Solve for $x$: $\left(\frac{1}{4}\right)^{2x+8} = (16)^{2x+5}$.
We begin by writing $\frac{1}{4}$ and $16$ as powers of $2$. \begin{align*}
(2^{-2})^{2x+8} & = (2^4)^{2x+5} \\
2^{-4x-16} & = 2^{8x + 20} \\
-4x - 16 & = 8x + 20 \\
x & = \boxed{-3}
\end{align*}
-3
Level 4
Algebra
Problem: Solve for $x$: $\left(\frac{1}{4}\right)^{2x+8} = (16)^{2x+5}$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the equation, first express both sides with the same base. The left side, $\left(\frac{1}{4}\right)^{2x+8}$, can be rewritten as $\left(4^{-1}\right)^{2x+8} = 4^{-(2x+8)}$. The right side, $(16)^{2x+5}$, can be expressed as $(4^2)^{2x+5} = 4^{2(2x+5)} = 4^{4x+10}$. 

Now, we have the equation $4^{-(2x+8)} = 4^{4x+10}$. Since the bases are the same, we can set the exponents equal to each other: 

$-(2x + 8) = 4x + 10$. 

Next, simplify the equation. Distributing the negative sign gives $-2x - 8 = 4x + 10$. 

Now, combine like terms by adding $2x$ to both sides: 

$-8 = 6x + 10$. 

Then, isolate the term with $x$ by subtracting $10$ from both sides: 

$-18 = 6x$. 

Finally, divide both sides by $6$ to solve for $x$: 

$x = -3$. 

Thus, the final answer is $-3$.<|END_OF_LATENT|> The final answer is $-3$.