How to solve $\log_2(8x) = 2^x$

Question

I am attempting to find the x intercepts for the two graphs, and would like to know how I would go about solving this equation;

$$\log_2(8x) = 2^x$$

If you could explain the steps in the process that would be great, as I will use it as a reference when solving harder variations of these questions.

Answer 1

$ \begin{align} &&\log_{2}(8x) &=2^x \\ &\Rightarrow &\frac{\ln(8x)}{\ln(2)}&=2^{x} \\ &\Rightarrow &\ln(8x) &= 2^x\ln(2) \\ &\Rightarrow &e^{\ln(8x)}&=e^{2^{x}\ln(2)} \\ &\Rightarrow &8x &=e^{\ln(2^{2^{x}})} \\ &\Rightarrow &8x &=2^{2^{x}} \\ &\Rightarrow &8 &= \frac{1}{x}4^{x} \end{align} $

Now this is where my intuition breaks down, introducing the Lambert W Function. Apparently For any real $a$ and $b$ , for $ \;a=\frac{1}{x}b^{x}\; $:

$ \begin{align} x=-\frac{W_{n}\left ( - \frac{\log(b)}{a} \right ) }{\log(b)} \end{align} \qquad \qquad $ Where $\log(b)\not=0\quad$ and $\quad abW_{n}\left ( - \frac{\log(b)}{a} \right ) \not =0 \quad $ , $n\in\mathbb{Z}$

Wolframe alpha computes $x$, using values 8 and 2 respectively, for $a$ and $b$:

$\begin{align} x&=2 \quad \text{or} \\ x&\approx 0.155 \end{align} $

Answer 2

The easy solution $x = 2$ you just have to be lucky and find (starting with $8x$ having to be a power of two where the exponent is itself a power of two narrows it down drastically).

I don't think you can find the other solution exactly. So you have to use a computer or calculator to solve it, either with or without drawing the graphs.

Usually, any solution to an equation like this is either "lucky" like the first one, or completely impossible to solve for, like the second. So this is one of those cases where just testing a few values of $x$ hoping that one of them solves the equation is actually one of the most viable ways to go.