The problem consists of three sub-problems: (a) Solve the exponential equation $8^{-x^2 + x} = 2^{5x-1}$. (b) Solve the logarithmic equation $log_a(2p+1) + log_a(3p-10) = log_a(11p)$ for $p>0$. (c) Solve the logarithmic equation $log_2(x) - 3 + log_2(x+3) = 0$.

1. Problem Description

The problem consists of three sub-problems:

(a) Solve the exponential equation $8^{-x^2 + x} = 2^{5x-1}$.

(b) Solve the logarithmic equation $log_a(2p+1) + log_a(3p-10) = log_a(11p)$ for $p>0$.

(c) Solve the logarithmic equation $log_2(x) - 3 + log_2(x+3) = 0$.

2. Solution Steps

(a) Solve $8^{-x^2 + x} = 2^{5x-1}$.

First, rewrite 8 as $2^3$:

$(2^3)^{-x^2 + x} = 2^{5x-1}$.

Then, using the property $(a^b)^c = a^{bc}$:

$2^{3(-x^2 + x)} = 2^{5x-1}$.

Since the bases are equal, we can equate the exponents:

$3(-x^2 + x) = 5x - 1$

$-3x^2 + 3x = 5x - 1$

$3x^2 + 2x - 1 = 0$

We can factor this quadratic equation as:

$(3x - 1)(x + 1) = 0$

Therefore, $3x - 1 = 0$ or $x + 1 = 0$.

$x = \frac{1}{3}$ or $x = -1$.

(b) Solve $log_a(2p+1) + log_a(3p-10) = log_a(11p)$ for $p>0$.

Using the logarithm property $log_a(b) + log_a(c) = log_a(bc)$:

$log_a((2p+1)(3p-10)) = log_a(11p)$

Since the logarithms have the same base, we can equate the arguments:

$(2p+1)(3p-10) = 11p$

$6p^2 - 20p + 3p - 10 = 11p$

$6p^2 - 17p - 10 = 11p$

$6p^2 - 28p - 10 = 0$

Divide the equation by 2:

$3p^2 - 14p - 5 = 0$

We can factor this quadratic equation as:

$(3p + 1)(p - 5) = 0$

Therefore, $3p + 1 = 0$ or $p - 5 = 0$.

$p = -\frac{1}{3}$ or $p = 5$.

Since $p>0$, we discard the solution $p = -\frac{1}{3}$. Also we have the condition that $3p-10>0$ so $p > \frac{10}{3}$. Since $5 > \frac{10}{3}$, $p=5$ is a valid solution.

(c) Solve $log_2(x) - 3 + log_2(x+3) = 0$.

Using the logarithm property $log_a(b) + log_a(c) = log_a(bc)$:

$log_2(x(x+3)) - 3 = 0$

$log_2(x(x+3)) = 3$

Using the property $log_a(b) = c \iff a^c = b$:

$x(x+3) = 2^3$

$x^2 + 3x = 8$

$x^2 + 3x - 8 = 0$

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-8)}}{2(1)}$

$x = \frac{-3 \pm \sqrt{9 + 32}}{2}$

$x = \frac{-3 \pm \sqrt{41}}{2}$

Since we must have $x>0$, we discard the negative root. Thus

$x = \frac{-3 + \sqrt{41}}{2}$

3. Final Answer

(a) $x = \frac{1}{3}, -1$

(b) $p = 5$

(c) $x = \frac{-3 + \sqrt{41}}{2}$