a basic question on logarithms (textbook's solution of logarithmic equation)

Question

In a textbook's solution of a logarithmic equation, I see the following lines:

$$\frac{1}{\log_x{3x}}=\frac{1}{\log_x{9x}}$$ It follows from this that $log_x{9x}=\log_x{3x}$, $\color{red}{\log_x{3}=0}$, $3=x^0$, $3=1$. This false equality shows that the equation has no solution.

Why exactly $\log_x{3}=0$?

Answer 1

Because, using a log-law, $$ \log_x 9x=\log_x(3\cdot 3x)=\log_x3+\log_x(3x). $$ Now, cancel a $\log_x(3x)$.

Answer 2

Another way to think of it is this:
$$\log_{x}{(9x)} = \log_{x}{(3x)}$$ $$\log_{x}{(9x)} - \log_{x}{(3x)} = 0$$ Now use the property that $\log_{a}{(x)} - \log_{a}{(y)} =\log_{a}{(\frac{x}{y})}$ and cancel.