Solve the equations:
Q. `2^(2x-1)-9xx2^(x-2)+1=0`.

Answer

To solve the equation \( 2^{2x-1} - 9 \cdot 2^{x-2} + 1 = 0 \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ 2^{2x-1} - 9 \cdot 2^{x-2} + 1 = 0 \]
We can rewrite \( 2^{2x-1} \) as \( \frac{2^{2x}}{2} = \frac{2^{2x}}{2} \) and \( 9 \cdot 2^{x-2} \) as \( 9 \cdot \frac{2^x}{4} = \frac{9 \cdot 2^x}{4} \). Thus, we have:
\[ \frac{2^{2x}}{2} - \frac{9 \cdot 2^x}{4} + 1 = 0 \]

### Step 2: Clear the fractions
To eliminate the fractions, we can multiply the entire equation by 4 (the least common multiple of the denominators):
\[ 4 \cdot \left( \frac{2^{2x}}{2} \right) - 4 \cdot \left( \frac{9 \cdot 2^x}{4} \right) + 4 \cdot 1 = 0 \]
This simplifies to:
\[ 2 \cdot 2^{2x} - 9 \cdot 2^x + 4 = 0 \]

### Step 3: Substitute \( t = 2^x \)
Let \( t = 2^x \). Then \( 2^{2x} = t^2 \). Substituting these into the equation gives:
\[ 2t^2 - 9t + 4 = 0 \]

### Step 4: Solve the quadratic equation
Now we need to solve the quadratic equation \( 2t^2 - 9t + 4 = 0 \). We can use the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 2 \), \( b = -9 \), and \( c = 4 \). Plugging in these values:
\[ t = \frac{9 \pm \sqrt{(-9)^2 - 4 \cdot 2 \cdot 4}}{2 \cdot 2} \]
\[ t = \frac{9 \pm \sqrt{81 - 32}}{4} \]
\[ t = \frac{9 \pm \sqrt{49}}{4} \]
\[ t = \frac{9 \pm 7}{4} \]

### Step 5: Calculate the values of \( t \)
Calculating the two possible values for \( t \):
1. \( t = \frac{16}{4} = 4 \)
2. \( t = \frac{2}{4} = \frac{1}{2} \)

### Step 6: Convert back to \( x \)
Recall that \( t = 2^x \). We now convert back to find \( x \):
1. For \( t = 4 \):
\[ 2^x = 4 \implies x = \log_2(4) = 2 \]
2. For \( t = \frac{1}{2} \):
\[ 2^x = \frac{1}{2} \implies x = \log_2\left(\frac{1}{2}\right) = -1 \]

### Final Answer
Thus, the solutions to the equation are:
\[ x = 2 \quad \text{and} \quad x = -1 \]