5.) \( 16^{x}-1 \) \( 5 \times 4^{x}-5 \)

To solve the equation \( 16^{x} - 1 = 5 \times 4^{x} - 5 \), we will first rewrite the terms in a more manageable form. 1. **Rewrite the bases**: Notice that \( 16 \) can be expressed as \( 4^2 \). Therefore, we can rewrite \( 16^x \) as \( (4^2)^x = 4^{2x} \). 2. **Substitute**: The equation now becomes: \[ 4^{2x} - 1 = 5 \times 4^{x} - 5 \] 3. **Rearrange the equation**: Move all terms to one side: \[ 4^{2x} - 5 \times 4^{x} + 4 = 0 \] 4. **Let \( y = 4^{x} \)**: This substitution simplifies our equation to: \[ y^2 - 5y + 4 = 0 \] 5. **Solve the quadratic equation**: We can use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = -5, c = 4 \). Now, let's calculate the discriminant and the roots. Calculating the discriminant: \[ b^2 - 4ac = (-5)^2 - 4 \cdot 1 \cdot 4 = 25 - 16 = 9 \] Now, applying the quadratic formula: \[ y = \frac{5 \pm \sqrt{9}}{2 \cdot 1} = \frac{5 \pm 3}{2} \] Calculating the two possible values for \( y \): 1. \( y = \frac{5 + 3}{2} = \frac{8}{2} = 4 \) 2. \( y = \frac{5 - 3}{2} = \frac{2}{2} = 1 \) 6. **Back substitute for \( x \)**: - For \( y = 4 \): \[ 4^{x} = 4 \implies x = 1 \] - For \( y = 1 \): \[ 4^{x} = 1 \implies x = 0 \] Thus, the solutions to the equation \( 16^{x} - 1 = 5 \times 4^{x} - 5 \) are: \[ \boxed{0 \text{ and } 1} \]