(d) \( \frac{10^{x} \cdot 25^{x-1} \cdot 2}{50^{x+1}} \)

To simplify the expression \( \frac{10^{x} \cdot 25^{x-1} \cdot 2}{50^{x+1}} \), we can first rewrite the bases in terms of prime factors. We know that \( 10 = 2 \cdot 5 \), \( 25 = 5^2 \), and \( 50 = 2 \cdot 5^2 \). We can rewrite the expression: - \( 10^x = (2 \cdot 5)^x = 2^x \cdot 5^x \) - \( 25^{x-1} = (5^2)^{x-1} = 5^{2(x-1)} = 5^{2x - 2} \) - \( 50^{x+1} = (2 \cdot 5^2)^{x+1} = 2^{x+1} \cdot 5^{2(x+1)} = 2^{x+1} \cdot 5^{2x + 2} \) Putting this all together, we have: \[ \frac{2^x \cdot 5^x \cdot 5^{2x - 2} \cdot 2}{2^{x+1} \cdot 5^{2x + 2}} \] Now combine the powers of 5 in the numerator: \[ = \frac{2^x \cdot 5^{x + 2x - 2} \cdot 2}{2^{x+1} \cdot 5^{2x + 2}} = \frac{2^{x + 1} \cdot 5^{3x - 2}}{2^{x + 1} \cdot 5^{2x + 2}} \] Now we can cancel \( 2^{x + 1} \) from the numerator and denominator: \[ = \frac{5^{3x - 2}}{5^{2x + 2}} = 5^{(3x - 2) - (2x + 2)} = 5^{x - 4} \] So the final simplified result is: \[ 5^{x - 4} \]