Solve For X

Here's an interesting little problem. The first time I saw it, I immediately thought Lambert W function and indeed that will yield the two real solutions but there is a simpler way to find one of the solutions. The following is taken from the Mind Your Decisions YouTube channel and I'm including the content here to practise my LaTeX skills and help consolidate what I learnt from the video. Here it is:$$ \begin{align} 3^x &= x^9 \\ (3^{x})^{1/9x} &= (x^{9})^{1/9x} \\ 3^{1/9} &= x^{1/x} \\ ({3^3})^{1/27} &= x^{1/x} \\ 27^{1/27} &= x^{1/x} \\ x &=27 \end{align}$$Quite a neat little trick but there are two solutions so how to we find the other one? That's where the Lambert W function comes in handy.$$ \begin{align} 3^x &= x^9 \\ e^{\ln{3^x}} &= x^9 \\ e^{x \ln{3}} &= x^9 \\ 1 &= \frac{x^9}{e^{x \ln{3}}} \\ 1 &= x^9 e^{-x \ln{3}} \\   (1)^{1/9} &= (x^9 e^{-x \ln{3}})^{1/9} \\ 1 &=xe^{(-x \ln{3})/9} \\ \frac{-\ln{3}}{9} &= x  \Big (\frac{-\ln{3}}{9}\Big ) e^{(-x \ln{3})/9} \\  W \Big (\frac{-\ln{3}}{9} \Big)&= W \Big (x \Big (\frac{-\ln{3}}{9} \Big) e^{(-x \ln{3})/9} \Big ) \\  W \Big (\frac{-\ln{3}}{9} \Big ) &= -(x \ln{3})/9 \\  \frac{-9}{\ln{3} }W \Big (\frac{-\ln{3}}{9} \Big ) &=x  \\ x_1 &= \frac{-9}{\ln{3} }W_{-1} \Big (\frac{-\ln{3}}{9} \Big )  = 27 \\ x_2 &= \frac{-9}{\ln{3} }W_{0} \Big (\frac{-\ln{3}}{9} \Big ) \approx 1.15 \end{align}$$The syntax for evaluation of the above in WolframAlpha uses \( \textbf{productlog} \) instead of W and goes like this:

• -9/ln 3 * productlog(-1, -ln(3)/9)
• -9/ln 3 * productlog(0, -ln(3)/9)

Because \( -\ln(3)/9 \approx -0.122 \), this is why we get two solutions. See Figure 1.

Figure 1: graph of \(x=y \, e^y\) and \(x=-\ln(3)/9\)
with intercepts A and B