(a) If \(9^{2x - 1} = \frac{81^{x - 2}}{3^{x}}\), find x.
(b) Without using Mathematical Tables, evaluate: \(\sqrt{\frac{0.81 \times 10^{-5}}{2.25 \times 10^{7}}}\)
(b) Without using Mathematical Tables, evaluate: \(\sqrt{\frac{0.81 \times 10^{-5}}{2.25 \times 10^{7}}}\)
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Correct Answer: Option n
Explanation:
(a) \(9^{2x - 1} = \frac{81^{x - 2}}{3^{x}}\)
\((3^{2})^{2x - 1} = \frac{(3^{4})^{x - 2}}{3^{x}}\)
\(3^{4x - 2} = \frac{3^{4x - 8}}{3^{x}}\)
\(3^{4x - 2} = 3^{4x - 8 - x}\)
\(3^{4x - 2} = 3^{3x - 8}\)
\(\therefore 4x - 2 = 3x - 8 \implies 4x - 3x = -8 + 2\)
\(x = -6\)
(b) \(\sqrt{\frac{0.81 \times 10^{-5}}{2.25 \times 10^{7}}}\)
= \(\sqrt{\frac{0.81}{2.25}} \times \sqrt{10^{-5 - 7}}\)
= \(\frac{0.9}{1.5} \times \sqrt{10^{-12}}\)
= \(0.6 \times 10^{-6}\)
= \(6 \times 10^{-7}\)
(a) \(9^{2x - 1} = \frac{81^{x - 2}}{3^{x}}\)
\((3^{2})^{2x - 1} = \frac{(3^{4})^{x - 2}}{3^{x}}\)
\(3^{4x - 2} = \frac{3^{4x - 8}}{3^{x}}\)
\(3^{4x - 2} = 3^{4x - 8 - x}\)
\(3^{4x - 2} = 3^{3x - 8}\)
\(\therefore 4x - 2 = 3x - 8 \implies 4x - 3x = -8 + 2\)
\(x = -6\)
(b) \(\sqrt{\frac{0.81 \times 10^{-5}}{2.25 \times 10^{7}}}\)
= \(\sqrt{\frac{0.81}{2.25}} \times \sqrt{10^{-5 - 7}}\)
= \(\frac{0.9}{1.5} \times \sqrt{10^{-12}}\)
= \(0.6 \times 10^{-6}\)
= \(6 \times 10^{-7}\)