The number of values of x satisfying the equation `log_(2)(9^(x-1)+7)=2+log_(2)(3^(x-1)+1)` is :

Options

A. 1
B. 2
C. 3
D. 0

Answer

To solve the equation \( \log_{2}(9^{x-1} + 7) = 2 + \log_{2}(3^{x-1} + 1) \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the original equation:
\[ \log_{2}(9^{x-1} + 7) = 2 + \log_{2}(3^{x-1} + 1) \]
We can rewrite \(2\) as \(\log_{2}(4)\) because \(2 = \log_{2}(2^2) = \log_{2}(4)\). Thus, we can rewrite the equation as:
\[ \log_{2}(9^{x-1} + 7) = \log_{2}(4) + \log_{2}(3^{x-1} + 1) \]

### Step 2: Use the properties of logarithms
Using the property of logarithms that states \(\log_{b}(A) + \log_{b}(C) = \log_{b}(A \cdot C)\), we can combine the right-hand side:
\[ \log_{2}(9^{x-1} + 7) = \log_{2}(4(3^{x-1} + 1)) \]

### Step 3: Set the arguments equal
Since the logarithms are equal, we can set their arguments equal to each other:
\[ 9^{x-1} + 7 = 4(3^{x-1} + 1) \]

### Step 4: Expand the equation
Expanding the right-hand side gives:
\[ 9^{x-1} + 7 = 4 \cdot 3^{x-1} + 4 \]
Rearranging this leads to:
\[ 9^{x-1} - 4 \cdot 3^{x-1} + 3 = 0 \]

### Step 5: Substitute \(3^{x-1}\)
Let \(T = 3^{x-1}\). Then \(9^{x-1} = (3^{x-1})^2 = T^2\). Substituting gives us:
\[ T^2 - 4T + 3 = 0 \]

### Step 6: Factor the quadratic equation
Now we can factor the quadratic:
\[ (T - 3)(T - 1) = 0 \]
Thus, we find:
\[ T = 3 \quad \text{or} \quad T = 1 \]

### Step 7: Solve for \(x\)
Recall that \(T = 3^{x-1}\):
1. For \(T = 3\):
\[ 3^{x-1} = 3 \implies x - 1 = 1 \implies x = 2 \]
2. For \(T = 1\):
\[ 3^{x-1} = 1 \implies x - 1 = 0 \implies x = 1 \]

### Conclusion
The values of \(x\) that satisfy the original equation are \(x = 1\) and \(x = 2\). Therefore, the number of values of \(x\) satisfying the equation is:
\[ \boxed{2} \]