Question paper, Set 3

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

$\displaystyle \sqrt[3]{5^{6}} = 5^{\frac{6}{3}} = 5^{2} = 25$

Answer: $25$

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

$\displaystyle a = 5^{\frac{6}{3}} = 5^{2} = 25$

$\displaystyle b = 2^{\frac{10}{5}} = 2^{2} = 4$

$ab = 25 \times 4 = 100$

Answer: $ab = 100$

$= (5^{3})^{2} \times 5^{-5}$

$= 5^{6} \times 5^{-5}$

$= 5^{6 - 5} = 5^{1}$

$= 5$

Answer: $5$

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

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

$\displaystyle = \frac{27 \times 4}{6}$

$= 18$

Answer: $18$

$\displaystyle = \left[\left(\frac{2}{5}\right)^{3}\right]^{-\frac{1}{3}} + 2^{2} - 3 \times (2^{4})^{\frac{1}{4}} \times 1$

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

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

$\displaystyle = \frac{5 + 8 - 12}{2}$

$\displaystyle = \frac{1}{2}$

Answer: $\displaystyle \frac{1}{2}$

$\displaystyle = \frac{8^{\frac{3}{5}} \times 5^{\frac{1}{4}} \times 27^{\frac{2}{3}}}{3^{\frac{9}{5}} \times 10^{\frac{1}{4}}} \times \frac{2^{\frac{1}{4}} \times 12^{\frac{4}{5}}}{4^{\frac{3}{5}} \times 8^{\frac{2}{5}}}$

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

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

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

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

$= 6$

Answer: $6$

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

$\displaystyle \frac{a-b}{ab} = \frac{1}{b} - \frac{1}{a}$, $\displaystyle \frac{b-c}{bc} = \frac{1}{c} - \frac{1}{b}$, $\displaystyle \frac{c-a}{ca} = \frac{1}{a} - \frac{1}{c}$

$\displaystyle = x^{\frac{1}{b} - \frac{1}{a} + \frac{1}{c} - \frac{1}{b} + \frac{1}{a} - \frac{1}{c}}$

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

Answer: Shown: LHS $= 1 =$ RHS

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

LHS $\displaystyle = x^{-\left(\frac{a^{2}}{bc} + \frac{b^{2}}{ca} + \frac{c^{2}}{ab}\right)}$

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

$\displaystyle = x^{-\frac{3abc}{abc}} = x^{-3} = \frac{1}{x^{3}} =$ RHS

Answer: Shown: LHS $\displaystyle = \frac{1}{x^{3}} =$ RHS

$5^{x + 2} = 5^{4}$

$\Rightarrow x + 2 = 4$

$\Rightarrow x = 2$

Answer: $x = 2$

$21 = 3^{1} \times 7^{1}$

$\Rightarrow 3^{2x - 5} \times 7^{x - 2} = 3^{1} \times 7^{1}$

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

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

$\therefore x = 3$

Answer: $x = 3$

$5^{x+2} = 5^{x} \times 5^{2}$

$= 15 \times 25 = 375$

Answer: $375$

$\displaystyle \left(\frac{2}{5}\right)^{x} = \left(\frac{5}{2}\right)^{-x}$

$\displaystyle \Rightarrow \left(\frac{5}{2}\right)^{-x} \cdot \left(\frac{5}{2}\right)^{3x} = \left(\frac{5}{2}\right)^{4}$

$\displaystyle \Rightarrow \left(\frac{5}{2}\right)^{2x} = \left(\frac{5}{2}\right)^{4}$

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

Answer: $x = 2$

$(2^{3})^{x - 1} = 2^{3x - 2} - 8$

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

$\displaystyle \Rightarrow \frac{2^{3x}}{8} = \frac{2^{3x}}{4} - 8$

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

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

$\Rightarrow 2^{3x} = 8 \times 8 = 64 = 2^{6}$

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

Answer: $x = 2$

$\displaystyle 3^{\frac{n}{2}} = 729 = 3^{6}$

$\displaystyle \Rightarrow \frac{n}{2} = 6 \Rightarrow n = 12$

$\displaystyle 2^{\frac{n}{4}} = 2^{\frac{12}{4}} = 2^{3} = 8$

Answer: $8$

$5 = 3^{q} = (2^{p})^{q} = 2^{pq}$

$7 = 5^{r} = (2^{pq})^{r} = 2^{pqr}$

$16 = 7^{t} = (2^{pqr})^{t} = 2^{pqrt}$

$2^{pqrt} = 16 = 2^{4}$

$\Rightarrow pqrt = 4$

Answer: $pqrt = 4$

Let $3^{x} = 5^{y} = 45^{-z} = k$ (say)

$\displaystyle \Rightarrow 3 = k^{\frac{1}{x}}$, $\displaystyle 5 = k^{\frac{1}{y}}$, $\displaystyle 45 = k^{-\frac{1}{z}}$

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

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

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

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

$\displaystyle \Rightarrow \frac{2}{x} + \frac{1}{y} + \frac{1}{z} = 0$

Answer: Shown: $\displaystyle \frac{2}{x} + \frac{1}{y} + \frac{1}{z} = 0$

$\displaystyle 0.0081 = \frac{81}{10000} = \left(\frac{3}{10}\right)^{4} = (0.3)^{4}$

$\displaystyle (0.0081)^{\frac{1}{4}} = \left[(0.3)^{4}\right]^{\frac{1}{4}} = 0.3$

Answer: $0.3$

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

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

$= 4 + 16 + 3 \times 4 \times x$

$x^{3} = 20 + 12x$

$\Rightarrow x^{3} - 12x = 20$

Answer: Shown: $x^{3} - 12x = 20$

$\displaystyle 0.5 = \frac{1}{2}$

$\displaystyle x^{2 - \frac{1}{2}} = 27 \Rightarrow x^{\frac{3}{2}} = 27$

$\displaystyle x = 27^{\frac{2}{3}} = (3^{3})^{\frac{2}{3}} = 3^{2} = 9$

Answer: $x = 9$

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

$= 5^{2}$

$= 25$

Answer: $25$