Solve: $-2^x-x/2=0$ analytically $-2^x-x/2=0$
I guess that the (only?) solution to this equation is $x=-1$. However, I do not know how to find this solution analytically.
Any help or hint is very much appreciated.
Edit: Thanks! Yes, there is only one solution that is $x=-1$. Now I am trying to understand how to derive this solution analytically. I basically want to, if possible, transform this equation so that it ends up saying $x=-1$. Probably using logarithms might help, but I have not found the way.
 A: If you rewrite the equation as $2^x=-{1\over2}x$, you can see that you are trying to find where the exponential curve $y=2^x$ and the straight line $y=-{1\over2}x$ intersect.  Since the exponential curve is increasing (as $x$ increases) while the straight line, which has negative slope, is decreasing, the curves have at most one point of intersection.  Thus $(x,y)=(-1,{1\over2})$ is the sole point of intersection, i.e., $x=-1$ is the only solution to the original equation.
A: Hint: This is equivalent to $2^{x+1}+x=0$. See what happens when you take the derivative...
A: HINT.-$2^{x+1}$ is increasing from $0$ till $\infty$ and $y=-x$. There are consequently just one solution to $2^{x+1}=-x$ which is obviously $x=-1$
A: $$-2^x-\frac{x}{2}=0$$
1.) Solution as series
To solve the equation analytically as a series, you can use Lagrange inversion.
2.) No solution with elementary inverses
The equation is a zeroing equation of an elementary function.
Powers with irrational exponents are transcendental functions. Therefore the equation isn't an algebraic equation, it's a transcendental equation:
$$-e^{\ln(2)x}-\frac{1}{2}x=0$$
Because $-e^{\ln(2)x}-\frac{1}{2}x$ is a polynomial of two algebraically independent monomials, the function $\mathbb{R}\to\mathbb{R}, x\mapsto -e^{\ln(2)x}-\frac{1}{2}x$ seems to have no elementary inverse. Therefore we possibly cannot solve the equation by rearranging it only by applying elementary functions (elementary operations) which we can read from the equation.
3.) Solution in closed form with Lambert W
The equation is solvable by applying Lambert W. By some simple rearrangings, we can bring the equation into a form that is solvable by Lambert W:
$$-e^{\ln(2)x}-\frac{1}{2}x=0\ \ \ \ \ \ |\cdot(-1)$$
$$e^{\ln(2)x}+\frac{1}{2}x=0\ \ \ \ \ \ |-\frac{1}{2}x$$
$$e^{\ln(2)x}=-\frac{1}{2}x\ \ \ \ \ \ |\cdot e^{-\ln(2)x}$$
$$-\frac{1}{2}xe^{-\ln(2)x}=1\ \ \ \ \ \ |\cdot 2\ln(2)$$
$$-\ln(2)xe^{-\ln(2)x}=2\ln(2)\ \ \ \ \ \ |\ W$$
$$-\ln(2)x=W(2\ln(2))\ \ \ \ \ \ |\ /(-\ln(2))$$
The only real solution is:
$$x=-\frac{W(2\ln(2))}{\ln(2)}.$$
$$x=-1,$$
because, if $a$ is a constant, $\ln(a)e^{\ln(a)}=a\ln(a)$ and therefore $\ln(a)=W(a\ln(a)),\frac{W(a\ln(a))}{\ln(a)}=1$.
