Exponents (Level 2): MCQ test, Set 2

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

$\sqrt[4]{5^{3}} =$
A $\displaystyle 5^{\frac{3}{4}}$ B $5^{-1}$ C $5^{12}$ D $\displaystyle 5^{\frac{4}{3}}$

Answer and solution

Answer: (A) $\displaystyle 5^{\frac{3}{4}}$

$\displaystyle \sqrt[n]{a^{m}} = a^{\frac{m}{n}}$ , so $\displaystyle \sqrt[4]{5^{3}} = 5^{\frac{3}{4}}$
If $a^{5} = 3^{20}$ and $a > 0$ , then $a =$
A $81$ B $3^{100}$ C $\displaystyle 3^{\frac{1}{4}}$ D $3^{15}$

Answer and solution

Answer: (A) $81$

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

$\displaystyle a = 3^{\frac{20}{5}} = 3^{4} = 81$
$8^{4} \times 2^{-11} =$
A $8^{-7}$ B $\displaystyle \frac{1}{2}$ C $2$ D $2^{23}$

Answer and solution

Answer: (C) $2$

$= (2^{3})^{4} \times 2^{-11}$

$= 2^{12} \times 2^{-11}$

$= 2^{12 - 11} = 2^{1}$

$= 2$
$\displaystyle (81)^{\frac{3}{4}} \div (32)^{\frac{3}{5}} =$
A $\displaystyle \frac{8}{27}$ B $\displaystyle \frac{27}{8}$ C $\displaystyle \frac{27}{2}$ D $\displaystyle \frac{3}{2}$

Answer and solution

Answer: (B) $\displaystyle \frac{27}{8}$

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

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

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

Answer and solution

Answer: (D) $\displaystyle \frac{1}{\sqrt{3}}$

$\sqrt[3]{27} = 3$

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

$\displaystyle = \frac{1}{3^{\frac{1}{2}}} = \frac{1}{\sqrt{3}}$
$\displaystyle \left(\frac{16}{81}\right)^{-\frac{3}{4}} =$
A $\displaystyle -\frac{27}{8}$ B $\displaystyle \frac{8}{27}$ C $\displaystyle \frac{27}{8}$ D $\displaystyle \frac{3}{2}$

Answer and solution

Answer: (C) $\displaystyle \frac{27}{8}$

$\displaystyle \left(\frac{16}{81}\right)^{-\frac{3}{4}} = \left[\left(\frac{2}{3}\right)^{4}\right]^{-\frac{3}{4}}$

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

$\displaystyle = \frac{27}{8}$
$\displaystyle \frac{8^{\frac{1}{2}} \times 25^{-\frac{1}{4}}}{10^{-\frac{1}{2}}} =$
A $4$ B $1$ C $\displaystyle \frac{1}{4}$ D $8$

Answer and solution

Answer: (A) $4$

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

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

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

$= 4$
$\displaystyle \left(\frac{5^{4}}{5^{2}}\right)^{3} \times \left(\frac{5^{2}}{5^{3}}\right)^{4} \times \left(\frac{5^{3}}{5^{4}}\right)^{2} =$
A $1$ B $0$ C $5^{12}$ D $5$

Answer and solution

Answer: (A) $1$

Powers: $(4 - 2)(3) + (2 - 3)(4) + (3 - 4)(2)$

$= 6 - 4 - 2 = 0$

Value $= 5^{0} = 1$
If $a + b + c = 0$ and $abc = 4$ , then $a^{3} + b^{3} + c^{3} =$
A $0$ B $12$ C $4$ D $64$

Answer and solution

Answer: (B) $12$

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

$= 3 \times 4 = 12$
If $\displaystyle \frac{2^{4x} \times 16^{2} \times 32}{2^{x}} = 2^{7}$ , then $x =$
A $-6$ B $\displaystyle -\frac{6}{5}$ C $-2$ D $\displaystyle \frac{20}{3}$

Answer and solution

Answer: (C) $-2$

$\displaystyle \frac{2^{4x}}{2^{x}} \times (2^{4})^{2} \times 2^{5} = 2^{7}$

$\Rightarrow 2^{3x} \times 2^{8} \times 2^{5} = 2^{7}$

$\Rightarrow 2^{3x + 13} = 2^{7}$

$\Rightarrow 3x + 13 = 7 \Rightarrow 3x = -6 \Rightarrow x = -2$
If $25^{x} = (5\sqrt{5})^{6}$ , then $x =$
A $9$ B $\displaystyle \frac{9}{2}$ C $3$ D $\displaystyle \frac{3}{2}$

Answer and solution

Answer: (B) $\displaystyle \frac{9}{2}$

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

$\displaystyle (5^{2})^{x} = \left(5^{\frac{3}{2}}\right)^{6}$

$\Rightarrow 5^{2x} = 5^{9}$

$\displaystyle \Rightarrow 2x = 9 \Rightarrow x = \frac{9}{2}$
If $3^{x - 3} \times 2^{2x - 4} = 48$ , then $x =$
A $4$ B $5$ C $\displaystyle \frac{5}{2}$ D $7$

Answer and solution

Answer: (A) $4$

$48 = 3^{1} \times 2^{4}$

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

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

$\Rightarrow x = 4$ and $x = 4$

$\therefore x = 4$
If $3^{x} = 7$ , then $9^{x} =$
A $63$ B $14$ C $\displaystyle \frac{7}{9}$ D $49$

Answer and solution

Answer: (D) $49$

$9^{x} = (3^{2})^{x} = (3^{x})^{2}$

$= 7^{2} = 49$
If $\displaystyle \left(\sqrt{\frac{3}{2}}\right)^{x - 2} = \frac{4}{9}$ , then $x =$
A $0$ B $6$ C $-2$ D $-6$

Answer and solution

Answer: (C) $-2$

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

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

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

$\Rightarrow x - 2 = -4$

$\Rightarrow x = -2$
If $9^{x} + 3^{2x + 1} = 324$ , then $x =$
A $2$ B $3$ C $4$ D $1$

Answer and solution

Answer: (A) $2$

$(3^{2})^{x} + 3^{2x + 1} = 324$

$\Rightarrow 3^{2x} + 3^{2x + 1} = 324$

$\Rightarrow 3^{2x} + 3 \times 3^{2x} = 324$

$\Rightarrow 3^{2x}\left(1 + 3\right) = 324$

$\Rightarrow 3^{2x} \times 4 = 324$

$\displaystyle \Rightarrow 3^{2x} = 324 \times \frac{1}{4} = 81 = 3^{4}$

$\Rightarrow 2x = 4 \Rightarrow x = 2$
If $\displaystyle 4^{x} = \frac{32}{2^{x}}$ , then $x =$
A $\displaystyle \frac{5}{2}$ B $\displaystyle \frac{5}{3}$ C $\displaystyle -\frac{5}{3}$ D $5$

Answer and solution

Answer: (B) $\displaystyle \frac{5}{3}$

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

$\displaystyle \Rightarrow 3x = 5 \Rightarrow x = \frac{5}{3}$
$\displaystyle \frac{1}{1 + x^{p-q}} + \frac{1}{1 + x^{q-p}} =$
A $x^{p-q}$ B $1$ C $0$ D $2$

Answer and solution

Answer: (B) $1$

$\displaystyle = \frac{x^{q}}{x^{q} + x^{p}} + \frac{x^{p}}{x^{p} + x^{q}}$

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

Answer and solution

Answer: (D) $x^{a^{3}-b^{3}}$

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

$= x^{a^{3}+a^{2}b+ab^{2}-a^{2}b-ab^{2}-b^{3}} = x^{a^{3}-b^{3}}$
If $2^{x} = 7$ , $7^{y} = 3$ and $3^{z} = 8$ , then $xyz =$
A $-3$ B $3$ C $8$ D $1$

Answer and solution

Answer: (B) $3$

$3 = 7^{y} = (2^{x})^{y} = 2^{xy}$

$8 = 3^{z} = (2^{xy})^{z} = 2^{xyz}$

$2^{xyz} = 8 = 2^{3}$

$\Rightarrow xyz = 3$
If $2^{x} = 5^{y} = 10^{-z}$ , then $\displaystyle \frac{1}{x} + \frac{1}{y} + \frac{1}{z} =$
A $-1$ B $2$ C $1$ D $0$

Answer and solution

Answer: (D) $0$

Let each $= k$ : $\displaystyle 2 = k^{\frac{1}{x}}$ , $\displaystyle 5 = k^{\frac{1}{y}}$ , $\displaystyle 10 = k^{-\frac{1}{z}}$

$10 = 2 \times 5$ , so $\displaystyle \frac{1}{x} + \frac{1}{y} = -\frac{1}{z}$ , so the sum is $0$
$\sqrt[3]{125^{-2}} =$
A $-25$ B $\displaystyle \frac{1}{5}$ C $25$ D $\displaystyle \frac{1}{25}$

Answer and solution

Answer: (D) $\displaystyle \frac{1}{25}$

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

$\displaystyle \sqrt[3]{125^{-2}} = 125^{-\frac{2}{3}} = (5^{3})^{-\frac{2}{3}}$

$\displaystyle = 5^{-2} = \frac{1}{5^{2}} = \frac{1}{25}$
$\displaystyle \left(\frac{2}{\sqrt{5}}\right)^{7} \div \left(\frac{2}{\sqrt{5}}\right)^{10} =$
A $\displaystyle \frac{125}{8}$ B $\displaystyle \frac{8 \sqrt{5}}{25}$ C $\displaystyle \frac{5 \sqrt{5}}{8}$ D $\displaystyle \frac{8}{125}$

Answer and solution

Answer: (C) $\displaystyle \frac{5 \sqrt{5}}{8}$

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

$\displaystyle \left(\frac{2}{\sqrt{5}}\right)^{7} \div \left(\frac{2}{\sqrt{5}}\right)^{10} = \frac{1}{\left(\frac{2}{\sqrt{5}}\right)^{3}}$

$\displaystyle = \frac{5 \sqrt{5}}{8}$
If $x = 2$ , then $\displaystyle (x^{\frac{1}{3}} + x^{-\frac{1}{3}})(x^{\frac{2}{3}} + x^{-\frac{2}{3}} - 1) =$
A $2$ B $\displaystyle \frac{17}{4}$ C $\displaystyle \frac{5}{2}$ D $\displaystyle \frac{3}{2}$

Answer and solution

Answer: (C) $\displaystyle \frac{5}{2}$

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

$\displaystyle = 2 + \frac{1}{2} = \frac{5}{2}$
$(81)^{0.3} \times (81)^{0.45} =$
A $3$ B $\displaystyle \frac{1}{27}$ C $9$ D $27$

Answer and solution

Answer: (D) $27$

$= (81)^{0.3 + 0.45} = (81)^{0.75}$

$\displaystyle 0.75 = \frac{3}{4}$

$\displaystyle = (3^{4})^{\frac{3}{4}} = 27$
$\displaystyle \left\{\left(2^{3}\right)^{-2}\right\}^{-\frac{1}{3}} =$
A $64$ B $2$ C $4$ D $\displaystyle \frac{1}{4}$

Answer and solution

Answer: (C) $4$

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

$= 2^{2}$

$= 4$

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MCQ test, Set 2