Logarithms: MCQ test, Set 9 | 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.

The logarithm of $2048$ to the base $2$ is
A $11$ B $\displaystyle \frac{1}{11}$ C $1024$ D $-11$

Answer and solution

Answer: (A) $11$

$2^{11} = 2048$

$\therefore \log_{2} 2048 = 11$
If $17^x = 289$ , then $x =$
A $-2$ B $\displaystyle \frac{1}{2}$ C $17$ D $2$

Answer and solution

Answer: (D) $2$

$17^{2} = 289$

$\therefore \log_{17} 289 = 2$
With base $3$ , which of these is not defined?
A $\log_{3} 0$ B $\log_{3} 1$ C $\displaystyle \log_{3} \frac{1}{3}$ D $\log_{3} 3$

Answer and solution

Answer: (A) $\log_{3} 0$

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

$3^x$ is never $0$ , so $\log_{3} 0$ is not defined.
The logarithmic form of $\displaystyle 32^{\frac{2}{5}} = 4$ is
A $\displaystyle 32 = \log_{\frac{2}{5}} 4$ B $\displaystyle 4 = \log_{32} \frac{2}{5}$ C $\displaystyle \frac{2}{5} = \log_{4} 32$ D $\displaystyle \frac{2}{5} = \log_{32} 4$

Answer and solution

Answer: (D) $\displaystyle \frac{2}{5} = \log_{32} 4$

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

The power $\displaystyle \frac{2}{5}$ is the logarithm of $4$ to the base $32$ .
The exponential form of $\log_{3} 6561 = 8$ is
A $6561 = 8^{3}$ B $6561 = 3^{8}$ C $3 = 6561^{8}$ D $6561 = 3 \times 8$

Answer and solution

Answer: (B) $6561 = 3^{8}$

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

So $6561 = 3^{8}$ .
$\log_{0.25} 1 + \log_{0.25} 0.25 =$
A $0.25$ B $2$ C $1$ D $0$

Answer and solution

Answer: (C) $1$

$\log_{0.25} 1 = 0,\ \log_{0.25} 0.25 = 1$

$= 0 + 1 = 1$
$\log_{5} 375 - \log_{5} 3 =$
A $\log_{5} 372$ B $3$ C $125$ D $\log_{5} 375 \div \log_{5} 3$

Answer and solution

Answer: (B) $3$

$\displaystyle = \log_{5} \frac{375}{3} = \log_{5} 125$

$= 3$ since $5^{3} = 125$
$\log_{2} 56 - \log_{2} 8 =$
A $\log_{2} 56 \div \log_{2} 8$ B $\log_{2} 448$ C $\log_{2} 7$ D $\log_{2} 48$

Answer and solution

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

$\displaystyle \log_{a} m - \log_{a} n = \log_{a} \frac{m}{n}$

$= \log_{2} 7$
$\log_{25} 3125 =$
A $\displaystyle \frac{5}{2}$ B $3$ C $125$ D $\displaystyle \frac{2}{5}$

Answer and solution

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

$\displaystyle = \log_{5^{2}} 5^{5} = \frac{5}{2} \log_{5} 5$

$\displaystyle = \frac{5}{2}$
$\log_{7} 4^{6} =$
A $4\log_{7} 6$ B $\log_{7} 24$ C $6\log_{7} 4$ D $(\log_{7} 4)^{6}$

Answer and solution

Answer: (C) $6\log_{7} 4$

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

$\therefore \log_{7} 4^{6} = 6\log_{7} 4$
$\log_{21} 5 \cdot \log_{5} 21 =$
A $105$ B $1$ C $0$ D $\displaystyle \frac{5}{21}$

Answer and solution

Answer: (B) $1$

$\log_{b} a \cdot \log_{a} b = 1$

So the product is $1$ .
$\log_{9} 32 \cdot \log_{6} 9 \cdot \log_{4} 6 =$
A $\displaystyle -\frac{5}{2}$ B $\displaystyle \frac{5}{2}$ C $\displaystyle \frac{7}{2}$ D $\displaystyle \frac{2}{5}$

Answer and solution

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

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

$\displaystyle = \log_{4} 32 = \frac{5}{2}$
$\displaystyle \frac{\log_{10} 512}{\log_{10} 8} =$
A $64$ B $\log_{10} 64$ C $\displaystyle \frac{1}{3}$ D $3$

Answer and solution

Answer: (D) $3$

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

$= \log_{8} 512 = 3$
$4^{\log_{4} 1.5} + 9^{\log_{9} 0.5} =$
A $1$ B $2$ C $13$ D $\displaystyle \frac{3}{4}$

Answer and solution

Answer: (B) $2$

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

$= 1.5 + 0.5 = 2$
$6^{\log_{7} 10} =$
A $6^{\log_{10} 7}$ B $7^{\log_{10} 6}$ C $10^{\log_{7} 6}$ D $10^{\log_{6} 7}$

Answer and solution

Answer: (C) $10^{\log_{7} 6}$

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

$\therefore 6^{\log_{7} 10} = 10^{\log_{7} 6}$
If $\log_{13} \sqrt{x} = \log_{13} 12$ , then $x =$
A $1728$ B $12$ C $24$ D $144$

Answer and solution

Answer: (D) $144$

$\Rightarrow \sqrt{x} = 12$

$\Rightarrow x = 12^{2} = 144$
If $\log_{10}(x - 3) = 2$ , then $x =$
A $23$ B $3$ C $97$ D $103$

Answer and solution

Answer: (D) $103$

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

$x - 3 = 10^{2} = 100$

$\Rightarrow x = 103$
If $\log_{7} x + \log_{7}(x - 48) = 2$ , then
A $x = 49$ B $x = 49$ (or) $x = -1$ C $x = 1$ D $x = -1$

Answer and solution

Answer: (A) $x = 49$

$\Rightarrow \log_{7} x(x - 48) = 2$

$\Rightarrow x^2 - 48x = 7^{2} = 49$

$\Rightarrow x^2 - 48x - 49 = 0$

$\Rightarrow (x + 1)(x - 49) = 0$

$\Rightarrow x = -1$ (or) $x = 49$

$x = -1$ is rejected: the log of a negative number is not defined.

$\therefore x = 49$
$\log_{49} 343 + \log_{343} 49 =$
A $\displaystyle \frac{5}{6}$ B $0$ C $\displaystyle \frac{13}{6}$ D $\displaystyle \frac{3}{2}$

Answer and solution

Answer: (C) $\displaystyle \frac{13}{6}$

$\displaystyle = \frac{3}{2} + \frac{2}{3}$ (writing both numbers as powers of $7$ )

$\displaystyle = \frac{13}{6}$
$\displaystyle \frac{1}{\log_{p} pqr} + \frac{1}{\log_{q} pqr} + \frac{1}{\log_{r} pqr} =$
A $1$ B $\log_{pqr} 3$ C $3$ D $0$

Answer and solution

Answer: (A) $1$

$= \log_{pqr} p + \log_{pqr} q + \log_{pqr} r$

$= \log_{pqr} pqr = 1$
$\log_{b^{3}} a^{2} \cdot \log_{a^{4}} b^{3} =$
A $\displaystyle \frac{1}{2}$ B $2$ C $\displaystyle \frac{3}{4}$ D $\displaystyle \frac{2}{3}$

Answer and solution

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

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

$\displaystyle = \frac{1}{2} \cdot \log_{b} a \log_{a} b = \frac{1}{2}(1) = \frac{1}{2}$
$10^{\log_{100} 81} =$
A $81$ B $\displaystyle \frac{81}{2}$ C $3$ D $9$

Answer and solution

Answer: (D) $9$

$\log_{100} 81 = \log_{10^2} 9^2 = \log_{10} 9$

$= 10^{\log_{10} 9} = 9$
If $\displaystyle \log \frac{x - y}{5} = \frac{1}{2}(\log x + \log y)$ , then $x^2 + y^2 =$
A $27xy$ B $25xy$ C $23xy$ D $5xy$

Answer and solution

Answer: (A) $27xy$

$\displaystyle \frac{x - y}{5} = \sqrt{xy} \Rightarrow (x - y)^2 = 25xy$

$\Rightarrow x^2 - 2xy + y^2 = 25xy \Rightarrow x^2 + y^2 = 27xy$
If $\displaystyle \log_{\frac{1}{x}} 16 = 4$ , then
A $\displaystyle x = \frac{1}{2}$ (or) $\displaystyle x = -\frac{1}{2}$ B $\displaystyle x = \frac{1}{2}$ C $x = 2$ D $\displaystyle x = -\frac{1}{2}$

Answer and solution

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

$\displaystyle 16 = \left(\frac{1}{x}\right)^{4} = x^{-4}$

$\displaystyle \Rightarrow x = \frac{1}{2}$ (the negative value is rejected: the base must be positive)
Which point lies on the graph of $y = \log_{2} x$ ?
A $(3, 8)$ B $(2, 0)$ C $(8, 3)$ D $(0, 1)$

Answer and solution

Answer: (C) $(8, 3)$

$x = 8:\ y = \log_{2} 8 = 3$

So $(8, 3)$ is on the graph.

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MCQ test, Set 9