Question paper, Set 6

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

$\displaystyle 6^{\frac{2}{7}} = \sqrt[7]{6^{2}} = \sqrt[7]{36}$

Answer: $\sqrt[7]{6^{2}} = \sqrt[7]{36}$

$\displaystyle \left(\frac{3}{2}\right)^{-2} = \left(\frac{2}{3}\right)^{2} = \frac{4}{9}$

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

$\displaystyle \frac{4}{9} + 4 = \frac{40}{9}$

Answer: $\displaystyle \frac{40}{9}$

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

$= 3^{8} \times 3^{-6}$

$= 3^{8 - 6} = 3^{2}$

$= 9$

Answer: $9$

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

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

$\displaystyle = \frac{125 \times 32}{100}$

$= 40$

Answer: $40$

$\displaystyle = \left[\left(\frac{2}{3}\right)^{4}\right]^{-\frac{3}{4}} + \frac{1}{\left[\left(\frac{3}{4}\right)^{3}\right]^{\frac{1}{3}}} - \frac{\sqrt{8^{2}}}{\sqrt[3]{4^{3}}}$

$\displaystyle = \left(\frac{3}{2}\right)^{3} + \frac{1}{\frac{3}{4}} - \frac{8}{4}$

$\displaystyle = \frac{27}{8} + \frac{4}{3} - 2$

$\displaystyle = \frac{81 + 32 - 48}{24}$

$\displaystyle = \frac{65}{24}$

Answer: $\displaystyle \frac{65}{24}$

$\displaystyle = \frac{27^{\frac{1}{3}} \times 5^{\frac{2}{5}} \times 6^{\frac{6}{5}}}{25^{\frac{4}{5}} \times 9^{-\frac{3}{5}}} \times \frac{4^{\frac{1}{2}} \times 25^{\frac{3}{5}}}{2^{\frac{6}{5}} \times 3^{\frac{7}{5}}}$

$\displaystyle = \frac{(3^{3})^{\frac{1}{3}} \times 5^{\frac{2}{5}} \times (2 \times 3)^{\frac{6}{5}}}{(5^{2})^{\frac{4}{5}} \times (3^{2})^{-\frac{3}{5}}} \times \frac{(2^{2})^{\frac{1}{2}} \times (5^{2})^{\frac{3}{5}}}{2^{\frac{6}{5}} \times 3^{\frac{7}{5}}}$

$\displaystyle = \frac{3 \times 5^{\frac{2}{5}} \times 2^{\frac{6}{5}} \times 3^{\frac{6}{5}}}{5^{\frac{8}{5}} \times 3^{-\frac{6}{5}}} \times \frac{2 \times 5^{\frac{6}{5}}}{2^{\frac{6}{5}} \times 3^{\frac{7}{5}}}$

$\displaystyle = 2^{\frac{6}{5} - \frac{6}{5} + 1} \times 3^{1 + \frac{6}{5} + \frac{6}{5} - \frac{7}{5}} \times 5^{\frac{2}{5} - \frac{8}{5} + \frac{6}{5}}$

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

$= 18$

Answer: $18$

LHS $\displaystyle = \frac{x^{(a+b)(a-b)}}{x^{2c(a-b)}} \cdot \frac{x^{(b+c)(b-c)}}{x^{2a(b-c)}} \cdot \frac{x^{(c+a)(c-a)}}{x^{2b(c-a)}}$

$\displaystyle = \frac{x^{a^{2}-b^{2}} \cdot x^{b^{2}-c^{2}} \cdot x^{c^{2}-a^{2}}}{x^{2ca-2cb} \cdot x^{2ab-2ac} \cdot x^{2bc-2ba}}$

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

$\displaystyle = \frac{x^{0}}{x^{0}} = 1 =$ RHS

Answer: Shown: LHS $= 1 =$ RHS

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

LHS $\displaystyle = 9^{\frac{a^{2}}{bc} + \frac{b^{2}}{ca} + \frac{c^{2}}{ab}}$

$\displaystyle = 9^{\frac{a^{3}+b^{3}+c^{3}}{abc}}$

$\displaystyle = 9^{\frac{3abc}{abc}} = 9^{3} = 729 =$ RHS

Answer: Shown: LHS $= 9^{3} = 729 =$ RHS

$6^{x + 1} = 6^{3}$

$\Rightarrow x + 1 = 3$

$\Rightarrow x = 2$

Answer: $x = 2$

$45 = 5^{1} \times 3^{2}$

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

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

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

$\therefore x = 3$

Answer: $x = 3$

$8^{x} = (2^{3})^{x} = (2^{x})^{3}$

$= 7^{3} = 343$

Answer: $343$

$\displaystyle \left(\frac{4}{3}\right)^{x} = \left(\frac{3}{4}\right)^{-x}$

$\displaystyle \Rightarrow \left(\frac{3}{4}\right)^{2x} \cdot \left(\frac{3}{4}\right)^{-x} = \left(\frac{3}{4}\right)^{3}$

$\displaystyle \Rightarrow \left(\frac{3}{4}\right)^{x} = \left(\frac{3}{4}\right)^{3}$

$\Rightarrow x = 3$

Answer: $x = 3$

$(6^{2})^{x - 1} = 6^{2x - 1} - 180$

$\Rightarrow 6^{2x - 2} = 6^{2x - 1} - 180$

$\displaystyle \Rightarrow \frac{6^{2x}}{36} = \frac{6^{2x}}{6} - 180$

$\displaystyle \Rightarrow 6^{2x}\left(\frac{1}{6} - \frac{1}{36}\right) = 180$

$\displaystyle \Rightarrow 6^{2x} \times \frac{5}{36} = 180$

$\displaystyle \Rightarrow 6^{2x} = 180 \times \frac{36}{5} = 1296 = 6^{4}$

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

Answer: $x = 2$

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

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

$\displaystyle 4^{x} = 4^{\frac{5}{2}} = (2^{2})^{\frac{5}{2}} = 2^{5} = 32$

Answer: $32$

$5 = 4^{y} = (3^{x})^{y} = 3^{xy}$

$\displaystyle \frac{1}{9} = 5^{z} = (3^{xy})^{z} = 3^{xyz}$

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

$\Rightarrow xyz = -2$

Answer: $xyz = -2$

Let $9^{x} = 4^{y} = 6^{z} = k$ (say)

$\displaystyle \Rightarrow 9 = k^{\frac{1}{x}}$, $\displaystyle 4 = k^{\frac{1}{y}}$, $\displaystyle 6 = k^{\frac{1}{z}}$

$\displaystyle 6 = 3 \times 2 = 9^{\frac{1}{2}} \times 4^{\frac{1}{2}}$

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

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

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

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

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

$\displaystyle \sqrt[5]{243^{-3}} = 243^{-\frac{3}{5}} = (3^{5})^{-\frac{3}{5}}$

$\displaystyle = 3^{-3} = \frac{1}{3^{3}} = \frac{1}{27}$

Answer: $\displaystyle \frac{1}{27}$

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

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

$= 9 + 81 + 3 \times 9 \times x$

$x^{3} = 90 + 27x$

$\Rightarrow x^{3} - 27x = 90$

Answer: Shown: $x^{3} - 27x = 90$

$= (1024)^{0.07 + 0.13} = (1024)^{0.2}$

$\displaystyle 0.2 = \frac{1}{5}$

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

Answer: $4$

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

$= 7^{1}$

$= 7$

Answer: $7$