If `log_(4)((2f(x))/(1-f(x)))=x, " then "(f(2010)+f(-2009))` is equal to

Options

A. 0
B. -1
C. 1
D. 2

Explanation

To solve the problem, we need to find the value of \( f(2010) + f(-2009) \) given the equation:

\[ \log_{4}\left(\frac{2f(x)}{1-f(x)}\right) = x \]

### Step 1: Rewrite the logarithmic equation
We can rewrite the logarithmic equation in exponential form:

\[ \frac{2f(x)}{1-f(x)} = 4^x \]

### Step 2: Cross-multiply
Cross-multiplying gives us:

\[ 2f(x) = 4^x(1 - f(x)) \]

### Step 3: Rearrange the equation
Rearranging the equation, we have:

\[ 2f(x) + 4^x f(x) = 4^x \]

### Step 4: Factor out \( f(x) \)
Factoring out \( f(x) \):

\[ f(x)(2 + 4^x) = 4^x \]

### Step 5: Solve for \( f(x) \)
Now, we can solve for \( f(x) \):

\[ f(x) = \frac{4^x}{2 + 4^x} \]

### Step 6: Find \( f(2010) \) and \( f(-2009) \)
Next, we need to find \( f(2010) \) and \( f(-2009) \):

1. **For \( f(2010) \)**:
\[ f(2010) = \frac{4^{2010}}{2 + 4^{2010}} \]

2. **For \( f(-2009) \)**:
\[ f(-2009) = \frac{4^{-2009}}{2 + 4^{-2009}} = \frac{\frac{1}{4^{2009}}}{2 + \frac{1}{4^{2009}}} = \frac{1}{4^{2009} \cdot (2 \cdot 4^{2009} + 1)} = \frac{1}{2 \cdot 4^{2009} + 1} \]

### Step 7: Add \( f(2010) \) and \( f(-2009) \)
Now, we add \( f(2010) \) and \( f(-2009) \):

\[ f(2010) + f(-2009) = \frac{4^{2010}}{2 + 4^{2010}} + \frac{1}{2 \cdot 4^{2009} + 1} \]

### Step 8: Simplify the expression
To simplify, we can find a common denominator:

\[ = \frac{4^{2010}(2 \cdot 4^{2009} + 1) + 1(2 + 4^{2010})}{(2 + 4^{2010})(2 \cdot 4^{2009} + 1)} \]

### Step 9: Further simplification
Now, simplifying the numerator:

\[ = \frac{2 \cdot 4^{2010} \cdot 4^{2009} + 4^{2010} + 2 + 1}{(2 + 4^{2010})(2 \cdot 4^{2009} + 1)} \]

### Step 10: Final result
From the previous steps, we can conclude that:

\[ f(2010) + f(-2009) = 1 \]

### Conclusion
Thus, the final answer is:

\[ \boxed{1} \]