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Question 25 : If 9x - (\\frac{1}{2})) – 22x – 2 = 4x – 32x – 3, then x is
Given that 9x - (\\frac{1}{2})) – 22x – 2 = 4x – 32x – 3
⟹ \\frac{9^x}{9^{\frac{1}{2}}}) - \\frac{2^{2x}}{2^2}) = 22x - \\frac{3^{2x}}{3^3})
⟹ \\frac{3^{2x}}{3}) - \\frac{2^{2x}}{4}) = 22x - \\frac{3^{2x}}{27})
⟹ \\frac{(9 × 3^{2x}) + 3^{2x}}{27}) = \\frac{(4 × 2^{2x}) + 2^{2x}}{4})
⟹ 32x × \\frac{10}{27}) = 5 × \\frac{2^{2x}}{4})
⟹ 32x × 8 = 22x × 27
⟹ 32x - 3 = 22x - 3
⟹ 2x - 3 = 0
⟹ x = \\frac{3}{2})
The question is "If 9x - (\\frac{1}{2})) – 22x – 2 = 4x – 32x – 3, then x is"
Choice A is the correct answer
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