Logarithms: MCQ test, Set 1 | 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.

$\log_{2} 64 =$
A $6$ B $\displaystyle \frac{1}{6}$ C $-6$ D $32$

Answer and solution

Answer: (A) $6$

$2^{6} = 64$

$\therefore \log_{2} 64 = 6$
The logarithm of $\displaystyle \frac{1}{9}$ to the base $3$ is
A $-2$ B $2$ C $\displaystyle -\frac{1}{2}$ D $\displaystyle \frac{1}{2}$

Answer and solution

Answer: (A) $-2$

$\displaystyle \frac{1}{9} = 3^{-2}$

$\displaystyle \therefore \log_{3} \frac{1}{9} = -2$
Which value of $a$ makes $\log_{a} 49$ defined?
A $a = 1$ B $a = -7$ C $a = 0$ D $a = 7$

Answer and solution

Answer: (D) $a = 7$

The base must satisfy $a > 0$ and $a \ne 1$ .

$-7$ , $1$ and $0$ are not allowed; $a = 7$ gives $\log_{7} 49 = 2$ .
The logarithmic form of $2^{9} = 512$ is
A $512 = \log_{2} 9$ B $2 = \log_{9} 512$ C $9 = \log_{2} 512$ D $9 = \log_{512} 2$

Answer and solution

Answer: (C) $9 = \log_{2} 512$

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

The power $9$ is the logarithm of $512$ to the base $2$ .
The exponential form of $\log_{9} 729 = 3$ is
A $729 = 9 \times 3$ B $729 = 9^{3}$ C $729 = 3^{9}$ D $9 = 729^{3}$

Answer and solution

Answer: (B) $729 = 9^{3}$

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

So $729 = 9^{3}$ .
$\log_{23} 23 =$
A $23$ B $0$ C $1$ D $-1$

Answer and solution

Answer: (C) $1$

$\log_{a} a = 1$ because $a^1 = a$
$\log_{4} 8 + \log_{4} 2 =$
A $\log_{4} 10$ B $\log_{4} 8 \times \log_{4} 2$ C $16$ D $2$

Answer and solution

Answer: (D) $2$

$= \log_{4}(8 \times 2) = \log_{4} 16$

$= 2$ since $4^{2} = 16$
$\log_{3} 54 - \log_{3} 2 =$
A $\log_{3} 52$ B $27$ C $\log_{3} 54 \div \log_{3} 2$ D $3$

Answer and solution

Answer: (D) $3$

$\displaystyle = \log_{3} \frac{54}{2} = \log_{3} 27$

$= 3$ since $3^{3} = 27$
$\log_{100} 10 =$
A $-1$ B $\displaystyle \frac{1}{2}$ C $2$ D $\displaystyle \frac{1}{10}$

Answer and solution

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

$\displaystyle = \log_{10^{2}} 10^{1} = \frac{1}{2} \log_{10} 10$

$\displaystyle = \frac{1}{2}$
$\log_{5} 7^{3} =$
A $\log_{5} 21$ B $7\log_{5} 3$ C $(\log_{5} 7)^{3}$ D $3\log_{5} 7$

Answer and solution

Answer: (D) $3\log_{5} 7$

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

$\therefore \log_{5} 7^{3} = 3\log_{5} 7$
$\log_{9} 4 \cdot \log_{4} 9 =$
A $1$ B $36$ C $0$ D $\displaystyle \frac{4}{9}$

Answer and solution

Answer: (A) $1$

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

So the product is $1$ .
$\log_{7} 2 \cdot \log_{16} 7 =$
A $\displaystyle \frac{1}{4}$ B $\displaystyle -\frac{1}{4}$ C $4$ D $\displaystyle \frac{1}{2}$

Answer and solution

Answer: (A) $\displaystyle \frac{1}{4}$

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

$\displaystyle = \log_{16} 2 = \frac{1}{4}$
$\displaystyle \frac{\log_{10} 729}{\log_{10} 27} =$
A $\log_{10} 27$ B $2$ C $27$ D $\displaystyle \frac{1}{2}$

Answer and solution

Answer: (B) $2$

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

$= \log_{27} 729 = 2$
$7^{\log_{7} 12} =$
A $1$ B $12$ C $\log_{7} 12$ D $7$

Answer and solution

Answer: (B) $12$

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

So the value is $12$ .
$4^{\log_{3} 7} =$
A $4^{\log_{7} 3}$ B $7^{\log_{4} 3}$ C $7^{\log_{3} 4}$ D $3^{\log_{7} 4}$

Answer and solution

Answer: (C) $7^{\log_{3} 4}$

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

$\therefore 4^{\log_{3} 7} = 7^{\log_{3} 4}$
If $\log_{4}(x + 5) = \log_{4}(2x - 1)$ , then $x =$
A $-2$ B $-6$ C $6$ D $-4$

Answer and solution

Answer: (C) $6$

$\Rightarrow x + 5 = 2x - 1$

$\Rightarrow x = 6$

$\Rightarrow x = 6$
If $\log_{7}(3x - 5) = 0$ , then $x =$
A $4$ B $2$ C $\displaystyle \frac{5}{3}$ D $\displaystyle -\frac{4}{3}$

Answer and solution

Answer: (B) $2$

$\log_{a} 1 = 0$

$3x - 5 = 1$

$\Rightarrow x = 2$
If $\log_{2} x - \log_{2}(x + 3) = 4$ , then
A $\displaystyle x = \frac{16}{5}$ B $\displaystyle x = -\frac{5}{16}$ C No real $x$ D $\displaystyle x = -\frac{16}{5}$

Answer and solution

Answer: (C) No real $x$

$\displaystyle \Rightarrow \log_{2} \frac{x}{x + 3} = 4$

$\displaystyle \Rightarrow \frac{x}{x + 3} = 2^{4} = 16$

$\Rightarrow x = 16x + 48$

$\displaystyle \Rightarrow x = -\frac{16}{5}$

For $\displaystyle x = -\frac{16}{5}$ , $\log_{2} x$ is not defined.

$\therefore$ there is no real $x$ .
$\log_{4} 8 + \log_{8} 4 =$
A $\displaystyle \frac{3}{2}$ B $\displaystyle \frac{13}{6}$ C $0$ D $\displaystyle \frac{5}{6}$

Answer and solution

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

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

$\displaystyle = \frac{13}{6}$
$\log_{b^{3}} a \cdot \log_{a^{2}} b =$
A $\displaystyle \frac{1}{2}$ B $6$ C $\displaystyle \frac{1}{6}$ D $\displaystyle \frac{1}{3}$

Answer and solution

Answer: (C) $\displaystyle \frac{1}{6}$

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

$\displaystyle = \frac{1}{6} \cdot \log_{b} a \log_{a} b = \frac{1}{6}(1) = \frac{1}{6}$
$\displaystyle \frac{1}{\log_{3} 6} + \frac{1}{\log_{2} 6} =$
A $\log_{6} 5$ B $1$ C $6$ D $5$

Answer and solution

Answer: (B) $1$

$\displaystyle \frac{1}{\log_{b} 6} = \log_{6} b$

$= \log_{6}(3 \times 2) = \log_{6} 6 = 1$
$5^{2 + \log_{5} 3} =$
A $75$ B $28$ C $15$ D $125$

Answer and solution

Answer: (A) $75$

$= 5^2 \times 5^{\log_{5} 3}$

$= 25 \times 3 = 75$
If $3\log x - 2\log y = \log k$ , then $k =$
A $\displaystyle \frac{x^3}{y^2}$ B $x^3 - y^2$ C $\displaystyle \frac{3x}{2y}$ D $\displaystyle \frac{x^2}{y^3}$

Answer and solution

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

$\log x^3 - \log y^2 = \log k$

$\displaystyle \Rightarrow \log \frac{x^3}{y^2} = \log k \Rightarrow k = \frac{x^3}{y^2}$
If $\log_{5} x = \log_{25} 81$ , then $x =$
A $81$ B $18$ C $\displaystyle \frac{81}{2}$ D $9$

Answer and solution

Answer: (D) $9$

$\displaystyle \log_{25} 81 = \log_{5^{2}} 9^{2} = \frac{2}{2}\log_{5} 9 = \log_{5} 9$

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

Answer and solution

Answer: (D) $-2$

$\displaystyle y = \log_{3} \frac{1}{9}$

$= -2$

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MCQ test, Set 1