Solving exponential equation using proper method

Question

Consider $\frac{1}{x} = 2^x$ . Find the number of solutions and also value of them.

My try : I tried to using lambert function but it was impossible to get the right answer for me. Because I'm not completely familiar with lambert function

Answer 1

To apply the Lambert W function, multiply both sides by $x\ln(2)$ to get

$$\ln(2)=x\ln(2)2^x$$

Since $2^x=e^{x\ln(2)}$,

$$\ln(2)=x\ln(2)e^{x\ln(2)}$$

Take the $W$ of both sides to get

$$x\ln(2)=W_k(\ln(2))$$

$$x=\frac{W_k(\ln(2))}{\ln(2)}\approx0.64118574450498598448620048$$

---

For the number of roots, notice that $x$ cannot be negative, since $2^x>0$. Over the interval $x\in(0,\infty)$, both functions are monotone and continuous, thus they have exactly one intersection. We can also discern that the intersection must occur at some $x\in(0,1)$ since both functions are monotone.

Answer 2

$2^x$ is continuous, greater than 0, and monotonically increasing.

$\frac 1x$ is not continuous at 0, but we only need to look to the right side of 0 since $\frac 1x <0$ when $x<0.$ It is monotonically decreasing.

These curves can cross at most one time.

Pick two values of $x$ one where $x$ is near $0,$ one where $x$ is large to show that they cross at least one time.