Solve $(2x-3)(2^x-4)+(x^2-3x+2)(2^x\cdot\log2)=0$ I want to find when this function equals to $0$:
$$f'(x)=(2x-3)(2^x-4)+(x^2-3x+2)(2^x\cdot\log2)=0$$
This is the derivative of $(x^2-3x+2)(2^x-4)$ and I set it equal to $0$ so I can get the "critical points" of the derivative. 
However it's not entirely clear to me how I would have to get the solutions. I developed it this way:
$$2x\cdot2^x-8x-3\cdot2^x+12+(x^2\cdot2^x\cdot\log2-3x\cdot2^x\cdot\log2+2^{x+1}\cdot\log2)=0$$
I understand this should be an exponential equation because there's $2^x$ in there although I'm not exactly sure what I need to group (maybe $2^x$). Any hints?
 A: Write $f(x)$ as,
$$f(x)=(x-1)(x-2)(2^x-4)$$
$$f'(x)=(x-2)(2^x-4)+(x-1)(2^x-4)+(x-1)(x-2)(2^x\log{2})$$
You can see $x=2$ as a solution.
The other solution has to be found numerically, as Dr.Sonhard stated.
A: Hint:$$x=2$$ is one solution, the other one can be obtained by a numerical method,
$$x\approx 1.3579195559851827513059400$$
A: Exponentials do not play nice with polynomials.  You might be able to see that $x=2$ is a root.  It is a zero of both $2^x-4$ and $x^2-3x+2$.  There is another one near $x=1.35792$ but Alpha does not give an exact form, so found it numerically.
A: Amazingly, we could find rather good approximations of the second solution approximating $f'(x)$ by a $[2,n]$ Padé approximant built at $x=2$ and get the appoximate solution at the price of a linear equation.
The second degree in numerator is required since $x=2$ is a known solution. So, we shall have, as an approximation
$$f'_{(n)}(x)=(x-2)\frac {a^{(n)}_0+a^{(n)}_1(x-2) } {1+\sum_{k=1}^n b_k (x-2)^k}$$ in which all coefficients are expressed in terms of higher derivatives. Moreover, for any $n$, we shall have $a^{(n)}_0=8 \log(2)$ and, then
$$x_{(n)}=2-\frac{a^{(n)}_0 } {a^{(n)}_1 }=2-\frac{8 \log(2) } {a^{(n)}_1 }$$ Moreover $a^{(n)}_1$ is itself a polynomial in $\log(2)$ with no constant term making the expression
$$x_{(n)}=2-\frac{8  } {\frac{a^{(n)}_1}{\log(2)} }=2-\frac 8 {c_{n}}$$ 
The first terms would be
$$\left(
\begin{array}{cc}
 n & c_n \\
 0 & {6 \left(2 +\log (2)\right)} \\
 1 & \frac{2 \left(108 +60 \log (2)+11 \log ^2(2)\right)}{9 
   (2+\log (2))} \\
 2 & \frac{2 \left(648 +396 \log (2)+86 \log ^2(2)+5 \log
   ^3(2)\right)}{ 108+60 \log (2)+11 \log ^2(2)} \\
 3 & \frac{2 \left(58320 +38880 \log (2)+9720 \log ^2(2)+840 \log
   ^3(2)-43 \log ^4(2)\right)}{15 \left(648+396 \log (2)+86 \log ^2(2)+5
   \log ^3(2)\right)}
\end{array}
\right)$$ and the decimal representation of the successive $x_{(n)}$ would then be
$$\left(
\begin{array}{cc}
 n & x_{(n)} \\
 0 &  1.50492 \\
 1 &  1.37399 \\
 2 &  1.35835 \\
 3 &  1.35785
\end{array}
\right)$$
