If `64^(2n+1)=16^(4n-1)`, what is the value of n?

Answer

To solve the equation \( 64^{(2n+1)} = 16^{(4n-1)} \), we can follow these steps:

### Step 1: Rewrite the bases
First, we express both sides of the equation using the same base. We know that:
- \( 64 = 4^3 \)
- \( 16 = 4^2 \)

Thus, we can rewrite the equation as:
\[ (4^3)^{(2n+1)} = (4^2)^{(4n-1)} \]

### Step 2: Apply the power of a power property
Using the property of exponents \((a^m)^n = a^{m \cdot n}\), we can simplify both sides:
\[ 4^{3(2n+1)} = 4^{2(4n-1)} \]

This simplifies to:
\[ 4^{(6n + 3)} = 4^{(8n - 2)} \]

### Step 3: Set the exponents equal to each other
Since the bases are the same, we can set the exponents equal to each other:
\[ 6n + 3 = 8n - 2 \]

### Step 4: Solve for \( n \)
Now, we will solve for \( n \). First, we can rearrange the equation:
\[ 6n + 3 + 2 = 8n \]
\[ 5 = 8n - 6n \]
\[ 5 = 2n \]

Now, divide both sides by 2:
\[ n = \frac{5}{2} \]

### Final Answer
Thus, the value of \( n \) is:
\[ \boxed{\frac{5}{2}} \]

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