We are given the equation $\frac{5^{n+3}}{25^{2n-2}} = 5^0$ and we need to find the value of $n$.

1. Problem Description

We are given the equation $\frac{5^{n+3}}{25^{2n-2}} = 5^0$ and we need to find the value of $n$.

2. Solution Steps

First, rewrite the equation using the fact that $25 = 5^2$:

$\frac{5^{n+3}}{(5^2)^{2n-2}} = 5^0$

Using the power of a power rule, $(a^m)^n = a^{mn}$, we have:

$\frac{5^{n+3}}{5^{2(2n-2)}} = 5^0$

$\frac{5^{n+3}}{5^{4n-4}} = 5^0$

Using the quotient of powers rule, $\frac{a^m}{a^n} = a^{m-n}$, we have:

$5^{(n+3)-(4n-4)} = 5^0$

$5^{n+3-4n+4} = 5^0$

$5^{-3n+7} = 5^0$

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

$-3n+7 = 0$

Now, solve for $n$:

$-3n = -7$

$n = \frac{-7}{-3}$

$n = \frac{7}{3}$

However, none of the given options match $n = \frac{7}{3}$. Let's check if there was an error in the problem statement or if one of the options is the correct answer.

Let's test the given values of $n$ in the original equation:

A. $n = 1$:

$\frac{5^{1+3}}{25^{2(1)-2}} = \frac{5^4}{25^0} = \frac{5^4}{1} = 5^4 \neq 5^0$

B. $n = 2$:

$\frac{5^{2+3}}{25^{2(2)-2}} = \frac{5^5}{25^2} = \frac{5^5}{(5^2)^2} = \frac{5^5}{5^4} = 5^{5-4} = 5^1 \neq 5^0$

C. $n = 3$:

$\frac{5^{3+3}}{25^{2(3)-2}} = \frac{5^6}{25^4} = \frac{5^6}{(5^2)^4} = \frac{5^6}{5^8} = 5^{6-8} = 5^{-2} \neq 5^0$

D. $n = 5$:

$\frac{5^{5+3}}{25^{2(5)-2}} = \frac{5^8}{25^8} = \frac{5^8}{(5^2)^8} = \frac{5^8}{5^{16}} = 5^{8-16} = 5^{-8} \neq 5^0$

It appears that there may be an error in the problem statement or the given options. However, if the question was $\frac{5^{n+3}}{25^{2n-2}} = 1$, then $5^0 = 1$ and $-3n+7 = 0$ gives $n = \frac{7}{3}$

3. Final Answer

None of the provided options are correct. The value of $n$ that satisfies the equation is $\frac{7}{3}$.

Final Answer: n = 7/3