Solve by factorisation \( 64^{y+1}=16 \) \( 3^{9 x-2}=27 \) \( \left(2.2^{x}-16\right)\left(3^{x+1}-9\right)=0 \)

**Problem 1** Solve \( 64^{y+1}=16 \). 1. Express 64 and 16 as powers of 2: \[ 64 = 2^6 \quad \text{and} \quad 16 = 2^4. \] 2. Substitute into the equation: \[ (2^6)^{y+1} = 2^4. \] 3. Simplify the left-hand side using the power of a power: \[ 2^{6(y+1)} = 2^4. \] 4. Equate the exponents: \[ 6(y+1) = 4. \] 5. Solve for \( y \): \[ y+1 = \frac{4}{6} = \frac{2}{3} \quad \Rightarrow \quad y = \frac{2}{3} - 1 = -\frac{1}{3}. \] --- **Problem 2** Solve \( 3^{9x-2}=27 \). 1. Express 27 as a power of 3: \[ 27 = 3^3. \] 2. Substitute into the equation: \[ 3^{9x-2} = 3^3. \] 3. Equate the exponents: \[ 9x-2 = 3. \] 4. Solve for \( x \): \[ 9x = 3+2 = 5 \quad \Rightarrow \quad x = \frac{5}{9}. \] --- **Problem 3** Solve \[ \left(2\cdot2^{x}-16\right)\left(3^{x+1}-9\right)=0. \] For a product to be zero, at least one factor must be zero. **Factor 1:** 1. Solve \( 2\cdot2^{x}-16 = 0 \). 2. Notice that \( 2\cdot2^{x} = 2^{x+1} \); therefore: \[ 2^{x+1} = 16. \] 3. Express 16 as a power of 2: \[ 16 = 2^4. \] 4. Equate the exponents: \[ x+1 = 4 \quad \Rightarrow \quad x = 3. \] **Factor 2:** 1. Solve \( 3^{x+1} - 9 = 0 \). 2. Express 9 as a power of 3: \[ 9 = 3^2. \] 3. Equate the equation: \[ 3^{x+1} = 3^2. \] 4. Equate the exponents: \[ x+1 = 2 \quad \Rightarrow \quad x = 1. \] Thus, the solutions for the third equation are \( x=3 \) and \( x=1 \).