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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

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    As an alternative to using the Lambert W function, you can solve it numerically, for example with the [Newton-Raphson method](https://en.wikipedia.org/wiki/Newton's_method). – 2017-02-14

2 Answers 2

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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.

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    What's the difference between $W_k(x)$ and $W(x)$ ? – 2017-02-14
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    @S.H.W Recall that $xe^x$ is not bijective from $(-\infty,0)$, thus, there are two such solutions when $x<0$ such that $$x=W(x)e^{W(x)}$$and thus we use $W_k(x)$ to discern which solution we are referencing. Here, we take $k=0$, which is the principal branch, but you could also note that $\ln(2)>0$, so we have no issues. Thought I'd use it since you brought up "how many solutions?" – 2017-02-14
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    Okay , in general can lambert function help us to find number of solutions ? – 2017-02-14
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    @S.H.W I can't say so. For example, it's not to hard to see that the solution to $$(-1)^x=x$$occurs at $x=-1$ by obvious checking, but the Lambert W function will not even tell you that. The most the Lambert W function will tell you is that there are infinite amount of complex solutions (depending on $k$), but not specifically anything about real solutions. You will need different arguments to find the amount of solutions. – 2017-02-14
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    Okay , Thank you a lot – 2017-02-14
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    No problem! :-) – 2017-02-14
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$2^x$ 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.