Find root of $x\cdot 2^x - 1$ function Is it possible to solve the equation: $x2^x - 1 = 0$ without using the function graph?
According to the function plot, its root is around 0.6.  
I need to get the numeric value of the function's root, preferably, using a method or algorithm that I can program (on the computer).
 A: A particularly elegant method to find approximations to the solution can be derived. Start with,
$$(1) \quad x \cdot 2^x-1=0$$
$$(2) \quad \Rightarrow x=2^{-x}$$
We can use the theory of dynamical systems to show that if we have,
$$(3) \quad x_{n+1}=2^{-x_n}$$
Then $x_n$ will converge to the solution of $(1)$ for suitable choice of $x_0$. Explicitly, the derivative of the right hand side of $(3)$ near the fixed point has an absolute value less than $1$. This means that the fixed point value of $x$ that we are seeking is “attractive”. This means we are free to write the solution as,
$$x=\lim_{n \to {\infty}} x_n=\lim_{n \to {\infty}} {}^n\left(\cfrac{1}{2}\right)= \left(\cfrac{1}{2} \right)^{\left(\cfrac{1}{2} \right)^{\left(\cfrac{1}{2} \right)^{\left(\cfrac{1}{2} \right)^{.^{.^{.}}}}}}$$
Which is referred to as infinite tetration.
A: Making it more general, let us consider the equation $$x^a \, b^x=c$$ As Travis commented, the solution(s) cannot express in terms of elementary function but they have solution in terms of Lambert function $W(z)$ which is such that $z=W(z)\,e^{W(z)}$.
The solution would be $$x=\frac{c }{\log (a)}W\left(\frac{\log (a)\, b^{\frac{1}{c}}}{c}\right)$$
If you do not want to use Lambert function, the only way is to use a numerical method such as Newton. But, if I may suggest, instead of considering $$f(x)=x^a \, b^x-c$$ it would be better to take logarithms of both sides of the initial equation and solve $$g(x)=a\log(x)+x\log(b)-\log(c)$$ which is much better conditioned.
To illustrate your specific case, let us try Newton $$g(x)=\log(x)+x \log(2)$$ $$g'(x)=\frac 1 x+\log(2)$$ $$g''(x)=-\frac 1 {x^2} <0$$ Newton iterates will be $$x_{n+1}=x_n-\frac{g(x_n)}{g'(x_n)}$$ which, in your case, will write $$x_{n+1}=\frac{x_n (1-\log (x_n))}{1+x_n \log (2)}$$ Being lazy, choosing $x_0=1$, the successive iterates will then be $$x_1=0.590616$$ $$x_2=0.639732$$ $$x_3=0.641185$$  $$x_4=0.641186$$ which is the solution for six significant figures.
Being less lazy, choosing $x_0=0.6$ as you noticed from the plot, the iterates would then be $$x_1=0.640231$$ $$x_2=0.641185$$ $$x_3=0.641186$$
You could notice that in the first case, we had a overshoot of the solution but this did not happen in the second case. The reason for that is that $g(1)>0$ while $g(0.6)<0$ and since $g''(x)<0$, by Darboux theorem, this had to happen. When you use Newton method for solving $f(x)=0$, it is always safe to start at a point $x_0$ such that $f(x_0)f''(x_0)>0$.
