Question paper, Set 2

$\displaystyle a^{m} = b^{n} \Rightarrow a = b^{\frac{n}{m}}$

$\displaystyle a = 2^{\frac{15}{3}} = 2^{5} = 32$

Answer: $a = 32$

$\displaystyle a^{m} = b^{n} \Rightarrow a = b^{\frac{n}{m}}$

$\displaystyle a = 4^{\frac{6}{3}} = 4^{2} = 16$

$\displaystyle b = 3^{\frac{6}{2}} = 3^{3} = 27$

$\displaystyle \frac{a}{b} = \frac{16}{27}$

Answer: $\displaystyle \frac{16}{27}$

$= (2^{4})^{3} \times 2^{-10}$

$= 2^{12} \times 2^{-10}$

$= 2^{12 - 10} = 2^{2}$

$= 4$

Answer: $4$

$\displaystyle = \frac{(2^{4})^{\frac{3}{4}} \times (5^{3})^{\frac{2}{3}}}{(10^{2})^{\frac{1}{2}}}$

$\displaystyle = \frac{2^{3} \times 5^{2}}{10}$

$\displaystyle = \frac{8 \times 25}{10}$

$= 20$

Answer: $20$

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

$\displaystyle = 27 - 2 \times 3^{2} + \frac{2}{5}$

$\displaystyle = 27 - 18 + \frac{2}{5}$

$\displaystyle = \frac{135 - 90 + 2}{5}$

$\displaystyle = \frac{47}{5}$

Answer: $\displaystyle \frac{47}{5}$

$\displaystyle = \frac{5^{\frac{6}{5}} \times 9^{\frac{6}{5}} \times 2^{\frac{4}{5}}}{15 \times 8^{\frac{1}{5}}} \times \frac{12^{\frac{2}{5}} \times 3^{\frac{4}{5}}}{15^{\frac{1}{5}} \times 9^{\frac{1}{5}}}$

$\displaystyle = \frac{5^{\frac{6}{5}} \times (3^{2})^{\frac{6}{5}} \times 2^{\frac{4}{5}}}{(3 \times 5) \times (2^{3})^{\frac{1}{5}}} \times \frac{(2^{2} \times 3)^{\frac{2}{5}} \times 3^{\frac{4}{5}}}{(3 \times 5)^{\frac{1}{5}} \times (3^{2})^{\frac{1}{5}}}$

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

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

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

$= 18$

Answer: $18$

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

$(a-b)(a+b-c) = a^{2}-b^{2}-ca+bc$

$(b-c)(b+c-a) = b^{2}-c^{2}-ab+ca$

$(c-a)(c+a-b) = c^{2}-a^{2}-bc+ab$

Adding these, every term cancels, so the power is $0$

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

Answer: Shown: LHS $= 1 =$ RHS

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

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

$\displaystyle = x^{\frac{2(a^{3}+b^{3}+c^{3})}{abc}}$

$\displaystyle = x^{\frac{2 \times 3abc}{abc}} = x^{6} =$ RHS

Answer: Shown: LHS $= x^{6} =$ RHS

$\displaystyle 10^{x} = 0.0001 = \frac{1}{10^{4}} = 10^{-4}$

$\Rightarrow x = -4$

$\Rightarrow x = -4$

Answer: $x = -4$

$800 = 5^{2} \times 2^{5}$

$\Rightarrow 5^{x - 1} \times 2^{3x - 4} = 5^{2} \times 2^{5}$

$\Rightarrow x - 1 = 2$ and $3x - 4 = 5$

$\Rightarrow x = 3$ and $x = 3$

$\therefore x = 3$

Answer: $x = 3$

$\displaystyle 2^{x-6} = \frac{2^{x}}{2^{6}}$

$\displaystyle = \frac{192}{64} = 3$

Answer: $3$

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

$\displaystyle \Rightarrow \left(\frac{2}{3}\right)^{\frac{x - 1}{2}} = \left(\frac{2}{3}\right)^{-3}$

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

$\Rightarrow x - 1 = -6$

$\Rightarrow x = -5$

Answer: $x = -5$

$(2^{2})^{x + 1} = 2^{2x + 3} - 64$

$\Rightarrow 2^{2x + 2} = 2^{2x + 3} - 64$

$\Rightarrow 4 \times 2^{2x} = 8 \times 2^{2x} - 64$

$\Rightarrow 2^{2x}\left(8 - 4\right) = 64$

$\Rightarrow 2^{2x} \times 4 = 64$

$\displaystyle \Rightarrow 2^{2x} = 64 \times \frac{1}{4} = 16 = 2^{4}$

$\Rightarrow 2x = 4 \Rightarrow x = 2$

Answer: $x = 2$

$(2^{2})^{x+3} = (2^{3})^{x+1}$

$\Rightarrow 2^{2x+6} = 2^{3x+3}$

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

$3^{x-1} = 3^{3-1} = 3^{2} = 9$

Answer: $9$

$7 = 5^{y} = (3^{x})^{y} = 3^{xy}$

$9 = 7^{z} = (3^{xy})^{z} = 3^{xyz}$

$3^{xyz} = 9 = 3^{2}$

$\Rightarrow xyz = 2$

Answer: $xyz = 2$

Let $2^{x} = 3^{y} = 72^{z} = k$ (say)

$\displaystyle \Rightarrow 2 = k^{\frac{1}{x}}$, $\displaystyle 3 = k^{\frac{1}{y}}$, $\displaystyle 72 = k^{\frac{1}{z}}$

$72 = 2^{3} \times 3^{2}$

$\displaystyle \Rightarrow k^{\frac{1}{z}} = \left(k^{\frac{1}{x}}\right)^{3} \times \left(k^{\frac{1}{y}}\right)^{2}$

$\displaystyle \Rightarrow k^{\frac{1}{z}} = k^{\frac{3}{x} + \frac{2}{y}}$

$\displaystyle \Rightarrow \frac{1}{z} = \frac{3}{x} + \frac{2}{y}$

Answer: Shown: $\displaystyle \frac{1}{z} = \frac{3}{x} + \frac{2}{y}$

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

$\displaystyle \sqrt[3]{27^{-2}} = 27^{-\frac{2}{3}} = (3^{3})^{-\frac{2}{3}}$

$\displaystyle = 3^{-2} = \frac{1}{3^{2}} = \frac{1}{9}$

Answer: $\displaystyle \frac{1}{9}$

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

$\displaystyle x^{3} = \left(5^{\frac{1}{3}}\right)^{3} + \left(25^{\frac{1}{3}}\right)^{3} + 3 \cdot 5^{\frac{1}{3}} \cdot 25^{\frac{1}{3}} \left(5^{\frac{1}{3}} + 25^{\frac{1}{3}}\right)$

$= 5 + 25 + 3 \times 5 \times x$

$x^{3} = 30 + 15x$

$\Rightarrow x^{3} - 15x = 30$

Answer: Shown: $x^{3} - 15x = 30$

$= (625)^{0.17 + 0.08} = (625)^{0.25}$

$\displaystyle 0.25 = \frac{1}{4}$

$\displaystyle = (5^{4})^{\frac{1}{4}} = 5$

Answer: $5$

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

$= 2^{3}$

$= 8$

Answer: $8$