To solve the equation

$\frac{1}{2} \log_7 x = 3 \log_7 3,$

we can follow these steps:

1. Remove the coefficients from each side:
   We can rewrite the equation by isolating $\log_7 x$ and $\log_7 3$ terms.

   $\log_7 x = 2 \cdot 3 \log_7 3 = 6 \log_7 3.$

2. Combine the logarithmic terms:
   Using the property $a \log_b c = \log_b (c^a)$, we rewrite $6 \log_7 3$ as:

   $\log_7 x = \log_7 (3^6).$

3. Simplify $3^6$:

   $3^6 = 729.$

   So, the equation becomes:

   $\log_7 x = \log_7 729.$

4. Set the arguments equal to each other:
   Since the bases and the logarithmic functions are equal, we conclude that:

   $x = 729.$

Thus, the solution is: $\text{Solution set: } \{729\}.$

Would you like further details or have any questions?

Here are 5 related questions for further practice:

1. Solve $\log_5 (2x) = 3 \log_5 4$.
2. If $\frac{1}{3} \log_2 x = \log_2 8$, find $x$.
3. Solve the equation $\log_7 (x + 1) = 2 \log_7 5$.
4. If $\log_{10} (5x) = 2 \log_{10} 3$, determine the value of $x$.
5. Find the solution to $\log_b (x + 2) = \log_b 9$.

Tip: Remember that if $\log_b A = \log_b B$, then $A = B$ (assuming $A$ and $B$ are within the domain of the logarithmic function).