Logarithms: MCQ test, Set 10 | Expresswala Academy

Download the PDF (with key sheet and solutions, 78.3 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.

$\displaystyle \log_{10} \frac{1}{1000} =$
A $\displaystyle -\frac{1}{3}$ B $\displaystyle \frac{1}{3}$ C $3$ D $-3$

Answer and solution

Answer: (D) $-3$

$\displaystyle \frac{1}{1000} = 10^{-3}$

$\displaystyle \therefore \log_{10} \frac{1}{1000} = -3$
The logarithm of $729$ to the base $3$ is
A $6$ B $243$ C $\displaystyle \frac{1}{6}$ D $-6$

Answer and solution

Answer: (A) $6$

$3^{6} = 729$

$\therefore \log_{3} 729 = 6$
The base $a$ in $\log_{a} N$ must satisfy
A $a < 0$ or $a > 1$ B $a > 0$ and $a \ne 1$ C $a \ne 0$ only D $a > 1$ only

Answer and solution

Answer: (B) $a > 0$ and $a \ne 1$

$a = 0$ , $a = 1$ and negative $a$ all lead to results with no meaning.

$\therefore a > 0$ , $a \ne 1$ (a base like $\displaystyle \frac{1}{2}$ is allowed).
The logarithmic form of $7^{4} = 2401$ is
A $2401 = \log_{7} 4$ B $7 = \log_{4} 2401$ C $4 = \log_{2401} 7$ D $4 = \log_{7} 2401$

Answer and solution

Answer: (D) $4 = \log_{7} 2401$

$a^x = N \Leftrightarrow x = \log_{a} N$

The power $4$ is the logarithm of $2401$ to the base $7$ .
The exponential form of $\displaystyle \log_{125} 25 = \frac{2}{3}$ is
A $\displaystyle 25 = 125^{\frac{2}{3}}$ B $\displaystyle 125 = 25^{\frac{2}{3}}$ C $\displaystyle 25 = 125 \times \frac{2}{3}$ D $\displaystyle 25 = (\frac{2}{3})^{125}$

Answer and solution

Answer: (A) $\displaystyle 25 = 125^{\frac{2}{3}}$

$\log_{a} N = x \Leftrightarrow N = a^x$

So $\displaystyle 25 = 125^{\frac{2}{3}}$ .
$\displaystyle \frac{\log_{9} 1}{\log_{9} 9} =$
A $\displaystyle \frac{1}{9}$ B $0$ C $1$ D $9$

Answer and solution

Answer: (B) $0$

$\log_{9} 1 = 0,\ \log_{9} 9 = 1$

$\displaystyle = \frac{0}{1} = 0$
$\log_{6} 18 + \log_{6} 12 =$
A $3$ B $216$ C $\log_{6} 18 \times \log_{6} 12$ D $\log_{6} 30$

Answer and solution

Answer: (A) $3$

$= \log_{6}(18 \times 12) = \log_{6} 216$

$= 3$ since $6^{3} = 216$
$\log_{9} 18 - \log_{9} 2 =$
A $\log_{9} 16$ B $9$ C $\log_{9} 18 \div \log_{9} 2$ D $1$

Answer and solution

Answer: (D) $1$

$\displaystyle = \log_{9} \frac{18}{2} = \log_{9} 9$

$= 1$ since $9^{1} = 9$
$\log_{64} 16 =$
A $\displaystyle \frac{3}{2}$ B $-2$ C $\displaystyle \frac{2}{3}$ D $\displaystyle \frac{1}{4}$

Answer and solution

Answer: (C) $\displaystyle \frac{2}{3}$

$\displaystyle = \log_{2^{6}} 2^{4} = \frac{4}{6} \log_{2} 2$

$\displaystyle = \frac{2}{3}$
$\displaystyle \log_{\frac{1}{10}} 10000 =$
A $\displaystyle \frac{1}{4}$ B $-4$ C $\displaystyle -\frac{1}{4}$ D $4$

Answer and solution

Answer: (B) $-4$

$\displaystyle = \log_{10^{-1}} 10^{4} = \frac{4}{-1}\log_{10} 10$

$= -4$
If $\log_{10} 7 = m$ , then $\log_{7} 10 =$
A $m$ B $\displaystyle \frac{1}{m}$ C $\displaystyle -\frac{1}{m}$ D $-m$

Answer and solution

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

$\displaystyle \log_{7} 10 = \frac{1}{\log_{10} 7} = \frac{1}{m}$
$\log_{11} 25 \cdot \log_{125} 11 =$
A $\displaystyle \frac{3}{2}$ B $\displaystyle \frac{4}{3}$ C $\displaystyle -\frac{2}{3}$ D $\displaystyle \frac{2}{3}$

Answer and solution

Answer: (D) $\displaystyle \frac{2}{3}$

$\log_{b} a \cdot \log_{c} b = \log_{c} a$

$\displaystyle = \log_{125} 25 = \frac{2}{3}$
$\displaystyle \frac{\log 121}{\log 1331} =$
A $\displaystyle \frac{2}{3}$ B $\displaystyle \frac{3}{2}$ C $\displaystyle \log \frac{1}{11}$ D $11$

Answer and solution

Answer: (A) $\displaystyle \frac{2}{3}$

$\displaystyle \frac{\log a}{\log b} = \log_{b} a$

$\displaystyle = \log_{1331} 121 = \frac{2}{3}$
$\displaystyle 8^{\log_{8} \frac{1}{4}} \times 5^{\log_{5} 12} =$
A $40$ B $48$ C $3$ D $\displaystyle \frac{1}{48}$

Answer and solution

Answer: (C) $3$

$a^{\log_{a} N} = N$

$\displaystyle = \frac{1}{4} \times 12 = 3$
$13^{\log_{2} 9} - 9^{\log_{2} 13} =$
A $1$ B $4$ C $-1$ D $0$

Answer and solution

Answer: (D) $0$

$13^{\log_{2} 9} = 9^{\log_{2} 13}$

So the difference is $0$ .
If $\log_{12}(2x + 9) = \log_{12}(5x)$ , then $x =$
A $-3$ B $\displaystyle -\frac{9}{7}$ C $4$ D $3$

Answer and solution

Answer: (D) $3$

$\Rightarrow 2x + 9 = 5x$

$\Rightarrow 3x = 9$

$\Rightarrow x = 3$
If $\log_{5}(3x + 4) = 2$ , then $x =$
A $2$ B $\displaystyle \frac{29}{3}$ C $7$ D $\displaystyle -\frac{4}{3}$

Answer and solution

Answer: (C) $7$

$\log_{a} N = x \Rightarrow N = a^x$

$3x + 4 = 5^{2} = 25$

$\Rightarrow x = 7$
If $\log_{5}(x + 13) - \log_{5}(x - 1) = 1$ , then
A No real $x$ B $\displaystyle x = \frac{11}{2}$ C $\displaystyle x = -\frac{9}{2}$ D $\displaystyle x = \frac{9}{2}$

Answer and solution

Answer: (D) $\displaystyle x = \frac{9}{2}$

$\displaystyle \Rightarrow \log_{5} \frac{x + 13}{x - 1} = 1$

$\displaystyle \Rightarrow \frac{x + 13}{x - 1} = 5$

$\Rightarrow x + 13 = 5x - 5$

$\displaystyle \Rightarrow x = \frac{9}{2}$

For $\displaystyle x = \frac{9}{2}$ both logarithms are defined.
$\log_{2} \log_{4} 256 =$
A $4$ B $8$ C $2$ D $1$

Answer and solution

Answer: (C) $2$

$\log_{4} 256 = \log_{4} 4^4 = 4$

$\log_{2} 4 = 2$
$\displaystyle \frac{1}{\log_{49} 7} + \frac{1}{\log_{343} 7} =$
A $6$ B $5$ C $\displaystyle \frac{5}{6}$ D $\displaystyle \frac{1}{5}$

Answer and solution

Answer: (B) $5$

$= \log_{7} 49 + \log_{7} 343$

$= 2 + 3 = 5$
$\displaystyle \frac{1}{1 + \log_{2} 15} + \frac{1}{1 + \log_{3} 10} + \frac{1}{1 + \log_{5} 6} =$
A $1$ B $0$ C $\displaystyle \frac{1}{3}$ D $3$

Answer and solution

Answer: (A) $1$

$1 + \log_{2} 15 = \log_{2} 2 + \log_{2} 15 = \log_{2} 30$ , and so on

$= \log_{30} 2 + \log_{30} 3 + \log_{30} 5 = \log_{30} 30 = 1$
$3^{-1 + \log_{3} 12} =$
A $36$ B $11$ C $4$ D $-12$

Answer and solution

Answer: (C) $4$

$= 3^{-1} \times 3^{\log_{3} 12}$

$\displaystyle = \frac{1}{3} \times 12 = 4$
If $2\log a + \log b - 3\log c = \log k$ , then $k =$
A $\displaystyle \frac{2ab}{3c}$ B $a^2 + b - c^3$ C $\displaystyle \frac{a^2 b}{c^3}$ D $a^2 b c^3$

Answer and solution

Answer: (C) $\displaystyle \frac{a^2 b}{c^3}$

$\log a^2 + \log b - \log c^3 = \log k$

$\displaystyle \Rightarrow \log \frac{a^2 b}{c^3} = \log k \Rightarrow k = \frac{a^2 b}{c^3}$
If $\displaystyle \log_{x} 32 = \frac{5}{3}$ , then $x =$
A $-8$ B $8$ C $\displaystyle \frac{1}{8}$ D $\displaystyle \frac{1}{32}$

Answer and solution

Answer: (B) $8$

$\displaystyle 32 = x^{\frac{5}{3}}$

$\displaystyle \Rightarrow x = 32^{\frac{3}{5}} = 8$
On the graph of $\displaystyle y = \log_{\frac{1}{4}} x$ , the value of $y$ at $x = 16$ is
A $-2$ B $2$ C $\displaystyle \frac{1}{2}$ D $\displaystyle -\frac{1}{2}$

Answer and solution

Answer: (A) $-2$

$\displaystyle y = \log_{\frac{1}{4}} 16$

$= -2$

Menu

MCQ test, Set 10