Logarithms: MCQ test, Set 7 | 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.

$\log_{3} 243 =$
A $5$ B $81$ C $-5$ D $\displaystyle \frac{1}{5}$

Answer and solution

Answer: (A) $5$

$3^{5} = 243$

$\therefore \log_{3} 243 = 5$
The logarithm of $\displaystyle \frac{1}{64}$ to the base $4$ is
A $-3$ B $3$ C $\displaystyle \frac{1}{3}$ D $\displaystyle -\frac{1}{3}$

Answer and solution

Answer: (A) $-3$

$\displaystyle \frac{1}{64} = 4^{-3}$

$\displaystyle \therefore \log_{4} \frac{1}{64} = -3$
Which logarithm of $81$ is defined?
A $\log_{-9} 81$ B $\log_{1} 81$ C $\log_{9} 81$ D $\log_{0} 81$

Answer and solution

Answer: (C) $\log_{9} 81$

The base must be positive and not $1$ .

Only $\log_{9} 81$ qualifies: $9^2 = 81$ , so it equals $2$ .
The logarithmic form of $6^{4} = 1296$ is
A $4 = \log_{1296} 6$ B $1296 = \log_{6} 4$ C $4 = \log_{6} 1296$ D $6 = \log_{4} 1296$

Answer and solution

Answer: (C) $4 = \log_{6} 1296$

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

The power $4$ is the logarithm of $1296$ to the base $6$ .
The exponential form of $\log_{0.1} 0.01 = 2$ is
A $0.01 = 2^{0.1}$ B $0.1 = 0.01^{2}$ C $0.01 = 0.1 \times 2$ D $0.01 = 0.1^{2}$

Answer and solution

Answer: (D) $0.01 = 0.1^{2}$

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

So $0.01 = 0.1^{2}$ .
$\log_{a} 1 = 0$ is true because
A $a^1 = a$ B $1^a = 1$ C $a^0 = 1$ D $0^a = 0$

Answer and solution

Answer: (C) $a^0 = 1$

$a^0 = 1 \Rightarrow 0 = \log_{a} 1$

(for $a > 0$ , $a \ne 1$ )
$\log_{10} 40 + \log_{10} 25 =$
A $1000$ B $\log_{10} 65$ C $3$ D $\log_{10} 40 \times \log_{10} 25$

Answer and solution

Answer: (C) $3$

$= \log_{10}(40 \times 25) = \log_{10} 1000$

$= 3$ since $10^{3} = 1000$
$\log_{6} 72 - \log_{6} 2 =$
A $2$ B $36$ C $\log_{6} 72 \div \log_{6} 2$ D $\log_{6} 70$

Answer and solution

Answer: (A) $2$

$\displaystyle = \log_{6} \frac{72}{2} = \log_{6} 36$

$= 2$ since $6^{2} = 36$
$\log_{343} 2401 =$
A $1$ B $7$ C $\displaystyle \frac{3}{4}$ D $\displaystyle \frac{4}{3}$

Answer and solution

Answer: (D) $\displaystyle \frac{4}{3}$

$\displaystyle = \log_{7^{3}} 7^{4} = \frac{4}{3} \log_{7} 7$

$\displaystyle = \frac{4}{3}$
$\log_{11} 2^{7} =$
A $2\log_{11} 7$ B $\log_{11} 14$ C $7\log_{11} 2$ D $(\log_{11} 2)^{7}$

Answer and solution

Answer: (C) $7\log_{11} 2$

$\log_{b} a^m = m\log_{b} a$

$\therefore \log_{11} 2^{7} = 7\log_{11} 2$
$\displaystyle \frac{1}{\log_{400} 20} =$
A $2$ B $-2$ C $20$ D $\displaystyle \frac{1}{2}$

Answer and solution

Answer: (A) $2$

$= \log_{20} 400 = \log_{20} 20^{2}$

$= 2$
$\log_{3} 81 \cdot \log_{7} 3 \cdot \log_{9} 7 =$
A $2$ B $3$ C $\displaystyle \frac{1}{2}$ D $-2$

Answer and solution

Answer: (A) $2$

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

$= \log_{9} 81 = 2$
$\displaystyle \frac{\log_{10} 1024}{\log_{10} 32} =$
A $\displaystyle \frac{1}{2}$ B $2$ C $32$ D $\log_{10} 32$

Answer and solution

Answer: (B) $2$

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

$= \log_{32} 1024 = 2$
$3 \times 6^{\log_{6} 5} =$
A $5$ B $18$ C $90$ D $15$

Answer and solution

Answer: (D) $15$

$6^{\log_{6} 5} = 5$

$= 3 \times 5 = 15$
$7^{\log_{2} 5} =$
A $2^{\log_{5} 7}$ B $5^{\log_{2} 7}$ C $5^{\log_{7} 2}$ D $7^{\log_{5} 2}$

Answer and solution

Answer: (B) $5^{\log_{2} 7}$

$a^{\log_{c} b} = b^{\log_{c} a}$ : the base and the number inside the log change places.

$\therefore 7^{\log_{2} 5} = 5^{\log_{2} 7}$
If $\log_{8}(4x + 3) = \log_{8} 23$ , then $x =$
A $-5$ B $5$ C $6$ D $\displaystyle \frac{13}{2}$

Answer and solution

Answer: (B) $5$

$\Rightarrow 4x + 3 = 23$

$\Rightarrow 4x = 20$

$\Rightarrow x = 5$
If $\log_{12}(x + 5) = 0$ , then $x =$
A $-5$ B $6$ C $-4$ D $7$

Answer and solution

Answer: (C) $-4$

$\log_{a} 1 = 0$

$x + 5 = 1$

$\Rightarrow x = -4$
If $\log_{4} x - \log_{4}(x + 1) = 1$ , then
A $\displaystyle x = \frac{4}{3}$ B No real $x$ C $\displaystyle x = -\frac{3}{4}$ D $\displaystyle x = -\frac{4}{3}$

Answer and solution

Answer: (B) No real $x$

$\displaystyle \Rightarrow \log_{4} \frac{x}{x + 1} = 1$

$\displaystyle \Rightarrow \frac{x}{x + 1} = 4$

$\Rightarrow x = 4x + 4$

$\displaystyle \Rightarrow x = -\frac{4}{3}$

For $\displaystyle x = -\frac{4}{3}$ , $\log_{4} x$ is not defined.

$\therefore$ there is no real $x$ .
$\log_{0.3} 0.027 =$
A $-3$ B $\displaystyle \frac{1}{3}$ C $9$ D $3$

Answer and solution

Answer: (D) $3$

$0.027 = (0.3)^3$

$\therefore \log_{0.3}(0.3)^3 = 3$
$\log_{b^{4}} a^{3} \cdot \log_{a^{2}} b =$
A $\displaystyle \frac{1}{2}$ B $\displaystyle \frac{8}{3}$ C $\displaystyle \frac{3}{4}$ D $\displaystyle \frac{3}{8}$

Answer and solution

Answer: (D) $\displaystyle \frac{3}{8}$

$\displaystyle = \frac{3}{4}\log_{b} a \cdot \frac{1}{2}\log_{a} b$

$\displaystyle = \frac{3}{8} \cdot \log_{b} a \log_{a} b = \frac{3}{8}(1) = \frac{3}{8}$
$\displaystyle \frac{1}{\log_{4} 144} + \frac{1}{\log_{3} 144} =$
A $\displaystyle \frac{1}{2}$ B $7$ C $\log_{144} 7$ D $144$

Answer and solution

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

$\displaystyle \frac{1}{\log_{b} 144} = \log_{144} b$

$\displaystyle = \log_{144}(4 \times 3) = \log_{144} 12 = \frac{1}{2}$
$3^{\log_{9} 49} =$
A $7$ B $\sqrt{7}$ C $49$ D $\displaystyle \frac{49}{9}$

Answer and solution

Answer: (A) $7$

$\log_{9} 49 = \log_{3^2} 7^2 = \log_{3} 7$

$= 3^{\log_{3} 7} = 7$
If $2\log(x - y) = \log x + \log y + 2\log 2$ , then $x^2 + y^2 =$
A $2xy$ B $6xy$ C $8xy$ D $4xy$

Answer and solution

Answer: (B) $6xy$

$\log(x - y)^2 = \log 4xy \Rightarrow (x - y)^2 = 4xy$

$\Rightarrow x^2 - 2xy + y^2 = 4xy \Rightarrow x^2 + y^2 = 6xy$
If $\log_{7} x = \log_{49} 36$ , then $x =$
A $12$ B $6$ C $18$ D $36$

Answer and solution

Answer: (B) $6$

$\displaystyle \log_{49} 36 = \log_{7^{2}} 6^{2} = \frac{2}{2}\log_{7} 6 = \log_{7} 6$

$\log_{7} x = \log_{7} 6 \Rightarrow x = 6$
The graph of $y = \log_{8} x$
A falls from left to right B lies only below the x-axis C cuts the y-axis at $(0, 1)$ D rises from left to right

Answer and solution

Answer: (D) rises from left to right

The base $8$ is more than $1$ .

For $a > 1$ the graph of $\log_{a} x$ rises from left to right.

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MCQ test, Set 7