If x satisfies `log_2(9^(x-1)+7)=2+log_2(3^(x-1)+1)`, then

Options

A. `xnotin Q`
B. `xnotinN`
C. `xnotin{x notinQ:xlt0}`
D. `xnotinN_e` (set of even natural number)

Answer

To solve the equation \( \log_2(9^{x-1} + 7) = 2 + \log_2(3^{x-1} + 1) \), we can follow these steps:

### Step 1: Rewrite the equation
We can rewrite the right-hand side of the equation using the property of logarithms:
\[ 2 + \log_2(3^{x-1} + 1) = \log_2(2^2) + \log_2(3^{x-1} + 1) = \log_2(4(3^{x-1} + 1)) \]
So, our equation becomes:
\[ \log_2(9^{x-1} + 7) = \log_2(4(3^{x-1} + 1)) \]

### Step 2: Eliminate the logarithm
Since the logarithms are equal, we can set the arguments equal to each other:
\[ 9^{x-1} + 7 = 4(3^{x-1} + 1) \]

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

### Step 4: Rearrange the equation
Rearranging the equation, we have:
\[ 9^{x-1} - 4 \cdot 3^{x-1} + 3 = 0 \]

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

### Step 6: Factor the quadratic equation
Now we can factor the quadratic:
\[ (t - 3)(t - 1) = 0 \]

### Step 7: Solve for \( t \)
Setting each factor to zero gives us:
\[ t - 3 = 0 \quad \Rightarrow \quad t = 3 \]
\[ t - 1 = 0 \quad \Rightarrow \quad t = 1 \]

### Step 8: Back-substitute for \( x \)
Recall that \( t = 3^{x-1} \):
1. If \( t = 3 \):
\[ 3^{x-1} = 3 \quad \Rightarrow \quad x - 1 = 1 \quad \Rightarrow \quad x = 2 \]
2. If \( t = 1 \):
\[ 3^{x-1} = 1 \quad \Rightarrow \quad x - 1 = 0 \quad \Rightarrow \quad x = 1 \]

### Final Solution
Thus, the values of \( x \) that satisfy the equation are:
\[ x = 1 \quad \text{and} \quad x = 2 \]