log2(3× +1) *log3X = 2*log2(3 × +1)

To solve the equation \( \log_2(3x + 1) \cdot \log_3 x = 2 \cdot \log_2(3x + 1) \), we can start by simplifying it step-by-step.

First, let's rewrite the equation for clarity:

\[ \log_2(3x + 1) \cdot \log_3 x - 2 \cdot \log_2(3x + 1) = 0 \]

Factoring out \(\log_2(3x + 1)\):

\[ \log_2(3x + 1) \left( \log_3 x - 2 \right) = 0 \]

This product equals zero if either one of the factors equals zero:

1. \(\log_2(3x + 1) = 0\)
2. \(\log_3 x - 2 = 0\)

For the first case:

\(\log_2(3x + 1) = 0\) implies:

\[ 3x + 1 = 2^0 \implies 3x + 1 = 1 \implies 3x = 0 \implies x = 0 \]

However, \(x = 0\) is not valid because logarithms are undefined for non-positive values.

For the second case:

\(\log_3 x - 2 = 0\) implies:

\[ \log_3 x = 2 \implies x = 3^2 \implies x = 9 \]

Conclusion:

The only potential solution to the equation is \(x = 9\).

To ensure that it is valid, let's check if it satisfies the original equation:

Calculating:

1. \(3x + 1 = 3(9) + 1 = 27 + 1 = 28\)
2. \(\log_2(28) \cdot \log_3(9) = \log_2(28) \cdot 2\) (since \(\log_3(9) = \log_3(3^2) = 2\))
3. The right side is \(2 \cdot \log_2(28)\).

Both sides are equal. Thus, the solution is confirmed.

The solution is:

\[ \boxed{9} \]