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Let's solve the given equation:

\[
2^{4n+3} = 4^{n-1}
\]

First, rewrite \(4^{n-1}\) in terms of base 2, since \(4 = 2^2\):

\[
4^{n-1} = (2^2)^{n-1} = 2^{2(n-1)} = 2^{2n-2}
\]

Now the equation becomes:

\[
2^{4n+3} = 2^{2n-2}
\]

Since the bases are the same, we can set the exponents equal to each other:

\[
4n + 3 = 2n - 2
\]

Now, solve for \(n\):

\[
4n - 2n = -2 - 3
\]
\[
2n = -5
\]
\[
n = -\frac{5}{2}
\]

Thus, the value of \(n\) is \(\boxed{-\frac{5}{2}}\), which corresponds to option B.

Would you like a further breakdown of this solution, or do you have any questions?

Here are 5 related questions to deepen your understanding:

1. What would happen if you changed the base of the exponents in this problem?
2. How would you solve if both sides of the equation had different bases, such as 3 and 5?
3. What are the properties of exponents that allow us to simplify such equations?
4. Can you solve a similar equation with fractional exponents?
5. How would the solution change if the powers were multiplied instead of added?

**Tip:** When solving equations involving exponents, converting all terms to the same base simplifies the problem significantly.

This problem involves solving an exponential equation where both sides are rewritten with the same base (2). Learn how to apply the properties of exponents to simplify the equation and solve for n. Suitable for high school students (Grades 9-12).

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