What value of x satisfies 8^x+1 = 32^x-2?

Rewrite both sides with the same base. Since $8 = 2^3$ and $32 = 2^5$, the equation becomes $2^{3(x+1)} = 2^{5(x-2)}$. Equal bases with equal values mean the exponents are equal: $3x + 3 = 5x - 10$. Subtract $3x$ and add 10: $13 = 2x$, so $x = \frac{13}{2}$, choice A. Choice B ($\frac{11}{2}$) results from a sign slip, using $3x + 3 = 5x - 8$ or subtracting 10 incorrectly. Choice C (2) is the number inside the parentheses. Choice D (5) comes from the same mistake, ignoring the factor of 3 and 5 on the exponents, or mismatching bases.