If \(\log_{3} x = \log_{9} 3\), find the value of x.
\(3^{2}\)
\(3^{\frac{1}{2}}\)
\(3^{\frac{1}{3}}\)
\(2^{13}\)
Correct answer is B
\(\log_{3} x = \log_{9} 3 \implies \log_{3} x = \log_{9} 9^{\frac{1}{2}} = \frac{1}{2}\log_{9} 9\)
\(\log_{3} x = \frac{1}{2} \)
\(\therefore x = 3^{\frac{1}{2}}\)
(1, 2)
(1, -2)
(-1, 2)
(-1, -2)
Correct answer is A
\(4(2^{x^2}) = 8^{x} \equiv (2^{2})(2^{x^2}) = (2^{3})^{x}\)
\(\implies 2^{2 + x^{2}} = 2^{3x}\)
Comparing bases, we have
\(2 + x^{2} = 3x \implies x^{2} - 3x + 2 = 0\)
\(x^{2} - 2x - x + 2 = 0 \)
\(x(x - 2) - 1(x - 2) = 0\)
\((x - 1) = 0\) or \((x - 2) = 0\)
\(x = \text{1 or 2}\)
\((\frac{2\pi}{3}, \frac{4\pi}{3})\)
\((\frac{\pi}{6}, \frac{5\pi}{6})\)
\((\frac{\pi}{5}, \frac{2\pi}{5})\)
\((\frac{\pi}{3}, \frac{5\pi}{3})\)
Correct answer is D
\(2\cos x - 1 = 0 \implies 2\cos x = 1\)
\(\cos x = \frac{1}{2}\)
\(x = \cos^{-1} (\frac{1}{2})\)
= \(\frac{\pi}{3}\) = \(\frac{5\pi}{3}\)
Simplify \(\frac{1 - 2\sqrt{5}}{2 + 3\sqrt{2}}\).
\(14(2\sqrt{2} + 6\sqrt{5} - 4\sqrt{10})\)
\(\frac{1}{14}(2 - 3\sqrt{2} - 4\sqrt{5} - 6\sqrt{10})\)
\(\frac{1}{14}(3\sqrt{2} + 4\sqrt{5} - 6\sqrt{10} - 2)\)
\(14(2 + 3\sqrt{2} - 6\sqrt{5} + 4\sqrt{10})\)
Correct answer is C
\(\frac{1 - 2\sqrt{5}}{2 + 3\sqrt{2}} = (\frac{1 - 2\sqrt{5}}{2 + 3\sqrt{2}})(\frac{2 - 3\sqrt{2}}{2 - 3\sqrt{2}})\)
= \(\frac{2 - 3\sqrt{2} - 4\sqrt{5} + 6\sqrt{10}}{4 - 6\sqrt{2} + 6\sqrt{2} - 18}\)
= \(\frac{2 - 3\sqrt{2} - 4\sqrt{5} + 6\sqrt{10}}{-14}\)
= \(\frac{1}{14}(3\sqrt{2} + 4\sqrt{5} - 2 - 6\sqrt{10})\) (dividing through with the minus sign)
9\(ms^{-1}\)
9.49\(ms^{-1}\)
13.42\(ms^{-1}\)
18.97\(ms^{-1}\)
Correct answer is D
Due to conservation of energy, K.E = P.E
\(\frac{1}{2}mv^{2} = mgh\)
\(\implies m \times 10 \times 18 = \frac{1}{2}mv^{2}\)
\(v^{2} = 2 \times 10 \times 18 = 360\)
\(v = \sqrt{360} = 18.97ms^{-1}\)