Show the sum of the roots of the equation `x+1=2log_2(2^x+3)-2log_4(1980-2^-x)` is `log_(2)11`.

Answer

To solve the equation \( x + 1 = 2 \log_2(2^x + 3) - 2 \log_4(1980 - 2^{-x}) \) and show that the sum of the roots is \( \log_2 11 \), we can follow these steps:

### Step 1: Rewrite the equation
Start with the given equation:
\[ x + 1 = 2 \log_2(2^x + 3) - 2 \log_4(1980 - 2^{-x}) \]

### Step 2: Change the base of the logarithm
Recall that \( \log_4(a) = \frac{1}{2} \log_2(a) \). Therefore, we can rewrite the second logarithm:
\[ x + 1 = 2 \log_2(2^x + 3) - 2 \cdot \frac{1}{2} \log_2(1980 - 2^{-x}) \]
This simplifies to:
\[ x + 1 = 2 \log_2(2^x + 3) - \log_2(1980 - 2^{-x}) \]

### Step 3: Combine the logarithms
Using the properties of logarithms, we can combine the right-hand side:
\[ x + 1 = \log_2((2^x + 3)^2) - \log_2(1980 - 2^{-x}) \]
This can be expressed as:
\[ x + 1 = \log_2\left(\frac{(2^x + 3)^2}{1980 - 2^{-x}}\right) \]

### Step 4: Exponentiate both sides
To eliminate the logarithm, we exponentiate both sides:
\[ 2^{x + 1} = \frac{(2^x + 3)^2}{1980 - 2^{-x}} \]

### Step 5: Clear the fraction
Multiply both sides by \( 1980 - 2^{-x} \):
\[ 2^{x + 1} (1980 - 2^{-x}) = (2^x + 3)^2 \]

### Step 6: Expand and rearrange
Expanding the left-hand side gives:
\[ 1980 \cdot 2^{x + 1} - 2^{x + 1 - x} = (2^x + 3)^2 \]
This simplifies to:
\[ 1980 \cdot 2^{x + 1} - 2 = (2^x + 3)^2 \]
Now, expand the right-hand side:
\[ 1980 \cdot 2^{x + 1} - 2 = 2^{2x} + 6 \cdot 2^x + 9 \]

### Step 7: Rearranging into a standard form
Rearranging gives:
\[ 2^{2x} + 6 \cdot 2^x + 9 - 1980 \cdot 2^{x + 1} + 2 = 0 \]
This can be rewritten as:
\[ 2^{2x} - 3954 \cdot 2^x + 11 = 0 \]

### Step 8: Let \( t = 2^x \)
Substituting \( t = 2^x \) transforms the equation into a quadratic:
\[ t^2 - 3954t + 11 = 0 \]

### Step 9: Find the sum of the roots
Using the quadratic formula, the sum of the roots \( t_1 + t_2 \) is given by:
\[ t_1 + t_2 = \frac{-b}{a} = \frac{3954}{1} = 3954 \]

### Step 10: Relate back to \( x \)
Since \( t = 2^x \), we have:
\[ t_1 t_2 = 11 \implies 2^{x_1} \cdot 2^{x_2} = 11 \implies 2^{x_1 + x_2} = 11 \]
Taking logarithm base 2 on both sides gives:
\[ x_1 + x_2 = \log_2 11 \]

### Conclusion
Thus, we have shown that the sum of the roots of the equation is \( \log_2 11 \).