Solve $(2x-3)(2^x-4)+(x^2-3x+2)(2^x\cdot\log2)=0$

Question

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?

Answer 1

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.

Answer 2

Hint:$$x=2$$ is one solution, the other one can be obtained by a numerical method, $$x\approx 1.3579195559851827513059400$$

Answer 3

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.

Answer 4

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

$$\left( \begin{array}{cc} n & x_{(n)} \\ 0 & 1.50492 \\ 1 & 1.37399 \\ 2 & 1.35835 \\ 3 & 1.35785 \end{array} \right)$$