Point intersection of $y=2^x$ and $y=1/x$ Im trying find the point intersection of the functions.
$$y=2^x$$
$$y=1/x$$
I know that to solve this, the equations are equated.
$$2^x=1/x$$
$$2^x*x=1$$
but, how solve this??
 A: \begin{align}x2^x&=1\\xe^{x\ln(2)}&=1\\x\ln(2)e^{x\ln(2)}&=\ln(2)\\x\ln(2)&=W(\ln(2))\\x&=\frac{W(\ln(2))}{\ln(2)}\end{align}
See Lambert $W$-function, defined by $y = xe^x \iff x=W(y)$.
A: If you do not want to use Lambert function, consider that you are looking for the zero of function
$$f(x)=2^x\,x-1$$ the derivatives of which being
$$f'(x)=2^x (x \log (2)+1) \qquad \text{and} \qquad f''(x)=2^x \log (2) (x \log (2)+2)$$ The first deivative cancels at
$$x_*=-\frac{1}{\log (2)}\implies f(x_*)=-1-\frac{1}{e \log (2)}\approx 0.255 \implies f''(x_*)=\frac{\log (2)}{e} >0$$ So, you have one root which $> x_*$.
By inspection, $f(0)=-1$ and $f(1)=1$. So, let us start Newton method using $x_0=\frac 12$ to get the following iterates
$$\left(
\begin{array}{cc}
 n & x_n \\
 0 & 0.5000000000 \\
 1 & 0.6538027945 \\
 2 & 0.6412785001 \\
 3 & 0.6411857496 \\
 4 & 0.6411857445
\end{array}
\right)$$ which is the solution for ten significant figures.
You can notice that we have, at the first iteration, an overshoot of the solution : this is normal (by Darboux-Fourier theorem) since
$$f\left(\frac{1}{2}\right)=\frac{1}{\sqrt{2}}-1 <0 \qquad \text{and} \qquad f''\left(\frac{1}{2}\right)=\frac{\log ^2(2)}{\sqrt{2}}+2 \sqrt{2} \log (2) >0$$
A: The intersection point cannot be represented by elementary functions. In fact, it is x=W(ln(2))/ln(2). Here, W is the product log function.
