If `4^(x)*n^(2)=4^(x+1)*n` and x and n are both positive integers, what is the value of n?

Options

A. `2`
B. `4`
C. `6`
D. `8`

Answer

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

### Step 1: Rewrite the equation
We start with the original equation:
\[ 4^x \cdot n^2 = 4^{x+1} \cdot n \]
We can rewrite \( 4^{x+1} \) as \( 4^x \cdot 4 \):
\[ 4^x \cdot n^2 = 4^x \cdot 4 \cdot n \]

### Step 2: Cancel \( 4^x \) from both sides
Since \( 4^x \) is a common factor on both sides and \( x \) is a positive integer (hence \( 4^x \neq 0 \)), we can safely divide both sides by \( 4^x \):
\[ n^2 = 4n \]

### Step 3: Rearrange the equation
Now, we can rearrange the equation to bring all terms to one side:
\[ n^2 - 4n = 0 \]

### Step 4: Factor the equation
We can factor the left-hand side:
\[ n(n - 4) = 0 \]

### Step 5: Solve for \( n \)
Setting each factor to zero gives us:
1. \( n = 0 \) (not a positive integer)
2. \( n - 4 = 0 \) which leads to \( n = 4 \)

Since \( n \) must be a positive integer, we conclude that:
\[ n = 4 \]

### Final Answer
The value of \( n \) is \( 4 \).

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