To solve the equation \( \log_3 x = 9 \log x^3 \), we can simplify and rearrange the equation step by step.
First, recognize that:
\[ \log x^3 = 3 \log x \]
So we can rewrite the equation:
\[ \log_3 x = 9 \cdot 3 \log x \]
which simplifies to:
\[ \log_3 x = 27 \log x \]
Next, we can use the change of base formula for logarithms, which states that \( \log_a b = \frac{\log b}{\log a} \). Applying this to \( \log_3 x \), we get:
\[ \log_3 x = \frac{\log x}{\log 3} \]
Substituting back into our equation gives:
\[ \frac{\log x}{\log 3} = 27 \log x \]
Now, if \( \log x \neq 0 \) (which we assume as \( x > 1 \)), we can divide both sides by \( \log x \):
\[ \frac{1}{\log 3} = 27 \]
From this, we can rearrange and solve for \( \log 3 \):
\[ \log 3 = \frac{1}{27} \]
This equality indicates actual numbers, however, it suggests there might have been an error in handling or an inconsistency as no \( x \) satisfies this statement traditionally with the logarithm definitions.
To summarize, we re-check:
- \( \log_3 x = \frac{\log x}{\log 3} \rightarrow \frac{1}{\log 3} = 27\) leads to an inconsistency with standard logarithm definitions.
- Must assume: \( \log x = 0\) giving \( x = 1\) is matched leading also to check back.
Thus ensuring back we can cross-validate if \(x = 1\) satisfies original formed:
Indeed both sides yield:
LHS: \[ \log_3 1 = 0 \] RHS: \[ 9 \cdot 3 \cdot 0 = 0 \]
Thus, the only solution for the equation \( \log_3 x = 9 \log x^3 \) is:
\[ \boxed{1} \]