Solve for $n$: $2^n\cdot 4^n=64^{n-36}$.
Since $4=2^2$, $4^n=2^{2n}$.  Since $64=2^6$, $64^{n-36}=2^{6(n-36)}$.  Thus,

$$2^{n+2n}=2^{6(n-36)}\Rightarrow 3n=6n-216$$

So $3n=216\Rightarrow n=\boxed{72}$.
72
Level 4
Algebra
Problem: Solve for $n$: $2^n\cdot 4^n=64^{n-36}$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the equation, first express all terms with the same base. Recognize that \(4\) can be rewritten as \(2^2\) and \(64\) as \(2^6\). Thus, the equation becomes:

\[
2^n \cdot (2^2)^n = (2^6)^{n-36}
\]

This simplifies to:

\[
2^n \cdot 2^{2n} = 2^{6(n-36)}
\]

Combine the left side:

\[
2^{n + 2n} = 2^{6(n-36)}
\]

This results in:

\[
2^{3n} = 2^{6(n-36)}
\]

Since the bases are the same, equate the exponents:

\[
3n = 6(n - 36)
\]

Distributing on the right side gives:

\[
3n = 6n - 216
\]

Rearranging the equation leads to:

\[
3n - 6n = -216
\]

This simplifies to:

\[
-3n = -216
\]

Dividing both sides by \(-3\) yields:

\[
n = 72
\]

Thus, the final answer is \(72\).<|END_OF_LATENT|> The final answer is $72$.