Question paper, Set 10

$\displaystyle \left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^{n}$

$\displaystyle \left(-\frac{1}{3}\right)^{-3} = (-3)^{3}$

$= -27$ (3 is odd)

Answer: $-27$

$\displaystyle \sqrt[5]{4^{10}} = 4^{\frac{10}{5}} = 4^{2} = 16$

$\displaystyle \sqrt[3]{6^{-3}} = 6^{\frac{-3}{3}} = 6^{-1} = \frac{1}{6}$

$\displaystyle 16 \times \frac{1}{6} = \frac{8}{3}$

Answer: $\displaystyle \frac{8}{3}$

$= (7^{3})^{2} \times (7^{2})^{-3}$

$= 7^{6} \times 7^{-6}$

$= 7^{6 - 6} = 7^{0}$

$= 1$

Answer: $1$

$\displaystyle = \frac{(2^{7})^{\frac{3}{7}} \times (3^{4})^{\frac{1}{4}}}{(12^{2})^{\frac{1}{2}}}$

$\displaystyle = \frac{2^{3} \times 3}{12}$

$\displaystyle = \frac{8 \times 3}{12}$

$= 2$

Answer: $2$

$\displaystyle = \left[\left(\frac{5}{6}\right)^{2}\right]^{-\frac{1}{2}} + \left[\left(\frac{2}{3}\right)^{3}\right]^{-\frac{2}{3}} - \frac{\sqrt{10^{2}}}{\sqrt[3]{5^{3}}}$

$\displaystyle = \frac{6}{5} + \left(\frac{3}{2}\right)^{2} - \frac{10}{5}$

$\displaystyle = \frac{6}{5} + \frac{9}{4} - 2$

$\displaystyle = \frac{24 + 45 - 40}{20}$

$\displaystyle = \frac{29}{20}$

Answer: $\displaystyle \frac{29}{20}$

$\displaystyle = \frac{5^{\frac{8}{5}} \times 8^{\frac{1}{4}} \times 3^{\frac{3}{5}}}{6^{-\frac{4}{5}} \times 15^{\frac{3}{5}}} \times \frac{27^{\frac{1}{5}} \times 12^{\frac{2}{5}}}{12^{\frac{4}{5}} \times 2^{\frac{3}{4}}}$

$\displaystyle = \frac{5^{\frac{8}{5}} \times (2^{3})^{\frac{1}{4}} \times 3^{\frac{3}{5}}}{(2 \times 3)^{-\frac{4}{5}} \times (3 \times 5)^{\frac{3}{5}}} \times \frac{(3^{3})^{\frac{1}{5}} \times (2^{2} \times 3)^{\frac{2}{5}}}{(2^{2} \times 3)^{\frac{4}{5}} \times 2^{\frac{3}{4}}}$

$\displaystyle = \frac{5^{\frac{8}{5}} \times 2^{\frac{3}{4}} \times 3^{\frac{3}{5}}}{2^{-\frac{4}{5}} \times 3^{-\frac{4}{5}} \times 3^{\frac{3}{5}} \times 5^{\frac{3}{5}}} \times \frac{3^{\frac{3}{5}} \times 2^{\frac{4}{5}} \times 3^{\frac{2}{5}}}{2^{\frac{8}{5}} \times 3^{\frac{4}{5}} \times 2^{\frac{3}{4}}}$

$\displaystyle = 2^{\frac{3}{4} + \frac{4}{5} - \frac{8}{5} - \frac{3}{4} + \frac{4}{5}} \times 3^{\frac{3}{5} + \frac{4}{5} - \frac{3}{5} - \frac{4}{5} + \frac{3}{5} + \frac{2}{5}} \times 5^{\frac{8}{5} - \frac{3}{5}}$

$= 2^{0} \times 3^{1} \times 5^{1}$

$= 15$

Answer: $15$

$= (x^{a-b})^{a+b} \cdot (x^{b-c})^{b+c}$

$= x^{a^{2}-b^{2}} \cdot x^{b^{2}-c^{2}}$

$= x^{a^{2}-b^{2}+b^{2}-c^{2}}$

$= x^{a^{2}-c^{2}}$

Answer: $x^{a^{2}-c^{2}}$

Given $p + q + r = 0 \Rightarrow p^{3} + q^{3} + r^{3} = 3pqr$

LHS $= 2^{p^{3}+q^{3}+r^{3}} = 2^{3pqr}$

$= (2^{3})^{pqr} = 8^{pqr} =$ RHS

Answer: Shown: both sides $= 2^{3pqr}$

$\displaystyle 5^{4x} = 0.04 = \frac{1}{5^{2}} = 5^{-2}$

$\Rightarrow 4x = -2$

$\displaystyle \Rightarrow x = -\frac{1}{2}$

Answer: $\displaystyle x = -\frac{1}{2}$

$56 = 7^{1} \times 2^{3}$

$\Rightarrow 7^{x - 1} \times 2^{2x - 1} = 7^{1} \times 2^{3}$

$\Rightarrow x - 1 = 1$ and $2x - 1 = 3$

$\Rightarrow x = 2$ and $x = 2$

$\therefore x = 2$

Answer: $x = 2$

$(10^{y})^{3} = 6^{3}$

$\Rightarrow 10^{y} = 6$

$\displaystyle 10^{-y} = \frac{1}{10^{y}} = \frac{1}{6}$

Answer: $\displaystyle \frac{1}{6}$

$\displaystyle (3^{3})^{2x} = \frac{1}{3^{4}}$

$\Rightarrow 3^{6x} = 3^{-4}$

$\Rightarrow 6x = -4$

$\displaystyle \Rightarrow x = -\frac{2}{3}$

Answer: $\displaystyle x = -\frac{2}{3}$

$(2^{2})^{x + 1} - 2^{2x + 1} = 128$

$\Rightarrow 2^{2x + 2} - 2^{2x + 1} = 128$

$\Rightarrow 4 \times 2^{2x} - 2 \times 2^{2x} = 128$

$\Rightarrow 2^{2x}\left(4 - 2\right) = 128$

$\Rightarrow 2^{2x} \times 2 = 128$

$\displaystyle \Rightarrow 2^{2x} = 128 \times \frac{1}{2} = 64 = 2^{6}$

$\Rightarrow 2x = 6 \Rightarrow x = 3$

Answer: $x = 3$

$(5^{3})^{x-1} = 5^{2}$

$\displaystyle \Rightarrow 5^{3x-3} = 5^{2} \Rightarrow 3x - 3 = 2 \Rightarrow x = \frac{5}{3}$

$\displaystyle 27^{x} = (3^{3})^{\frac{5}{3}} = 3^{5} = 243$

Answer: $243$

$5 = 3^{y} = (10^{x})^{y} = 10^{xy}$

$0.001 = 5^{z} = (10^{xy})^{z} = 10^{xyz}$

$10^{xyz} = 0.001 = 10^{-3}$

$\Rightarrow xyz = -3$

Answer: $xyz = -3$

Let $2^{x} = 5^{y} = 1000^{z} = k$ (say)

$\displaystyle \Rightarrow 2 = k^{\frac{1}{x}}$, $\displaystyle 5 = k^{\frac{1}{y}}$, $\displaystyle 1000 = k^{\frac{1}{z}}$

$1000 = 2^{3} \times 5^{3}$

$\displaystyle \Rightarrow k^{\frac{1}{z}} = \left(k^{\frac{1}{x}}\right)^{3} \times \left(k^{\frac{1}{y}}\right)^{3}$

$\displaystyle \Rightarrow k^{\frac{1}{z}} = k^{\frac{3}{x} + \frac{3}{y}}$

$\displaystyle \Rightarrow \frac{1}{z} = \frac{3}{x} + \frac{3}{y}$

$\displaystyle \Rightarrow \frac{1}{z} = \frac{3y + 3x}{xy} = \frac{3(x + y)}{xy}$

$\displaystyle \Rightarrow z = \frac{xy}{3(x + y)}$

Answer: Shown: $\displaystyle z = \frac{xy}{3(x + y)}$

$\displaystyle \sqrt[n]{a^{m}} = a^{\frac{m}{n}}$

$\displaystyle \sqrt[6]{64^{-5}} = 64^{-\frac{5}{6}} = (2^{6})^{-\frac{5}{6}}$

$\displaystyle = 2^{-5} = \frac{1}{2^{5}} = \frac{1}{32}$

Answer: $\displaystyle \frac{1}{32}$

$(a + b)(a^{2} + b^{2} - ab) = a^{3} + b^{3}$ with $\displaystyle a = 2^{\frac{1}{3}}$, $\displaystyle b = 3^{\frac{1}{3}}$

$\displaystyle ab = 2^{\frac{1}{3}} \cdot 3^{\frac{1}{3}} = 6^{\frac{1}{3}}$

Value $\displaystyle = (2^{\frac{1}{3}})^{3} + (3^{\frac{1}{3}})^{3} = 2 + 3 = 5$

Answer: $5$

$\displaystyle x^{\frac{1}{2} - 2} = \frac{1}{8} \Rightarrow x^{-\frac{3}{2}} = 2^{-3}$

$\displaystyle \Rightarrow x^{\frac{3}{2}} = 2^{3} = 8$

$\displaystyle x = 8^{\frac{2}{3}} = (2^{3})^{\frac{2}{3}} = 2^{2} = 4$

Answer: $x = 4$

$\displaystyle = 10^{(-3) \times 2 \times (-\frac{1}{2})}$

$= 10^{3}$

$= 1000$

Answer: $1000$