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

Options

A. `1/3`
B. `1/9`
C. `1/2`
D. 1

Answer

To solve the equation \( 16^{(n+1)} = 64 \times 4^{-n} \), we will express all terms with the same base.

### Step 1: Rewrite the bases
We know that:
- \( 16 = 4^2 \)
- \( 64 = 4^3 \)

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

### Step 2: Simplify the left-hand side
Using the power of a power property \((a^m)^n = a^{m \cdot n}\), we can simplify the left-hand side:
\[ 4^{2(n+1)} = 4^3 \times 4^{-n} \]

### Step 3: Simplify the right-hand side
Using the property of exponents \(a^m \times a^n = a^{m+n}\), we can simplify the right-hand side:
\[ 4^{2(n+1)} = 4^{3 - n} \]

### Step 4: Set the exponents equal to each other
Since the bases are the same, we can set the exponents equal to each other:
\[ 2(n+1) = 3 - n \]

### Step 5: Solve for n
Now we will solve for \(n\):
1. Distribute on the left:
\[ 2n + 2 = 3 - n \]
2. Add \(n\) to both sides:
\[ 2n + n + 2 = 3 \]
\[ 3n + 2 = 3 \]
3. Subtract 2 from both sides:
\[ 3n = 1 \]
4. Divide by 3:
\[ n = \frac{1}{3} \]

### Final Answer
The value of \(n\) is \(\frac{1}{3}\).

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