Question paper, Set 4

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

$\displaystyle \left(\frac{2}{5}\right)^{-3} = \left(\frac{5}{2}\right)^{3} = \frac{125}{8}$

Answer: $\displaystyle \frac{125}{8}$

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

$\displaystyle \sqrt[3]{2^{9}} = 2^{\frac{9}{3}} = 2^{3} = 8$

$\displaystyle \sqrt[4]{3^{8}} = 3^{\frac{8}{4}} = 3^{2} = 9$

$8 + 9 = 17$

Answer: $17$

$= (7^{2})^{3} \times 7^{-4}$

$= 7^{6} \times 7^{-4}$

$= 7^{6 - 4} = 7^{2}$

$= 49$

Answer: $49$

$\displaystyle = \frac{(7^{2})^{\frac{3}{2}} \times (2^{6})^{\frac{1}{6}}}{(7^{3})^{\frac{2}{3}}}$

$\displaystyle = \frac{7^{3} \times 2}{7^{2}}$

$\displaystyle = \frac{343 \times 2}{49}$

$= 14$

Answer: $14$

$\displaystyle = \left[\left(\frac{3}{2}\right)^{2}\right]^{-\frac{3}{2}} + \frac{1}{\left[\left(\frac{1}{2}\right)^{3}\right]^{\frac{1}{3}}} + \frac{\sqrt{4^{2}}}{\sqrt[3]{2^{3}}}$

$\displaystyle = \left(\frac{2}{3}\right)^{3} + \frac{1}{\frac{1}{2}} + \frac{4}{2}$

$\displaystyle = \frac{8}{27} + 2 + 2$

$\displaystyle = \frac{8 + 54 + 54}{27}$

$\displaystyle = \frac{116}{27}$

Answer: $\displaystyle \frac{116}{27}$

$\displaystyle = \frac{9^{-\frac{1}{2}} \times 8^{\frac{2}{3}} \times 25^{\frac{3}{4}}}{3^{\frac{9}{5}} \times 2^{\frac{8}{5}}} \times \frac{12^{\frac{4}{5}} \times 27^{\frac{2}{3}}}{8^{\frac{1}{3}} \times 5^{\frac{3}{2}}}$

$\displaystyle = \frac{(3^{2})^{-\frac{1}{2}} \times (2^{3})^{\frac{2}{3}} \times (5^{2})^{\frac{3}{4}}}{3^{\frac{9}{5}} \times 2^{\frac{8}{5}}} \times \frac{(2^{2} \times 3)^{\frac{4}{5}} \times (3^{3})^{\frac{2}{3}}}{(2^{3})^{\frac{1}{3}} \times 5^{\frac{3}{2}}}$

$\displaystyle = \frac{3^{-1} \times 2^{2} \times 5^{\frac{3}{2}}}{3^{\frac{9}{5}} \times 2^{\frac{8}{5}}} \times \frac{2^{\frac{8}{5}} \times 3^{\frac{4}{5}} \times 3^{2}}{2 \times 5^{\frac{3}{2}}}$

$\displaystyle = 2^{2 - \frac{8}{5} - 1 + \frac{8}{5}} \times 3^{-1 - \frac{9}{5} + \frac{4}{5} + 2} \times 5^{\frac{3}{2} - \frac{3}{2}}$

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

$= 2$

Answer: $2$

LHS $\displaystyle = x^{\frac{a}{(a-b)(a-c)}} \cdot x^{\frac{b}{(b-c)(b-a)}} \cdot x^{\frac{c}{(c-a)(c-b)}}$

Power $\displaystyle = \frac{a}{(a-b)(a-c)} + \frac{b}{(b-c)(b-a)} + \frac{c}{(c-a)(c-b)}$

$\displaystyle = \frac{-a(b-c) - b(c-a) - c(a-b)}{(a-b)(b-c)(c-a)}$

$\displaystyle = \frac{-ab+ac-bc+ab-ca+bc}{(a-b)(b-c)(c-a)} = 0$

LHS $= x^{0} = 1 =$ RHS

Answer: Shown: LHS $= 1 =$ RHS

Given $a + b + c = 0 \Rightarrow a^{3} + b^{3} + c^{3} = 3abc$

LHS $\displaystyle = x^{\frac{a}{b} \cdot \frac{a}{c}} \cdot x^{\frac{b}{c} \cdot \frac{b}{a}} \cdot x^{\frac{c}{a} \cdot \frac{c}{b}}$

$\displaystyle = x^{\frac{a^{2}}{bc} + \frac{b^{2}}{ca} + \frac{c^{2}}{ab}}$

$\displaystyle = x^{\frac{a^{3}+b^{3}+c^{3}}{abc}} = x^{\frac{3abc}{abc}} = x^{3} =$ RHS

Answer: Shown: LHS $= x^{3} =$ RHS

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

$\Rightarrow 2x - 1 = 5$

$\Rightarrow x = 3$

Answer: $x = 3$

$50 = 2^{1} \times 5^{2}$

$\Rightarrow 2^{2x - 3} \times 5^{x} = 2^{1} \times 5^{2}$

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

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

$\therefore x = 2$

Answer: $x = 2$

$(10^{y})^{2} = 8^{2}$

$\Rightarrow 10^{y} = 8$

$\displaystyle 10^{-y} = \frac{1}{10^{y}} = \frac{1}{8}$

Answer: $\displaystyle \frac{1}{8}$

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

$\Rightarrow 2^{3x} = 2^{-4}$

$\Rightarrow 3x = -4$

$\displaystyle \Rightarrow x = -\frac{4}{3}$

Answer: $\displaystyle x = -\frac{4}{3}$

$(7^{2})^{x} = 7^{2x + 1} - 294$

$\Rightarrow 7^{2x} = 7^{2x + 1} - 294$

$\Rightarrow 7^{2x} = 7 \times 7^{2x} - 294$

$\Rightarrow 7^{2x}\left(7 - 1\right) = 294$

$\Rightarrow 7^{2x} \times 6 = 294$

$\displaystyle \Rightarrow 7^{2x} = 294 \times \frac{1}{6} = 49 = 7^{2}$

$\Rightarrow 2x = 2 \Rightarrow x = 1$

Answer: $x = 1$

$(3^{2})^{x} \cdot 3^{x} = 243$

$\Rightarrow 3^{2x} \cdot 3^{x} = 3^{5}$

$\Rightarrow 3^{3x} = 3^{5}$

$\displaystyle \Rightarrow 3x = 5 \Rightarrow x = \frac{5}{3}$

Answer: $\displaystyle x = \frac{5}{3}$

$7 = 5^{y} = (2^{x})^{y} = 2^{xy}$

$32 = 7^{z} = (2^{xy})^{z} = 2^{xyz}$

$2^{xyz} = 32 = 2^{5}$

$\Rightarrow xyz = 5$

Answer: $xyz = 5$

Let $5^{x} = 2^{y} = 40^{z} = k$ (say)

$\displaystyle \Rightarrow 5 = k^{\frac{1}{x}}$, $\displaystyle 2 = k^{\frac{1}{y}}$, $\displaystyle 40 = k^{\frac{1}{z}}$

$40 = 5 \times 2^{3}$

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

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

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

Answer: Shown: $\displaystyle \frac{1}{z} = \frac{1}{x} + \frac{3}{y}$

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

$\displaystyle \sqrt[4]{16^{-3}} = 16^{-\frac{3}{4}} = (2^{4})^{-\frac{3}{4}}$

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

Answer: $\displaystyle \frac{1}{8}$

$(a + b)^{3} = a^{3} + b^{3} + 3ab(a + b)$

$\displaystyle x^{3} = \left(7^{\frac{1}{3}}\right)^{3} + \left(7^{\frac{2}{3}}\right)^{3} + 3 \cdot 7^{\frac{1}{3}} \cdot 7^{\frac{2}{3}} \left(7^{\frac{1}{3}} + 7^{\frac{2}{3}}\right)$

$= 7 + 49 + 3 \times 7 \times x$

$x^{3} = 56 + 21x$

$\Rightarrow x^{3} - 21x = 56$

Answer: $56$

$= (16)^{0.35 + 0.4} = (16)^{0.75}$

$\displaystyle 0.75 = \frac{3}{4}$

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

Answer: $8$

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

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

$= 2^{-1}$

$\Rightarrow m = -1$

Answer: $m = -1$