If log3x=log93, find the value of x.
32
312
313
213
Correct answer is B
log3x=log93⟹log3x=log9912=12log99
log3x=12
∴
(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)
9ms^{-1}
9.49ms^{-1}
13.42ms^{-1}
18.97ms^{-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}