Simplify the following: (a) \( \frac{10^{x} \cdot 25^{x+1}}{5^{x} \cdot 50^{x-1}} \) (d) \( \frac{8^{n} \cdot 6^{n-3} \cdot 9^{1-n}}{16^{n-1} \cdot 3^{-n}} \) (g) \( \frac{3 \cdot 2^{x}-2^{x-1}}{2^{x}+2^{x+2}} \)

To simplify the given expressions, let's break down each one step-by-step. (a) Start with the expression: \[ \frac{10^{x} \cdot 25^{x+1}}{5^{x} \cdot 50^{x-1}} \] We can rewrite \(10\), \(25\), and \(50\) in terms of base \(5\): \[ 10^x = (2 \cdot 5)^x = 2^x \cdot 5^x \] \[ 25^{x+1} = (5^2)^{x+1} = 5^{2(x+1)} = 5^{2x + 2} \] \[ 50^{x-1} = (5^2 \cdot 2)^{x-1} = 5^{2(x-1)} \cdot 2^{x-1} = 5^{2x - 2} \cdot 2^{x-1} \] Now substitute back into the original expression: \[ \frac{2^x \cdot 5^{x} \cdot 5^{2x+2}}{5^{x} \cdot (5^{2x-2} \cdot 2^{x-1})} = \frac{2^x \cdot 5^{3x + 2}}{5^{3x - 2} \cdot 2^{x - 1}} \] This simplifies to: \[ \frac{2^x \cdot 5^{3x + 2}}{5^{3x - 2} \cdot 2^{x - 1}} = 2^{x - (x - 1)} \cdot 5^{(3x + 2) - (3x - 2)} = 2^{1} \cdot 5^{4} = 2 \cdot 625 = 1250 \] Thus the final simplified result for (a) is: \[ 1250 \] (d) For the expression: \[ \frac{8^{n} \cdot 6^{n-3} \cdot 9^{1-n}}{16^{n-1} \cdot 3^{-n}} \] Rewrite the bases: \[ 8^{n} = (2^3)^{n} = 2^{3n} \] \[ 6^{n-3} = (2 \cdot 3)^{n-3} = 2^{n-3} \cdot 3^{n-3} \] \[ 9^{1-n} = (3^2)^{1-n} = 3^{2(1-n)} = 3^{2 - 2n} \] \[ 16^{n-1} = (2^4)^{n-1} = 2^{4(n-1)} = 2^{4n - 4} \] \[ 3^{-n} = \frac{1}{3^n} \] Substitute these: \[ \frac{2^{3n} \cdot (2^{n-3} \cdot 3^{n-3}) \cdot 3^{2-2n}}{2^{4n-4} \cdot \frac{1}{3^n}} = \frac{2^{4n - 3} \cdot 3^{n - 3 + 2 - 2n}}{2^{4n - 4} \cdot 3^{-n}} \] Now simplifying: \[ = \frac{2^{4n - 3}}{2^{4n - 4}} \cdot 3^{(-n + n - 3 + 2)} = 2^{1} \cdot 3^{-1} = \frac{2}{3} \] Thus the simplified result for (d) is: \[ \frac{2}{3} \] (g) The expression to simplify is: \[ \frac{3 \cdot 2^{x}-2^{x-1}}{2^{x}+2^{x+2}} \] Start by factoring out \(2^{x-1}\) in the numerator: \[ = \frac{2^{x-1}(3 \cdot 2 - 1)}{2^x + 2^{x+2}} = \frac{2^{x-1}(6 - 1)}{2^x + 4 \cdot 2^x} = \frac{5 \cdot 2^{x