We are asked to solve two separate problems. (a) Solve the equation $8^{-x^2+x} = 2^{5x-1}$ for $x$. (b) Solve the equation $\log_a(2p+1) + \log_a(3p-10) = \log_a(11p)$ for $p$, given that $p>0$.

1. Problem Description

We are asked to solve two separate problems.

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

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

2. Solution Steps

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

We can rewrite $8$ as $2^3$, so the equation becomes $(2^3)^{-x^2+x} = 2^{5x-1}$.

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

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

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

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

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

Now, we can solve this quadratic equation for $x$. We can factor the quadratic:

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

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

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

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

Using the logarithm property $\log_a(x) + \log_a(y) = \log_a(xy)$, we have:

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

Since the bases of the logarithms are equal, 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 by 2:

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

Now, we can solve this quadratic equation for $p$. We can factor the quadratic:

$3p^2 - 14p - 5 = (3p+1)(p-5) = 0$

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

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

Since we are given that $p>0$, we must discard the negative solution $p = -\frac{1}{3}$.

Also, we need to check if $2p+1>0$ and $3p-10>0$ for the logarithmic terms to be defined. If $p=5$, we have $2p+1 = 2(5)+1 = 11 > 0$ and $3p-10 = 3(5)-10 = 15-10 = 5 > 0$. So $p=5$ is a valid solution.

3. Final Answer

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

(b) $p = 5$