How do you solve $2^{x-1}=\frac{1}{x}$?

Question

$2^{x-1}=\frac{1}{x}$

Clearly by substitute $x=1$ we were able to solve this problem but how do we really solve it using calculus?

Answer 1

Symbolic solution, using the Lambert W function. Recall: $xe^x = y \Longleftrightarrow W(x) = y$.

$$ 2^{x-1} = \frac{1}{x} \\ x2^x=2 \\ xe^{x\log 2} = 2 \\ x\log 2 e^{x\log 2} = 2\log 2 \\ x\log 2 = W(2\log 2) \\ x = \frac{W(2\log 2)}{\log 2} $$

An easily proved identity is: $$ W(z\log z) = \log z, \qquad z>1 $$ (where we choose the only real branch of $W$ in that region).

Answer 2

For $x > 0$, $1/x$ is monotonically decreasing and $2^{x-1}$ is monotonically increasing, and as you point out they are equal when $x = 1$. So that's the only solution for $x>0$.

For $x < 0$, $1/x < 0 $ and $2^{x-1} > 0$, so they are never equal.

So $x = 1$ is the only solution.