Solving $\frac{2x^2}{x-1}\cdot 2^x+8=0$ So recently, a friend of mine in grade $12$ got this question on her homework.
$$\left(\frac{2x^2}{x-1}\right)\left(2^x\right)+8=0$$
I tried rearranging this expression into $$\frac{-4x+4}{x^2}=2^x$$
and taking natural log on both sides and see if I can isolate the x, and in the end, I got
$$\ln\left(-4x+4\right)+2\ln\left(x\right)=x\ln(2).$$
However, I am not sure what I can do next. Is there something I am missing, or is this not the way I should approach this question?
I also searched how to do equations with an exponential function and polynomial on the Math Stack Exchange before, but they say you will need to employ Lambert's W function; I am not sure how you would rearrange this equation to use it.
 A: My answer is for symbolic solutions (closed-form solutions). It's probably beyond high school level. In high school, they instead learn solving equations also numerically.
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$$\frac{2x^2}{x-1}2^x+8=0$$
$$\frac{2x^2}{x-1}e^{\ln(2)x}+8=0$$
Because your equation contains two algebraically independent monomials ($x,e^{\ln(2)x}$), there is possibly no non-constant (partial) inverse that is an elementary function.
But if the equation can have solutions that are elementary numbers is a different mathematical problem.
For applying Lambert W, we want to try to rearrange your equation to the form
$$F(f(x)e^{f(x)})=F(y),$$
where $F$ and $f$ are unary functions in the complex numbers, and $y$ is independent of $x$.
$$\frac{2x^2}{x-1}2^x+8=0$$
$$\frac{-4x+4}{x^2}=2^x$$
$$\frac{-4x+4}{x^2}=e^{\ln(2)x}$$
$$\frac{x^2}{-4x+4}e^{\ln(2)x}=1$$
We see, the general equation isn't in a form that allows to apply Lambert W.
But it can be solved by Generalized Lambert W:
$$x=\frac{1}{\ln(2)}W\left(^{0,\ \ 0}_{\ln(2)};-4\ln(2)\right)=-\frac{1}{\ln(2)}W\left(^{-\ln(2)}_{0,\ \ 0};-\frac{1}{4\ln(2)}\right)$$
$-$ see the references below.
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[Mezö 2017] Mezö, I.: On the structure of the solution set of a generalized Euler-Lambert equation. J. Math. Anal. Appl. 455 (2017) (1) 538-553
[Mezö/Baricz 2017] Mezö, I.; Baricz, A.: On the generalization of the Lambert W function. Transact. Amer. Math. Soc. 369 (2017) (11) 7917–7934 (On the generalization of the Lambert W function with applications in theoretical physics. 2015)
[Castle 2018] Castle, P.: Taylor series for generalized Lambert W functions. 2018
A: Beside the generalized Lambert function, only numerical methods would solve the problem.
However, we can make decent approximation since, graphing, you notced that the solution is close to $\frac 34$.
Consider that you look for the zero of function
$$f(x)=\frac{2^{x+1} x^2}{1-x}-8$$ which is far from linearity. So, instead, consider $$g(x)=\log \left(\frac{2^{x+1} x^2}{1-x}\right)-\log(8)$$ Using Taylor series
$$\log \left(\frac{2^{x+1} x^2}{1-x}\right)=\log \left(\frac{9}{\sqrt[4]{2}}\right)+\left(x-\frac{3}{4}\right)
   \left(\frac{20}{3}+\log (2)\right)+\frac{56}{9}
   \left(x-\frac{3}{4}\right)^2+\frac{1856}{81}
   \left(x-\frac{3}{4}\right)^3+O\left(\left(x-\frac{3}{4}\right)^4\right)$$
Now, using series reversion
$$x=\frac{3}{4}+t-\frac{56 }{3 (20+\log (8))}t^2-\frac{ (73216+22272 \log
   (2))}{108 (20+\log (8))^2}t^3+O\left(t^4\right)$$ where $$t=\frac{3 \left(g(x)+\log(8)+\log \left(\frac{\sqrt[4]{2}}{9}\right)\right)}{20+\log (8)}\tag 1$$
Since we want $g(x)=0$, $t=\frac{3 \log \left(\frac{8192}{6561}\right)}{4 (20+\log (8))}$. Plugged in $(1)$, you have a (nasty) explicit expression for the estimate; converted to decimals, this gives
$x=\color{red}{0.7574926}57$ while the solution given by Newton method is
$x=\color{red}{0.757492663}$.
