Exponential equation problem $$2^{x-3}=\frac{1}{x}$$
So far, I've only managed to solve it graphically. I was wondering if there is any other method available?
I know about the $\ln$ method of course. 
 A: Since the left hand side is positive, $x$ must be positive.
The problem is equivalent to 
 $$x2^x=8$$
Notice that product of positive increasing function is increasing and hence it has a unique solution.
$$x2^x=2^3=2(2^2)$$
Hence $2$ is the unique solution.
A: A non-iterative approach is to use Lambert's W function. 
$$\begin{align} &2^{x-3}=\frac{1}{x}\\
\implies &x2^x=8 \\
\implies&xe^{x\ln 2}=8 \\
\implies&x\ln2\cdot e^{x\ln 2}=8\ln2 \\
\implies&x=\frac{W(8\ln2)}{\ln2} \\
\text{(Thanks}&\text{ to projectilemotion for the following:)} \\
\implies&x=\frac{W(4\ln4)}{\ln2} \text{ (Using identity: } W(x\ln x) = \ln x)\\
\implies&x=\frac{\ln4}{\ln2} = 2\\
\end{align}$$
A: The other obvious method is bisection -- find a value for which the left side is less than the right, a value for which it's greater, and then check the midpoint. Repeat. 
Of course, a first step is to guesstimate and try some integers; I did so and found $x = 2$ on my third try (after I'd checked $x = 1, 3$ becasue they were algebraically really easy). 
