Logarithms: MCQ test, Set 8 | 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.

If $\displaystyle 7^x = \frac{1}{49}$ , then $x =$
A $-2$ B $\displaystyle -\frac{1}{2}$ C $\displaystyle \frac{1}{2}$ D $2$

Answer and solution

Answer: (A) $-2$

$\displaystyle \frac{1}{49} = 7^{-2}$

$\displaystyle \therefore \log_{7} \frac{1}{49} = -2$
$\log_{15} 225 =$
A $15$ B $2$ C $-2$ D $\displaystyle \frac{1}{2}$

Answer and solution

Answer: (B) $2$

$15^{2} = 225$

$\therefore \log_{15} 225 = 2$
$\log_{a} 1$ is not defined when $a =$
A $\displaystyle \frac{1}{2}$ B $6$ C $1$ D $10$

Answer and solution

Answer: (C) $1$

The base cannot be $1$ .

For $a = 10$ , $\displaystyle \frac{1}{2}$ or $6$ , $\log_{a} 1 = 0$ ; for $a = 1$ it is not defined.
The logarithmic form of $\displaystyle 3^{-5} = \frac{1}{243}$ is
A $\displaystyle 3 = \log_{-5} \frac{1}{243}$ B $\displaystyle \frac{1}{243} = \log_{3}(-5)$ C $\displaystyle -5 = \log_{\frac{1}{243}} 3$ D $\displaystyle -5 = \log_{3} \frac{1}{243}$

Answer and solution

Answer: (D) $\displaystyle -5 = \log_{3} \frac{1}{243}$

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

The power $-5$ is the logarithm of $\displaystyle \frac{1}{243}$ to the base $3$ .
The exponential form of $\displaystyle \log_{49} 7 = \frac{1}{2}$ is
A $\displaystyle 7 = 49^{\frac{1}{2}}$ B $\displaystyle 49 = 7^{\frac{1}{2}}$ C $\displaystyle 7 = 49 \times \frac{1}{2}$ D $\displaystyle 7 = (\frac{1}{2})^{49}$

Answer and solution

Answer: (A) $\displaystyle 7 = 49^{\frac{1}{2}}$

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

So $\displaystyle 7 = 49^{\frac{1}{2}}$ .
$\log_{50} 50 \times \log_{7} 7 =$
A $2$ B $1$ C $350$ D $0$

Answer and solution

Answer: (B) $1$

$\log_{50} 50 = 1,\ \log_{7} 7 = 1$

$= 1 \times 1 = 1$
$\log_{11} 6 + \log_{11} 4 =$
A $\log_{11} 6 \times \log_{11} 4$ B $\log_{11} 2$ C $\log_{11} 24$ D $\log_{11} 10$

Answer and solution

Answer: (C) $\log_{11} 24$

$\log_{a} m + \log_{a} n = \log_{a} mn$

$= \log_{11} 24$
$\displaystyle \log_{2} 40 + \log_{2} \frac{1}{5} =$
A $\displaystyle \log_{2} 40 \times \log_{2} \frac{1}{5}$ B $8$ C $\displaystyle \log_{2} \frac{201}{5}$ D $3$

Answer and solution

Answer: (D) $3$

$\displaystyle = \log_{2}(40 \times \frac{1}{5}) = \log_{2} 8$

$= 3$ since $2^{3} = 8$
If $\log_{n^{5}} m^{10} = k\log_{n} m$ , then $k =$
A $2$ B $5$ C $50$ D $\displaystyle \frac{1}{2}$

Answer and solution

Answer: (A) $2$

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

$\displaystyle k = \frac{10}{5} = 2$
$\displaystyle \log_{\frac{1}{7}} 343 =$
A $\displaystyle -\frac{1}{3}$ B $3$ C $-3$ D $\displaystyle \frac{1}{3}$

Answer and solution

Answer: (C) $-3$

$\displaystyle = \log_{7^{-1}} 7^{3} = \frac{3}{-1}\log_{7} 7$

$= -3$
$\displaystyle \frac{1}{\log_{8} 3} =$
A $\log_{8} 3$ B $\log_{3} 8$ C $\displaystyle \log_{3} \frac{1}{8}$ D $-\log_{8} 3$

Answer and solution

Answer: (B) $\log_{3} 8$

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

$\displaystyle \therefore \frac{1}{\log_{8} 3} = \log_{3} 8$
$\log_{13} 64 \cdot \log_{2} 13 =$
A $-6$ B $12$ C $6$ D $\displaystyle \frac{1}{6}$

Answer and solution

Answer: (C) $6$

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

$= \log_{2} 64 = 6$
$\displaystyle \frac{\log 16}{\log 32} =$
A $\displaystyle \log \frac{1}{2}$ B $2$ C $\displaystyle \frac{5}{4}$ D $\displaystyle \frac{4}{5}$

Answer and solution

Answer: (D) $\displaystyle \frac{4}{5}$

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

$\displaystyle = \log_{32} 16 = \frac{4}{5}$
$(17^{\log_{17} 2})^3 =$
A $8$ B $6$ C $17$ D $4913$

Answer and solution

Answer: (A) $8$

$17^{\log_{17} 2} = 2$

$= 2^3 = 8$
$8^{\log_{5} 3} - 3^{\log_{5} 8} =$
A $-1$ B $1$ C $0$ D $5$

Answer and solution

Answer: (C) $0$

$8^{\log_{5} 3} = 3^{\log_{5} 8}$

So the difference is $0$ .
If $\log_{9}(7x - 3) = \log_{9}(4x + 9)$ , then $x =$
A $2$ B $4$ C $-4$ D $\displaystyle \frac{12}{11}$

Answer and solution

Answer: (B) $4$

$\Rightarrow 7x - 3 = 4x + 9$

$\Rightarrow 3x = 12$

$\Rightarrow x = 4$
If $\log_{3} \log_{2} x = 2$ , then $x =$
A $9$ B $4$ C $18$ D $512$

Answer and solution

Answer: (D) $512$

$\log_{2} x = 3^{2} = 9$

$\Rightarrow x = 2^{9} = 512$
If $\log_{2}(x^2 + 3x + 9) + \log_{2}(x - 3) = 3$ , then
A $x = \sqrt[3]{35}$ B $x = 2$ C $x = 35$ D $x = \sqrt[3]{33}$

Answer and solution

Answer: (A) $x = \sqrt[3]{35}$

$\log_{2}(x^3 - 27) = 3$

$x^3 = 27 + 8 = 35 \Rightarrow x = \sqrt[3]{35}$
$\log_{7} \sqrt[3]{49} =$
A $\displaystyle \frac{2}{3}$ B $\displaystyle \frac{3}{2}$ C $2$ D $\displaystyle \frac{1}{3}$

Answer and solution

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

$\displaystyle \sqrt[3]{49} = (7^2)^{\frac{1}{3}} = 7^{\frac{2}{3}}$

$\displaystyle \therefore \log_{7} 7^{\frac{2}{3}} = \frac{2}{3}$
If $\log_{x} y = t$ , then $\displaystyle \log_{\frac{1}{y}} x^3 =$
A $-3t$ B $\displaystyle -\frac{1}{3t}$ C $\displaystyle -\frac{3}{t}$ D $\displaystyle \frac{3}{t}$

Answer and solution

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

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

$\displaystyle = -3 \cdot \frac{1}{\log_{x} y} = -\frac{3}{t}$
If $\log_{x^2} y \cdot \log_{\sqrt[3]{z}} x = k\log_{z} y$ , then $k =$
A $\displaystyle \frac{1}{6}$ B $6$ C $\displaystyle \frac{3}{2}$ D $\displaystyle \frac{2}{3}$

Answer and solution

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

$\displaystyle \log_{x^2} y = \frac{1}{2}\log_{x} y,\ \log_{z^{\frac{1}{3}}} x = 3\log_{z} x$

$\displaystyle = \frac{3}{2}\log_{z} y \Rightarrow k = \frac{3}{2}$
$2^{\log_{5} 3} - 3^{\log_{5} 2} + 6^{\log_{6} 4} =$
A $0$ B $4$ C $3$ D $5$

Answer and solution

Answer: (B) $4$

$2^{\log_{5} 3} = 3^{\log_{5} 2}$ , so the first two cancel

$6^{\log_{6} 4} = 4$
If $\displaystyle \frac{1}{2}\log a + \frac{1}{3}\log b = \log k$ , then $k =$
A $a^2 b^3$ B $\displaystyle \frac{a}{2} + \frac{b}{3}$ C $\displaystyle \frac{ab}{6}$ D $\sqrt{a}\sqrt[3]{b}$

Answer and solution

Answer: (D) $\sqrt{a}\sqrt[3]{b}$

$\displaystyle \log a^{\frac{1}{2}} + \log b^{\frac{1}{3}} = \log k$

$\Rightarrow \log \sqrt{a}\sqrt[3]{b} = \log k \Rightarrow k = \sqrt{a}\sqrt[3]{b}$
If $\displaystyle \log_{x} 9 = -\frac{2}{3}$ , then $x =$
A $\displaystyle \frac{1}{9}$ B $\displaystyle -\frac{1}{27}$ C $27$ D $\displaystyle \frac{1}{27}$

Answer and solution

Answer: (D) $\displaystyle \frac{1}{27}$

$\displaystyle 9 = x^{-\frac{2}{3}}$

$\displaystyle \Rightarrow x = 9^{-\frac{3}{2}} = \frac{1}{27}$
On the graph of $y = \log_{5} x$ , the value of $y$ at $\displaystyle x = \frac{1}{5}$ is
A $1$ B $-1$ C $-2$ D $0$

Answer and solution

Answer: (B) $-1$

$\displaystyle y = \log_{5} \frac{1}{5}$

$= -1$

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MCQ test, Set 8