Exponents (Level 2): MCQ test, Set 6

Download the PDF (with key sheet and solutions, 95.2 KB)

Choose one answer for each question, then check your score at the end. Each correct answer carries 1 mark; there is no negative marking.

$\sqrt{5} \times \sqrt[3]{5} =$
A $\displaystyle 5^{\frac{1}{6}}$ B $\displaystyle 5^{\frac{2}{5}}$ C $\displaystyle 5^{\frac{5}{6}}$ D $\displaystyle 25^{\frac{5}{6}}$

Answer and solution

Answer: (C) $\displaystyle 5^{\frac{5}{6}}$

$\displaystyle \sqrt{5} \times \sqrt[3]{5} = 5^{\frac{1}{2}} \times 5^{\frac{1}{3}}$

$\displaystyle = 5^{\frac{1}{2} + \frac{1}{3}} = 5^{\frac{5}{6}}$
$\displaystyle \left(\frac{1}{10}\right)^{-3} =$
A $1000$ B $\displaystyle \frac{1}{1000}$ C $-30$ D $-1000$

Answer and solution

Answer: (A) $1000$

$\displaystyle \left(\frac{1}{10}\right)^{-3} = 10^{3} = 1000$
$100^{2} \times 10^{-5} =$
A $\displaystyle \frac{1}{10}$ B $10$ C $\displaystyle \frac{1}{1000000}$ D $10^{9}$

Answer and solution

Answer: (A) $\displaystyle \frac{1}{10}$

$= (10^{2})^{2} \times 10^{-5}$

$= 10^{4} \times 10^{-5}$

$= 10^{4 - 5} = 10^{-1}$

$\displaystyle = \frac{1}{10^{1}} = \frac{1}{10}$
$\displaystyle (625)^{\frac{3}{4}} \div (27)^{\frac{2}{3}} =$
A $\displaystyle \frac{125}{3}$ B $\displaystyle \frac{125}{9}$ C $\displaystyle \frac{9}{125}$ D $\displaystyle \frac{5}{3}$

Answer and solution

Answer: (B) $\displaystyle \frac{125}{9}$

$\displaystyle (5^{4})^{\frac{3}{4}} \div (3^{3})^{\frac{2}{3}}$

$= 5^{3} \div 3^{2}$

$\displaystyle = 125 \div 9 = \frac{125}{9}$
$\displaystyle \left(\frac{1}{4}\right)^{-\frac{1}{2}} + \left(\frac{8}{27}\right)^{-\frac{1}{3}} =$
A $\displaystyle \frac{8}{3}$ B $\displaystyle -\frac{7}{2}$ C $\displaystyle \frac{7}{6}$ D $\displaystyle \frac{7}{2}$

Answer and solution

Answer: (D) $\displaystyle \frac{7}{2}$

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

$\displaystyle \left(\frac{8}{27}\right)^{-\frac{1}{3}} = \left[\left(\frac{2}{3}\right)^{3}\right]^{-\frac{1}{3}} = \frac{3}{2}$

$\displaystyle 2 + \frac{3}{2} = \frac{7}{2}$
$\displaystyle \left(\frac{9}{4}\right)^{-\frac{1}{2}} + \left(\frac{1}{27}\right)^{-\frac{1}{3}} =$
A $\displaystyle -\frac{11}{3}$ B $\displaystyle \frac{11}{3}$ C $1$ D $\displaystyle \frac{11}{6}$

Answer and solution

Answer: (B) $\displaystyle \frac{11}{3}$

$\displaystyle \left(\frac{9}{4}\right)^{-\frac{1}{2}} = \left[\left(\frac{3}{2}\right)^{2}\right]^{-\frac{1}{2}} = \frac{2}{3}$

$\displaystyle \left(\frac{1}{27}\right)^{-\frac{1}{3}} = \left[\left(\frac{1}{3}\right)^{3}\right]^{-\frac{1}{3}} = 3$

$\displaystyle \frac{2}{3} + 3 = \frac{11}{3}$
$\displaystyle \frac{12^{\frac{4}{3}} \times 9^{\frac{1}{3}}}{2^{\frac{2}{3}}} =$
A $9$ B $4$ C $72$ D $36$

Answer and solution

Answer: (D) $36$

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

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

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

$= 36$
$\displaystyle \left(\frac{2^{6}}{2^{2}}\right)^{\frac{1}{12}} \times \left(\frac{2^{2}}{2^{3}}\right)^{\frac{1}{6}} \times \left(\frac{2^{3}}{2^{6}}\right)^{\frac{1}{18}} =$
A $1$ B $\displaystyle 2^{\frac{2}{3}}$ C $0$ D $2$

Answer and solution

Answer: (A) $1$

Powers: $\displaystyle (6 - 2)(\frac{1}{12}) + (2 - 3)(\frac{1}{6}) + (3 - 6)(\frac{1}{18})$

$\displaystyle = \frac{1}{3} - \frac{1}{6} - \frac{1}{6} = 0$

Value $= 2^{0} = 1$
If $a + b + c = 0$ , then $5^{a^{3}} \times 5^{b^{3}} \times 5^{c^{3}} =$
A $1$ B $5^{abc}$ C $15^{abc}$ D $125^{abc}$

Answer and solution

Answer: (D) $125^{abc}$

$a^{3} + b^{3} + c^{3} = 3abc$

$5^{a^{3}+b^{3}+c^{3}} = 5^{3abc} = (5^{3})^{abc} = 125^{abc}$
If $\displaystyle \frac{3^{6x} \times 9^{2} \times 27}{3^{2x}} = 3^{3}$ , then $x =$
A $\displaystyle -\frac{1}{2}$ B $-4$ C $\displaystyle \frac{5}{2}$ D $-1$

Answer and solution

Answer: (D) $-1$

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

$\Rightarrow 3^{4x} \times 3^{4} \times 3^{3} = 3^{3}$

$\Rightarrow 3^{4x + 7} = 3^{3}$

$\Rightarrow 4x + 7 = 3 \Rightarrow 4x = -4 \Rightarrow x = -1$
If $5^{2x} = 625$ and $10^{y} = 0.0001$ , then $x + y =$
A $0$ B $2$ C $-2$ D $6$

Answer and solution

Answer: (C) $-2$

$5^{2x} = 625 = 5^{4} \Rightarrow 2x = 4 \Rightarrow x = 2$

$10^{y} = 0.0001 = 10^{-4} \Rightarrow y = -4$

$x + y = 2 + (-4) = -2$
If $2^{x + 1} \times 5^{x - 2} = 80$ , then $x =$
A $6$ B $3$ C $4$ D $0$

Answer and solution

Answer: (B) $3$

$80 = 2^{4} \times 5^{1}$

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

$\Rightarrow x + 1 = 4$ and $x - 2 = 1$

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

$\therefore x = 3$
If $10^{2y} = 81$ , then $10^{y} =$
A $81$ B $\displaystyle \frac{81}{2}$ C $\displaystyle \frac{1}{9}$ D $9$

Answer and solution

Answer: (D) $9$

$(10^{y})^{2} = 9^{2}$

$\Rightarrow 10^{y} = 9$
If $\displaystyle \left(\frac{2}{7}\right)^{3x} \left(\frac{7}{2}\right)^{x} = \frac{16}{2401}$ , then $x =$
A $2$ B $1$ C $4$ D $-2$

Answer and solution

Answer: (A) $2$

$\displaystyle \left(\frac{7}{2}\right)^{x} = \left(\frac{2}{7}\right)^{-x}$

$\displaystyle \Rightarrow \left(\frac{2}{7}\right)^{3x} \cdot \left(\frac{2}{7}\right)^{-x} = \left(\frac{2}{7}\right)^{4}$

$\displaystyle \Rightarrow \left(\frac{2}{7}\right)^{2x} = \left(\frac{2}{7}\right)^{4}$

$\Rightarrow 2x = 4 \Rightarrow x = 2$
If $3^{x + 2} - 3^{x} = 216$ , then $x =$
A $4$ B $3$ C $2$ D $-3$

Answer and solution

Answer: (B) $3$

$9 \times 3^{x} - 3^{x} = 216$

$\Rightarrow 3^{x}\left(9 - 1\right) = 216$

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

$\displaystyle \Rightarrow 3^{x} = 216 \times \frac{1}{8} = 27 = 3^{3}$

$\Rightarrow x = 3$
If $\displaystyle 125^{x} = \frac{25}{5^{x}}$ , then $x =$
A $\displaystyle -\frac{1}{2}$ B $\displaystyle \frac{1}{2}$ C $\displaystyle \frac{2}{3}$ D $1$

Answer and solution

Answer: (B) $\displaystyle \frac{1}{2}$

$(5^{3})^{x} \cdot 5^{x} = 5^{2} \Rightarrow 5^{4x} = 5^{2}$

$\displaystyle \Rightarrow 4x = 2 \Rightarrow x = \frac{1}{2}$
$\displaystyle \frac{1}{1 + x^{a-b}} - \frac{1}{1 + x^{b-a}} =$
A $\displaystyle \frac{x^{b} - x^{a}}{x^{a} + x^{b}}$ B $0$ C $\displaystyle \frac{x^{a} - x^{b}}{x^{a} + x^{b}}$ D $1$

Answer and solution

Answer: (A) $\displaystyle \frac{x^{b} - x^{a}}{x^{a} + x^{b}}$

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

$\displaystyle = \frac{x^{b} - x^{a}}{x^{a} + x^{b}}$
$\displaystyle \left(\frac{x^{a^{2}+b^{2}}}{x^{ab}}\right)^{a+b} \div x^{b^{3}} =$
A $1$ B $x^{a^{3}+2b^{3}}$ C $x^{a^{3}}$ D $x^{a^{3}-b^{3}}$

Answer and solution

Answer: (C) $x^{a^{3}}$

$\displaystyle \left(\frac{x^{a^{2}+b^{2}}}{x^{ab}}\right)^{a+b} = x^{(a+b)(a^{2}-ab+b^{2})} = x^{a^{3}+b^{3}}$

$x^{a^{3}+b^{3}} \div x^{b^{3}} = x^{a^{3}}$
If $7^{x} = 2$ , $2^{y} = 5$ and $\displaystyle 5^{z} = \frac{1}{49}$ , then $xyz =$
A $1$ B $\displaystyle \frac{1}{49}$ C $-2$ D $2$

Answer and solution

Answer: (C) $-2$

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

$\displaystyle \frac{1}{49} = 5^{z} = (7^{xy})^{z} = 7^{xyz}$

$\displaystyle 7^{xyz} = \frac{1}{49} = 7^{-2}$

$\Rightarrow xyz = -2$
If $3^{x} = 4^{y} = 12^{z}$ , then $\displaystyle \frac{1}{x} + \frac{1}{y} =$
A $\displaystyle \frac{1}{z}$ B $0$ C $\displaystyle \frac{2}{z}$ D $z$

Answer and solution

Answer: (A) $\displaystyle \frac{1}{z}$

Let each $= k$ : $\displaystyle 3 = k^{\frac{1}{x}}$ , $\displaystyle 4 = k^{\frac{1}{y}}$ , $\displaystyle 12 = k^{\frac{1}{z}}$

$12 = 3 \times 4$ , so $\displaystyle \frac{1}{z} = \frac{1}{x} + \frac{1}{y}$
$\sqrt[5]{32^{-4}} =$
A $\displaystyle \frac{1}{2}$ B $16$ C $\displaystyle \frac{1}{16}$ D $-16$

Answer and solution

Answer: (C) $\displaystyle \frac{1}{16}$

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

$\displaystyle \sqrt[5]{32^{-4}} = 32^{-\frac{4}{5}} = (2^{5})^{-\frac{4}{5}}$

$\displaystyle = 2^{-4} = \frac{1}{2^{4}} = \frac{1}{16}$
$\displaystyle \left(\frac{\sqrt{2}}{3}\right)^{9} \div \left(\frac{\sqrt{2}}{3}\right)^{12} =$
A $\displaystyle \frac{2 \sqrt{2}}{27}$ B $\displaystyle \frac{8}{27}$ C $\displaystyle \frac{27}{8}$ D $\displaystyle \frac{27 \sqrt{2}}{4}$

Answer and solution

Answer: (D) $\displaystyle \frac{27 \sqrt{2}}{4}$

$\displaystyle \frac{a^{m}}{a^{n}} = \frac{1}{a^{n-m}}$

$\displaystyle \left(\frac{\sqrt{2}}{3}\right)^{9} \div \left(\frac{\sqrt{2}}{3}\right)^{12} = \frac{1}{\left(\frac{\sqrt{2}}{3}\right)^{3}}$

$\displaystyle = \frac{27 \sqrt{2}}{4}$
$\displaystyle (3^{\frac{1}{3}} + 5^{\frac{1}{3}})(3^{\frac{2}{3}} - 15^{\frac{1}{3}} + 5^{\frac{2}{3}}) =$
A $2$ B $15$ C $8$ D $34$

Answer and solution

Answer: (C) $8$

$(a + b)(a^{2} + b^{2} - ab) = a^{3} + b^{3}$ with $\displaystyle a = 3^{\frac{1}{3}}$ , $\displaystyle b = 5^{\frac{1}{3}}$

$= 3 + 5 = 8$
If $x^{1.5} = 0.008$ and $x > 0$ , then $x =$
A $0.2$ B $0.4$ C $0.04$ D $0.004$

Answer and solution

Answer: (C) $0.04$

$\displaystyle x^{\frac{3}{2}} = 0.008 = (0.2)^{3}$

$\displaystyle x = \left[(0.2)^{3}\right]^{\frac{2}{3}} = (0.2)^{2} = 0.04$
If $\displaystyle \left[\left\{\left(\frac{1}{3^{2}}\right)^{-3}\right\}^{\frac{1}{2}}\right]^{\frac{1}{3}} = 3^{m}$ , then $m =$
A $-2$ B $1$ C $3$ D $-1$

Answer and solution

Answer: (B) $1$

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

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

$= 3^{1}$

$\Rightarrow m = 1$

Menu

MCQ test, Set 6