Question: Solve $$2(2^x-1)x^2+(2^{2^x}-2)x=2^{x+1}+2\tag1$$ For $x$
I don't know where to begin. By trial and error, I've found two possible solutions for $x$ as $x=0,x=1$. But I'm not too sure since there could be other solutions than those two.
I do see that the terms go in $x^2,x,x^0$, but I'm not too sure what to do about the numbers raised to the $x^{\text{th}}$ power.
EDIT: According to the comments, $x=0,x=1$ are not solutions. But the question still is, how would you solve this.
Alpha finds only one solution with $x \approx 1.16048$. Usually when you mix exponentials and polynomials you get something that can only be solved numerically. Are you sure you copied the problem correctly? With an algebra-precalculus tag I would expect there to be a magic solution.
To find the zero of the equation
$$f(x)=2(2^x-1)x^2+(2^{2^x}-2)x-2^{x+1}-2$$ inspection shows that there is one root between $x=1$ $(f(1)=-2)$ and $x=2$ $(f(2)=42)$ and numerical methods should be used. The simplest would be Newton method which, starting from a "reasonable" guess $x_0$ will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ In the present case $$f'(x)=2^{x+1} x^2 \log (2)+4 \left(2^x-1\right) x+2^{2^x}+2^{x+2^x} x \log ^2(2)-2^{x+1} \log (2)-2$$ So, the iterates would be $$\left( \begin{array}{cc} n & x_n \\ 0 & 1 \\ 1 & 1.2031772015445846237 \\ 2 & 1.1627628813498884797 \\ 3 & 1.1604898005315350440 \\ 4 & 1.1604829696992279420 \\ 5 & 1.1604829696377328539 \end{array} \right)$$
But, as Ross Millikan wrote, this problem seems quite surprizing for an algebra-precalculus assignment. Moreover, the question is : is this the only root ? Difficult to prove it (even if it is true).
