Solving for $x$ in a log equation

Question

Given $(\log_3 x)^3 = 9 \log x$, solve for $x$.

Here is what I have so far: $$(\log_3 x)^3 = \frac{9\log_3 x}{\log_3 10}$$ $$let a = \log_3 x$$ $$a^3=\frac{9a}{\log_3 10}$$ $$a^3-\frac{9a}{\log_3 10} = 0$$ $$a(a^2-\frac{9}{log_3 10}$$ $$\log_3 x = 0, \log_3 x = \pm\sqrt{\frac{9}{\log_3 10}}$$

I solved the first part of that to give $x=1$, which I plugged back in and worked. But for the second part of the solution, $x$ could equal roughly $9.743156891$ or $0.1026361385$. Plugging them both into the original equation, I get the same on both sides. Yet, when I graphed it, the only solution, as far as I could see, is $1$.

I guess my real question is, are $9.74$ and $0.10$ actual solutions to the equation? Or are the extraneous for some reason?

Answer 1

All the three are solutions. The $x$ axis in your plot covers only a limited range.

Below is the plot of $$y = \left(\log_3(x) \right)^3 - 9 \log(x)$$

The plot was made using the software grapher on mac osx.

Answer 2

$x=1$ is one solution.

$$(\log_3(x))^3=9\log(x)$$ $$\left(\frac{\ln(x)}{\ln(3)}\right)^3=9\ln(x)$$ $$\frac{\ln(x)^3}{\ln(3)^3}=9\ln(x)$$ $$\frac{\ln(x)^2}{\ln(3)^3}=9$$ $$\ln(x)^2=9\ln(3)^3$$ for real $x$: $$\ln(x)=\pm 3\ln(3)^\frac{3}{2}$$ $$x=e^{\pm 3\ln(3)^\frac{3}{2}}$$