If \(\frac{27^x \times 3^{1 - x}}{9^{2x}} = 1\), find the value of x.
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Correct Answer: Option B
Explanation:
\(\frac{27^x \times 3^{1 - x}}{9^{2x}} = 1\)
\(\frac{3^{3x} \times 3^{1 - x}}{3^{2(2 - x)}} = 3^0\)
\(3^{3x} \times 3^{1 - x} \div 3^{4x} = 3^0\)
\(3^{(3x + 1 - x - 4x)} = 3^0\)
\(3^{(1 - 2x)} = 3^0\)
since the bases are equal,
1 - 2x = 0
- 2x = -1
x = \(\frac{1}{2}\)
\(\frac{27^x \times 3^{1 - x}}{9^{2x}} = 1\)
\(\frac{3^{3x} \times 3^{1 - x}}{3^{2(2 - x)}} = 3^0\)
\(3^{3x} \times 3^{1 - x} \div 3^{4x} = 3^0\)
\(3^{(3x + 1 - x - 4x)} = 3^0\)
\(3^{(1 - 2x)} = 3^0\)
since the bases are equal,
1 - 2x = 0
- 2x = -1
x = \(\frac{1}{2}\)