$3^n = 3 \cdot 9^3 \cdot 81^2$. What is the value of $n$?
We want to write everything in terms of powers of 3. Doing so gives us $3^n = 3 \cdot (3^2)^3 \cdot (3^4)^2$. This simplifies to $3^n = 3 \cdot 3^6 \cdot 3^8$, so $3^n = 3^{15}$. Therefore, $n = \boxed{15}$.
15
Level 2
Algebra
Problem: $3^n = 3 \cdot 9^3 \cdot 81^2$. What is the value of $n$?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve for \( n \) in the equation \( 3^n = 3 \cdot 9^3 \cdot 81^2 \), we first need to express all terms on the right side in terms of base 3. 

We know that \( 9 \) can be rewritten as \( 3^2 \) and \( 81 \) can be rewritten as \( 3^4 \). Therefore, we can substitute these values into the equation:

1. Rewrite \( 9^3 \):
   \[
   9^3 = (3^2)^3 = 3^{2 \cdot 3} = 3^6
   \]

2. Rewrite \( 81^2 \):
   \[
   81^2 = (3^4)^2 = 3^{4 \cdot 2} = 3^8
   \]

Now substitute these back into the original equation:
\[
3^n = 3 \cdot 3^6 \cdot 3^8
\]

Next, we can combine the terms on the right side using the property of exponents that states \( a^m \cdot a^n = a^{m+n} \):
\[
3^n = 3^{1 + 6 + 8}
\]

Now, calculate the exponent on the right side:
\[
1 + 6 + 8 = 15
\]

Thus, we have:
\[
3^n = 3^{15}
\]

Since the bases are the same, we can equate the exponents:
\[
n = 15
\]

The final answer is \( 15 \).<|END_OF_LATENT|> The final answer is $15$.