Exponential equation problem

Question

$$2^{x-3}=\frac{1}{x}$$

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.

Answer 1

A non-iterative approach is to use Lambert's W function.

$$\begin{align} &2^{x-3}=\frac{1}{x}\\ \implies &x2^x=8 \\ \implies&xe^{x\ln 2}=8 \\ \implies&x\ln2\cdot e^{x\ln 2}=8\ln2 \\ \implies&x=\frac{W(8\ln2)}{\ln2} \\ \text{(Thanks}&\text{ to projectilemotion for the following:)} \\ \implies&x=\frac{W(4\ln4)}{\ln2} \text{ (Using identity: } W(x\ln x) = \ln x)\\ \implies&x=\frac{\ln4}{\ln2} = 2\\ \end{align}$$

Answer 2

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).

Answer 3

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.