The sum of the roots of the equation `2^((33x-2))+2^((11x+2))=2^((22x+1))+1` is

Options

A. `(1)/(11)`
B. `(2)/(11)`
C. `(3)/(11)`
D. `(4)/(11)`

Answer

To solve the equation \(2^{(33x-2)} + 2^{(11x+2)} = 2^{(22x+1)} + 1\), we will follow these steps:

### Step 1: Substitution
Let \(t = 2^{11x}\). Then we can rewrite the equation in terms of \(t\):
- \(2^{33x} = (2^{11x})^3 = t^3\)
- \(2^{(11x + 2)} = 2^2 \cdot 2^{11x} = 4t\)
- \(2^{(22x + 1)} = 2 \cdot (2^{11x})^2 = 2t^2\)

Substituting these into the original equation gives:
\[ t^3 \cdot 2^{-2} + 4t = 2t^2 + 1 \]
This simplifies to:
\[ \frac{t^3}{4} + 4t = 2t^2 + 1 \]

### Step 2: Clear the Fraction
Multiply the entire equation by 4 to eliminate the fraction:
\[ t^3 + 16t = 8t^2 + 4 \]

### Step 3: Rearranging the Equation
Rearranging the equation gives:
\[ t^3 - 8t^2 + 16t - 4 = 0 \]

### Step 4: Finding the Sum of the Roots
For a cubic equation of the form \(at^3 + bt^2 + ct + d = 0\), the sum of the roots \(t_1, t_2, t_3\) can be found using the formula:
\[ t_1 + t_2 + t_3 = -\frac{b}{a} \]
Here, \(a = 1\) and \(b = -8\), so:
\[ t_1 + t_2 + t_3 = -\frac{-8}{1} = 8 \]

### Step 5: Relating Back to \(x\)
Since we set \(t = 2^{11x}\), we have:
\[ t_1 + t_2 + t_3 = 8 \]
This means:
\[ 2^{11x_1} + 2^{11x_2} + 2^{11x_3} = 8 \]
Since \(8 = 2^3\), we can equate the powers:
\[ 11x_1 + 11x_2 + 11x_3 = 3 \]
Thus:
\[ x_1 + x_2 + x_3 = \frac{3}{11} \]

### Final Answer
The sum of the roots of the equation is:
\[ \boxed{\frac{3}{11}} \]