$ and $x=4$
The non trivial solution can be computed with numerical methods or analyticaly, thanks to the Lambert W function : $$x^2=2^x=e^{x\ln(2)}$$ $$x e^{-\frac{\ln(2)}{2}x}=1$$ with $x=-\frac{2}{\ln(2)}X$ $$Xe^X=-\frac{\ln(2)}{2}$$ $$X=W\left(-\frac{\ln(2)}{2}\right)$$ $$x=-\frac{2}{\ln(2)}W\left(-\frac{\ln(2)}{2}\right)\simeq 0.766665$$
Hint: You have a "2" in one of the exponents and a "2" in one of the bases, and the rest of the symbols are "x". So what should you plug in for $x$ to make all the symbols equal?