If `2log_(16)(x^(2)+x)-log_(4)(x+1)=2`, then x=

Options

A. `-1`
B. `16`
C. `2`
D. none of these

Answer

To solve the equation \( 2\log_{16}(x^2 + x) - \log_{4}(x + 1) = 2 \), we will follow these steps:

### Step 1: Change the base of the logarithms
We can express \( \log_{16} \) in terms of \( \log_{4} \) since \( 16 = 4^2 \).

Using the change of base formula:
\[ \log_{16}(a) = \frac{\log_{4}(a)}{\log_{4}(16)} = \frac{\log_{4}(a)}{2} \]
Thus, we can rewrite \( 2\log_{16}(x^2 + x) \) as:
\[ 2\log_{16}(x^2 + x) = 2 \cdot \frac{\log_{4}(x^2 + x)}{2} = \log_{4}(x^2 + x) \]

### Step 2: Rewrite the equation
Substituting back into the original equation gives us:
\[ \log_{4}(x^2 + x) - \log_{4}(x + 1) = 2 \]

### Step 3: Use the property of logarithms
Using the property \( \log_{a}(m) - \log_{a}(n) = \log_{a}\left(\frac{m}{n}\right) \), we can combine the logarithms:
\[ \log_{4}\left(\frac{x^2 + x}{x + 1}\right) = 2 \]

### Step 4: Exponentiate both sides
To eliminate the logarithm, we exponentiate both sides:
\[ \frac{x^2 + x}{x + 1} = 4^2 \]
This simplifies to:
\[ \frac{x^2 + x}{x + 1} = 16 \]

### Step 5: Cross-multiply
Cross-multiplying gives:
\[ x^2 + x = 16(x + 1) \]
Expanding the right side:
\[ x^2 + x = 16x + 16 \]

### Step 6: Rearrange the equation
Rearranging the equation to one side gives:
\[ x^2 + x - 16x - 16 = 0 \]
This simplifies to:
\[ x^2 - 15x - 16 = 0 \]

### Step 7: Factor the quadratic equation
Now we need to factor the quadratic:
\[ (x - 16)(x + 1) = 0 \]

### Step 8: Solve for x
Setting each factor to zero gives us:
\[ x - 16 = 0 \quad \Rightarrow \quad x = 16 \]
\[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \]

### Step 9: Check for valid solutions
Since logarithms are defined for positive arguments, we check:
- For \( x = 16 \): \( x^2 + x = 256 + 16 = 272 \) (valid)
- For \( x = -1 \): \( x^2 + x = 1 - 1 = 0 \) (invalid)

Thus, the only valid solution is:
\[ \boxed{16} \]