Question paper, Set 5

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

$\displaystyle x = 3^{\frac{10}{5}} = 3^{2} = 9$

Answer: $x = 9$

$\displaystyle \frac{12^{5}}{4^{5}} = \left(\frac{12}{4}\right)^{5} = 3^{5}$

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

$= 9$

Answer: $9$

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

$= 2^{10} \times 2^{-9}$

$= 2^{10 - 9} = 2^{1}$

$= 2$

Answer: $2$

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

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

$\displaystyle = \frac{32 \times 27}{8}$

$= 108$

Answer: $108$

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

$\displaystyle = 25 - 4 \times 2^{2} + \frac{3}{4}$

$\displaystyle = 25 - 16 + \frac{3}{4}$

$\displaystyle = \frac{100 - 64 + 3}{4}$

$\displaystyle = \frac{39}{4}$

Answer: $\displaystyle \frac{39}{4}$

$\displaystyle = \frac{9 \times 2^{\frac{7}{5}} \times 5^{\frac{9}{5}}}{15^{\frac{4}{5}} \times 4^{\frac{3}{5}}} \times \frac{12^{\frac{2}{5}} \times 3^{\frac{4}{5}}}{27^{\frac{2}{3}} \times 9^{\frac{1}{5}}}$

$\displaystyle = \frac{(3^{2}) \times 2^{\frac{7}{5}} \times 5^{\frac{9}{5}}}{(3 \times 5)^{\frac{4}{5}} \times (2^{2})^{\frac{3}{5}}} \times \frac{(2^{2} \times 3)^{\frac{2}{5}} \times 3^{\frac{4}{5}}}{(3^{3})^{\frac{2}{3}} \times (3^{2})^{\frac{1}{5}}}$

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

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

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

$= 10$

Answer: $10$

LHS $= x^{(a-b)(a^{2}+ab+b^{2})} \cdot x^{(b-c)(b^{2}+bc+c^{2})} \cdot x^{(c-a)(c^{2}+ca+a^{2})}$

$(a-b)(a^{2}+ab+b^{2}) = a^{3}+a^{2}b+ab^{2}-a^{2}b-ab^{2}-b^{3} = a^{3}-b^{3}$

In the same way the other powers are $b^{3}-c^{3}$ and $c^{3}-a^{3}$

LHS $= x^{a^{3}-b^{3}+b^{3}-c^{3}+c^{3}-a^{3}} = x^{0} = 1 =$ RHS

Answer: Shown: LHS $= 1 =$ RHS

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

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

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

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

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

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

$\Rightarrow x - 3 = -3$

$\Rightarrow x = 0$

Answer: $x = 0$

$54 = 3^{3} \times 2^{1}$

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

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

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

$\therefore x = 2$

Answer: $x = 2$

$\displaystyle 7^{x-2} = \frac{7^{x}}{7^{2}}$

$\displaystyle = \frac{98}{49} = 2$

Answer: $2$

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

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

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

$\Rightarrow x + 3 = -4$

$\Rightarrow x = -7$

Answer: $x = -7$

$(3^{3})^{x - 1} = 3^{3x - 1} - 216$

$\Rightarrow 3^{3x - 3} = 3^{3x - 1} - 216$

$\displaystyle \Rightarrow \frac{3^{3x}}{27} = \frac{3^{3x}}{3} - 216$

$\displaystyle \Rightarrow 3^{3x}\left(\frac{1}{3} - \frac{1}{27}\right) = 216$

$\displaystyle \Rightarrow 3^{3x} \times \frac{8}{27} = 216$

$\displaystyle \Rightarrow 3^{3x} = 216 \times \frac{27}{8} = 729 = 3^{6}$

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

Answer: $x = 2$

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

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

$\Rightarrow 4x - 2 = 3x + 3 \Rightarrow x = 5$

$2^{x-2} = 2^{5-2} = 2^{3} = 8$

Answer: $8$

$c = b^{y} = (a^{x})^{y} = a^{xy}$

$a^{2} = c^{z} = (a^{xy})^{z} = a^{xyz}$

$\Rightarrow xyz = 2$

Answer: $xyz = 2$

Let $4^{x} = 6^{y} = 24^{z} = k$ (say)

$\displaystyle \Rightarrow 4 = k^{\frac{1}{x}}$, $\displaystyle 6 = k^{\frac{1}{y}}$, $\displaystyle 24 = k^{\frac{1}{z}}$

$24 = 4 \times 6$

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

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

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

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

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

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

$\displaystyle 0.343 = \frac{343}{1000} = \left(\frac{7}{10}\right)^{3} = (0.7)^{3}$

$\displaystyle \sqrt[3]{0.343} = (0.343)^{\frac{1}{3}} = \left[(0.7)^{3}\right]^{\frac{1}{3}} = 0.7$

Answer: $0.7$

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

$\displaystyle ab = x^{\frac{1}{3}} \cdot x^{-\frac{1}{3}} = x^{0} = 1$

So the value $\displaystyle = (x^{\frac{1}{3}})^{3} + (x^{-\frac{1}{3}})^{3} = x + \frac{1}{x}$

$\displaystyle = 5 + \frac{1}{5} = \frac{26}{5}$

Answer: $\displaystyle \frac{26}{5}$

$\displaystyle 1.25 = \frac{5}{4}$

$\displaystyle \frac{x}{x^{\frac{5}{4}}} = \frac{8}{x} \Rightarrow \frac{x^{2}}{x^{\frac{5}{4}}} = 8$

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

$\displaystyle \Rightarrow x = 8^{\frac{4}{3}} = (2^{3})^{\frac{4}{3}} = 2^{4} = 16$

Answer: $x = 16$

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

$\displaystyle = \left\{\frac{3^{-2}}{3^{-4}}\right\}^{\frac{1}{2}} = \left\{3^{2}\right\}^{\frac{1}{2}}$

$= 3^{1} = 3$

Answer: $3$