Logarithms: MCQ test, Set 4 | Expresswala Academy

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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.

$\log_{18} 324 =$
A $\displaystyle \frac{1}{2}$ B $2$ C $18$ D $-2$

Answer and solution

Answer: (B) $2$

$18^{2} = 324$

$\therefore \log_{18} 324 = 2$
The logarithm of $\displaystyle \frac{1}{81}$ to the base $9$ is
A $\displaystyle -\frac{1}{2}$ B $\displaystyle \frac{1}{2}$ C $2$ D $-2$

Answer and solution

Answer: (D) $-2$

$\displaystyle \frac{1}{81} = 9^{-2}$

$\displaystyle \therefore \log_{9} \frac{1}{81} = -2$
Which value of $N$ makes $\log_{8} N$ defined?
A $N = 0$ B $N = 64$ C $N = -64$ D $N = -8$

Answer and solution

Answer: (B) $N = 64$

The number must be positive: $N > 0$ .

Only $N = 64$ is positive; $\log_{8} 64 = 2$ .
The logarithmic form of $\displaystyle 8^{\frac{1}{3}} = 2$ is
A $\displaystyle 8 = \log_{\frac{1}{3}} 2$ B $\displaystyle 2 = \log_{8} \frac{1}{3}$ C $\displaystyle \frac{1}{3} = \log_{8} 2$ D $\displaystyle \frac{1}{3} = \log_{2} 8$

Answer and solution

Answer: (C) $\displaystyle \frac{1}{3} = \log_{8} 2$

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

The power $\displaystyle \frac{1}{3}$ is the logarithm of $2$ to the base $8$ .
The exponential form of $\log_{5} 625 = 4$ is
A $5 = 625^{4}$ B $625 = 5^{4}$ C $625 = 5 \times 4$ D $625 = 4^{5}$

Answer and solution

Answer: (B) $625 = 5^{4}$

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

So $625 = 5^{4}$ .
$\log_{16} 1 + \log_{16} 16 =$
A $16$ B $0$ C $2$ D $1$

Answer and solution

Answer: (D) $1$

$\log_{16} 1 = 0,\ \log_{16} 16 = 1$

$= 0 + 1 = 1$
$\log_{14} 7 + \log_{14} 28 =$
A $2$ B $\log_{14} 7 \times \log_{14} 28$ C $196$ D $\log_{14} 35$

Answer and solution

Answer: (A) $2$

$= \log_{14}(7 \times 28) = \log_{14} 196$

$= 2$ since $14^{2} = 196$
$\log_{4} 192 - \log_{4} 3 =$
A $\log_{4} 192 \div \log_{4} 3$ B $\log_{4} 189$ C $64$ D $3$

Answer and solution

Answer: (D) $3$

$\displaystyle = \log_{4} \frac{192}{3} = \log_{4} 64$

$= 3$ since $4^{3} = 64$
If $\log_{q^{3}} p^{12} = k\log_{q} p$ , then $k =$
A $36$ B $4$ C $\displaystyle \frac{1}{4}$ D $9$

Answer and solution

Answer: (B) $4$

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

$\displaystyle k = \frac{12}{3} = 4$
$\displaystyle \log_{\frac{1}{4}} 64 =$
A $-3$ B $3$ C $\displaystyle \frac{1}{3}$ D $\displaystyle -\frac{1}{3}$

Answer and solution

Answer: (A) $-3$

$\displaystyle = \log_{4^{-1}} 4^{3} = \frac{3}{-1}\log_{4} 4$

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

Answer and solution

Answer: (C) $3$

$= \log_{4} 64 = \log_{4} 4^{3}$

$= 3$
$\log_{10} 9 \cdot \log_{27} 10 =$
A $\displaystyle \frac{3}{2}$ B $\displaystyle \frac{2}{3}$ C $\displaystyle -\frac{2}{3}$ D $\displaystyle \frac{4}{3}$

Answer and solution

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

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

$\displaystyle = \log_{27} 9 = \frac{2}{3}$
$\displaystyle \frac{\log 8}{\log 128} =$
A $\displaystyle \frac{7}{3}$ B $16$ C $\displaystyle \frac{3}{7}$ D $\displaystyle \log \frac{1}{16}$

Answer and solution

Answer: (C) $\displaystyle \frac{3}{7}$

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

$\displaystyle = \log_{128} 8 = \frac{3}{7}$
$9^{\log_{9} 20} - 6^{\log_{6} 11} =$
A $9$ B $14$ C $3$ D $31$

Answer and solution

Answer: (A) $9$

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

$= 20 - 11 = 9$
$9^{\log_{4} 7} - 7^{\log_{4} 9} =$
A $-1$ B $1$ C $0$ D $2$

Answer and solution

Answer: (C) $0$

$9^{\log_{4} 7} = 7^{\log_{4} 9}$

So the difference is $0$ .
If $\log_{6}(5x - 4) = \log_{6}(3x + 6)$ , then $x =$
A $\displaystyle \frac{5}{4}$ B $1$ C $5$ D $-5$

Answer and solution

Answer: (C) $5$

$\Rightarrow 5x - 4 = 3x + 6$

$\Rightarrow 2x = 10$

$\Rightarrow x = 5$
If $\log_{2} \log_{3} x = 1$ , then $x =$
A $8$ B $9$ C $6$ D $3$

Answer and solution

Answer: (B) $9$

$\log_{3} x = 2^{1} = 2$

$\Rightarrow x = 3^{2} = 9$
If $\log_{3}(x + 8) - \log_{3}(x - 2) = 1$ , then
A $x = 7$ B $x = 8$ C No real $x$ D $x = -7$

Answer and solution

Answer: (A) $x = 7$

$\displaystyle \Rightarrow \log_{3} \frac{x + 8}{x - 2} = 1$

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

$\Rightarrow x + 8 = 3x - 6$

$\Rightarrow x = 7$

For $x = 7$ both logarithms are defined.
$\log_{5} \sqrt{125} =$
A $\displaystyle \frac{3}{2}$ B $3$ C $\displaystyle \frac{2}{3}$ D $\displaystyle \frac{5}{2}$

Answer and solution

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

$\displaystyle \sqrt{125} = (5^3)^{\frac{1}{2}} = 5^{\frac{3}{2}}$

$\displaystyle \therefore \log_{5} 5^{\frac{3}{2}} = \frac{3}{2}$
$\log_{b^{2}} a \cdot \log_{a^{5}} b =$
A $\displaystyle \frac{1}{5}$ B $\displaystyle \frac{1}{2}$ C $10$ D $\displaystyle \frac{1}{10}$

Answer and solution

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

$\displaystyle = \frac{1}{2}\log_{b} a \cdot \frac{1}{5}\log_{a} b$

$\displaystyle = \frac{1}{10} \cdot \log_{b} a \log_{a} b = \frac{1}{10}(1) = \frac{1}{10}$
$\displaystyle \frac{1}{\log_{2} 30} + \frac{1}{\log_{3} 30} + \frac{1}{\log_{5} 30} =$
A $\log_{30} 10$ B $10$ C $30$ D $1$

Answer and solution

Answer: (D) $1$

$\displaystyle \frac{1}{\log_{b} 30} = \log_{30} b$

$= \log_{30}(2 \times 3 \times 5) = \log_{30} 30 = 1$
$8^{\log_{64} 25} =$
A $5$ B $625$ C $\displaystyle \frac{25}{8}$ D $25$

Answer and solution

Answer: (A) $5$

$\log_{64} 25 = \log_{8^2} 5^2 = \log_{8} 5$

$= 8^{\log_{8} 5} = 5$
If $\displaystyle \log \frac{p + q}{6} = \frac{1}{2}(\log p + \log q)$ , then $p^2 + q^2 =$
A $36pq$ B $38pq$ C $32pq$ D $34pq$

Answer and solution

Answer: (D) $34pq$

$\displaystyle \frac{p + q}{6} = \sqrt{pq} \Rightarrow (p + q)^2 = 36pq$

$\Rightarrow p^2 + 2pq + q^2 = 36pq \Rightarrow p^2 + q^2 = 34pq$
If $\log_{2} x = \log_{8} 125$ , then $x =$
A $\displaystyle \frac{125}{3}$ B $125$ C $375$ D $5$

Answer and solution

Answer: (D) $5$

$\displaystyle \log_{8} 125 = \log_{2^{3}} 5^{3} = \frac{3}{3}\log_{2} 5 = \log_{2} 5$

$\log_{2} x = \log_{2} 5 \Rightarrow x = 5$
On the graph of $y = \log_{4} x$ , the value of $y$ at $x = 64$ is
A $\displaystyle \frac{1}{3}$ B $\displaystyle -\frac{1}{3}$ C $3$ D $-3$

Answer and solution

Answer: (C) $3$

$y = \log_{4} 64$

$= 3$

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MCQ test, Set 4