解方程 $2^{4p+1} + 20(4^{p-1}) = 3$。 解释为什么当 $k < -3\frac{1}{8}$ 时，方程 $2^{4p+1} + 20(4^{p-1}) = k$ 无解。

1. Express the terms using a common base power, such as $4^p$.

$2^{4p+1} = 2^1 \cdot (2^2)^{2p} = 2(4^{2p}) = 2(4^p)^2$
$20(4^{p-1}) = 20 \cdot \frac{4^p}{4} = 5(4^p)$

2. Substitute $u = 4^p$ to form a quadratic equation.

Let $u = 4^p$.
The equation becomes:
$2u^2 + 5u = 3$
$2u^2 + 5u - 3 = 0$

3. Solve the quadratic equation for $u$.

$(2u - 1)(u + 3) = 0$
$u = \frac{1}{2}$ or $u = -3$

4. Solve for $p$ and reject the invalid solution.

Substitute back $u = 4^p$:
$4^p = \frac{1}{2}$ or $4^p = -3$

For $4^p = -3$, there is no real solution since $4^p > 0$ for all real $p$.

For $4^p = \frac{1}{2}$:
$2^{2p} = 2^{-1}$
$2p = -1$
$p = -\frac{1}{2}$

1. Express the equation as a quadratic in terms of $4^p$.

Let $x = 4^p$.
$2^{4p+1} + 20(4^{p-1}) = k$
$2(2^{4p}) + 20\left(\frac{4^p}{4}\right) = k$
$2(4^{2p}) + 5(4^p) = k$
$2x^2 + 5x - k = 0$

2. Find the discriminant of the quadratic equation.

For the quadratic equation $2x^2 + 5x - k = 0$ to have real solutions for $x$, its discriminant must be non-negative.
$\Delta = b^2 - 4ac$
$\Delta = 5^2 - 4(2)(-k)$
$\Delta = 25 + 8k$

3. Explain the condition for no solution.

If $k < -3\frac{1}{8}$, then $k < -\frac{25}{8}$.
$8k < -25 \implies 25 + 8k < 0$.
Since $\Delta < 0$, the quadratic equation has no real solutions for $x$.
Consequently, there are no real solutions for $p$.