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

Options

A. 0
B. 1
C. 2
D. both (b) and ©

Answer

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

### Step 1: Rewrite the equation
We start with the equation:
\[ \log_2(1 + 3^{x-1}) + 2 = \log_2(3^{2x-2} + 7) \]
We can rewrite \(2\) as \(\log_2(4)\) since \(2 = \log_2(4)\). Thus, we have:
\[ \log_2(1 + 3^{x-1}) + \log_2(4) = \log_2(3^{2x-2} + 7) \]

### Step 2: Use the property of logarithms
Using the property of logarithms that states \(\log_a(b) + \log_a(c) = \log_a(b \cdot c)\), we can combine the left-hand side:
\[ \log_2(4(1 + 3^{x-1})) = \log_2(3^{2x-2} + 7) \]

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

### Step 4: Expand and simplify
Expanding the left-hand side gives:
\[ 4 + 4 \cdot 3^{x-1} = 3^{2x-2} + 7 \]
Now, rearranging the equation:
\[ 4 \cdot 3^{x-1} = 3^{2x-2} + 7 - 4 \]
\[ 4 \cdot 3^{x-1} = 3^{2x-2} + 3 \]

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

### Step 6: Rearrange into a standard quadratic equation
Rearranging gives:
\[ t^2 - 4t + 3 = 0 \]

### Step 7: Factor the quadratic equation
Factoring the quadratic:
\[ (t - 3)(t - 1) = 0 \]

### Step 8: Solve for \(t\)
Setting each factor to zero gives:
1. \(t - 3 = 0 \Rightarrow t = 3\)
2. \(t - 1 = 0 \Rightarrow t = 1\)

### Step 9: Back-substitute for \(x\)
Recall that \(t = 3^{x-1}\):
1. If \(t = 3\):
\[ 3^{x-1} = 3 \Rightarrow x - 1 = 1 \Rightarrow x = 2 \]
2. If \(t = 1\):
\[ 3^{x-1} = 1 \Rightarrow x - 1 = 0 \Rightarrow x = 1 \]

### Final Answer
Thus, the solutions for \(x\) are:
\[ x = 1 \quad \text{and} \quad x = 2 \]