Question:

**Step 1:** Rewrite the given equation using logarithm properties. $$ \log_2(9^{x-1} + 7) = 2 + \log_2(3^{x-1} + 1) $$ $$ \log_2(9^{x-1} + 7) = \log_2(2^2) + \log_2(3^{x-1} + 1) $$ $$ \log_2(9^{x-1} + 7) = \log_2(4) + \log_2(3^{x-1} + 1) $$ $$ \log_2(9^{x-1} + 7) = \log_2[4(3^{x-1} + 1)] $$ **Step 2:** Since the logarithm function is one-to-one, equate the arguments. $$ 9^{x-1} + 7 = 4(3^{x-1} + 1) $$ $$ 9^{x-1} + 7 = 4 \cdot 3^{x-1} + 4 $$ **Step 3:** Express $9^{x-1}$ in terms of $3^{x-1}$. Since $9^{x-1} = (3^2)^{x-1} = (3^{x-1})^2$, let $y = 3^{x-1}$. Note that $y > 0$ for any real $x$. Substituting $y$ into the equation: $$ y^2 + 7 = 4y + 4 $$ $$ y^2 - 4y + 3 = 0 $$ **Step 4:** Solve the quadratic equation for $y$. $$ (y - 1)(y - 3) = 0 $$ This yields two possible values for $y$: $$ y = 1 \quad \text{or} \quad y = 3 $$ **Step 5:** Substitute back $y = 3^{x-1}$ to find the values of $x$. Case 1: $3^{x-1} = 1$ $$ 3^{x-1} = 3^0 $$ $$ x - 1 = 0 $$ $$ x = 1 $$ Case 2: $3^{x-1} = 3$ $$ 3^{x-1} = 3^1 $$ $$ x - 1 = 1 $$ $$ x = 2 $$ **Step 6:** Verify both solutions satisfy the domain restrictions of the original logarithmic equation (arguments of logarithms must be positive). For $x = 1$: The argument of the first logarithm is $9^{1-1} + 7 = 9^0 + 7 = 1 + 7 = 8 > 0$. The argument of the second logarithm is $3^{1-1} + 1 = 3^0 + 1 = 1 + 1 = 2 > 0$. Both are positive, so $x=1$ is a valid candidate. Substituting $x=1$ into the original equation: $\log_2(8) = 2 + \log_2(2) \Rightarrow 3 = 2 + 1$, which is true. For $x = 2$: The argument of the first logarithm is $9^{2-1} + 7 = 9^1 + 7 = 9 + 7 = 16 > 0$. The argument of the second logarithm is $3^{2-1} + 1 = 3^1 + 1 = 3 + 1 = 4 > 0$. Both are positive, so $x=2$ is a valid candidate. Substituting $x=2$ into the original equation: $\log_2(16) = 2 + \log_2(4) \Rightarrow 4 = 2 + 2$, which is true. **Step 7:** Both solutions $x=1$ and $x=2$ satisfy the original equation and its domain restrictions. Therefore, there are two values of $x$ satisfying the equation.