\( 1.1(x+9)\left(3^{x-2}-27\right)=0 \)

To solve the equation: \[ 1.1(x + 9)\left(3^{x - 2} - 27\right) = 0 \] We can use the **Zero Product Property**, which states that if a product of factors equals zero, then at least one of the factors must be zero. Let's break down the equation into its factors: 1. **First Factor:** \(1.1\) Since \(1.1 \neq 0\), this factor does not provide any solutions. 2. **Second Factor:** \(x + 9\) Setting this equal to zero: \[ x + 9 = 0 \implies x = -9 \] 3. **Third Factor:** \(3^{x - 2} - 27\) Setting this equal to zero: \[ 3^{x - 2} - 27 = 0 \implies 3^{x - 2} = 27 \] Recall that \(27\) can be expressed as a power of \(3\): \[ 27 = 3^3 \] So, \[ 3^{x - 2} = 3^3 \] Since the bases are the same, we can set the exponents equal to each other: \[ x - 2 = 3 \implies x = 5 \] **Final Solutions:** \[ x = -9 \quad \text{and} \quad x = 5 \]