Logarithms: MCQ test, Set 3 | 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 $216$ to the base $6$ is
A $3$ B $\displaystyle \frac{1}{3}$ C $36$ D $-3$

Answer and solution

Answer: (A) $3$

$6^{3} = 216$

$\therefore \log_{6} 216 = 3$
If $7^x = 343$ , then $x =$
A $49$ B $\displaystyle \frac{1}{3}$ C $3$ D $-3$

Answer and solution

Answer: (C) $3$

$7^{3} = 343$

$\therefore \log_{7} 343 = 3$
Which of the following logarithms of $5$ is not defined?
A $\log_{1} 5$ B $\log_{25} 5$ C $\log_{5} 5$ D $\displaystyle \log_{\frac{1}{5}} 5$

Answer and solution

Answer: (A) $\log_{1} 5$

The base cannot be $1$ : $1^x = 1$ for every $x$ , never $5$ .

So $\log_{1} 5$ is not defined.
The logarithmic form of $19^{2} = 361$ is
A $2 = \log_{19} 361$ B $361 = \log_{19} 2$ C $2 = \log_{361} 19$ D $19 = \log_{2} 361$

Answer and solution

Answer: (A) $2 = \log_{19} 361$

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

The power $2$ is the logarithm of $361$ to the base $19$ .
The exponential form of $\displaystyle \log_{64} 4 = \frac{1}{3}$ is
A $\displaystyle 64 = 4^{\frac{1}{3}}$ B $\displaystyle 4 = 64^{\frac{1}{3}}$ C $\displaystyle 4 = 64 \times \frac{1}{3}$ D $\displaystyle 4 = (\frac{1}{3})^{64}$

Answer and solution

Answer: (B) $\displaystyle 4 = 64^{\frac{1}{3}}$

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

So $\displaystyle 4 = 64^{\frac{1}{3}}$ .
$5\log_{8} 8 + 3\log_{8} 1 =$
A $5$ B $3$ C $15$ D $8$

Answer and solution

Answer: (A) $5$

$\log_{8} 8 = 1,\ \log_{8} 1 = 0$

$= 5(1) + 3(0) = 5$
$\log_{9} 27 + \log_{9} 3 =$
A $2$ B $\log_{9} 27 \times \log_{9} 3$ C $\log_{9} 30$ D $81$

Answer and solution

Answer: (A) $2$

$= \log_{9}(27 \times 3) = \log_{9} 81$

$= 2$ since $9^{2} = 81$
$\log_{5} 36 - \log_{5} 4 =$
A $\log_{5} 144$ B $\log_{5} 9$ C $\log_{5} 32$ D $\log_{5} 36 \div \log_{5} 4$

Answer and solution

Answer: (B) $\log_{5} 9$

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

$= \log_{5} 9$
$\log_{49} 343 =$
A $7$ B $1$ C $\displaystyle \frac{3}{2}$ D $\displaystyle \frac{2}{3}$

Answer and solution

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

$\displaystyle = \log_{7^{2}} 7^{3} = \frac{3}{2} \log_{7} 7$

$\displaystyle = \frac{3}{2}$
$\log_{3} 10^{4} =$
A $10\log_{3} 4$ B $(\log_{3} 10)^{4}$ C $\log_{3} 40$ D $4\log_{3} 10$

Answer and solution

Answer: (D) $4\log_{3} 10$

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

$\therefore \log_{3} 10^{4} = 4\log_{3} 10$
$\displaystyle \frac{1}{\log_{6} 5} =$
A $-\log_{6} 5$ B $\displaystyle \log_{5} \frac{1}{6}$ C $\log_{6} 5$ D $\log_{5} 6$

Answer and solution

Answer: (D) $\log_{5} 6$

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

$\displaystyle \therefore \frac{1}{\log_{6} 5} = \log_{5} 6$
$\log_{3} 125 \cdot \log_{5} 3 =$
A $-3$ B $3$ C $\displaystyle \frac{1}{3}$ D $6$

Answer and solution

Answer: (B) $3$

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

$= \log_{5} 125 = 3$
$\displaystyle \frac{\log_{10} 6561}{\log_{10} 9} =$
A $729$ B $\log_{10} 729$ C $4$ D $\displaystyle \frac{1}{4}$

Answer and solution

Answer: (C) $4$

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

$= \log_{9} 6561 = 4$
$3^{\log_{3} 6} \times 4^{\log_{4} 5} =$
A $11$ B $12$ C $30$ D $20$

Answer and solution

Answer: (C) $30$

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

$= 6 \times 5 = 30$
$2^{\log_{5} 6} =$
A $5^{\log_{6} 2}$ B $6^{\log_{5} 2}$ C $2^{\log_{6} 5}$ D $6^{\log_{2} 5}$

Answer and solution

Answer: (B) $6^{\log_{5} 2}$

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

$\therefore 2^{\log_{5} 6} = 6^{\log_{5} 2}$
If $\log_{3}(3x + 1) = \log_{3} 19$ , then $x =$
A $-6$ B $7$ C $\displaystyle \frac{20}{3}$ D $6$

Answer and solution

Answer: (D) $6$

$\Rightarrow 3x + 1 = 19$

$\Rightarrow 3x = 18$

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

Answer and solution

Answer: (C) $5$

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

$2x - 1 = 3^{2} = 9$

$\Rightarrow x = 5$
If $\log_{5} x + \log_{5}(x + 4) = 1$ , then
A $x = -5$ B $x = 5$ C $x = 1$ (or) $x = -5$ D $x = 1$

Answer and solution

Answer: (D) $x = 1$

$\Rightarrow \log_{5} x(x + 4) = 1$

$\Rightarrow x^2 + 4x = 5$

$\Rightarrow x^2 + 4x - 5 = 0$

$\Rightarrow (x - 1)(x + 5) = 0$

$\Rightarrow x = 1$ (or) $x = -5$

$x = -5$ is rejected: the log of a negative number is not defined.

$\therefore x = 1$
$\log_{0.2} 0.0016 =$
A $-4$ B $4$ C $\displaystyle \frac{1}{4}$ D $8$

Answer and solution

Answer: (B) $4$

$0.0016 = (0.2)^4$

$\therefore \log_{0.2}(0.2)^4 = 4$
$\displaystyle \frac{\log 2\sqrt{2} + \log 5\sqrt{5}}{\log 10} =$
A $\displaystyle \frac{2}{3}$ B $\displaystyle \frac{1}{2}$ C $1$ D $\displaystyle \frac{3}{2}$

Answer and solution

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

$\displaystyle 2\sqrt{2} \times 5\sqrt{5} = (2 \times 5)^{\frac{3}{2}} = 10^{\frac{3}{2}}$

$\displaystyle = \log_{10} 10^{\frac{3}{2}} = \frac{3}{2}$
$\log_{2} 9 \cdot \log_{3} 4 =$
A $6$ B $2$ C $1$ D $4$

Answer and solution

Answer: (D) $4$

$\log_{2} 9 = 2\log_{2} 3,\ \log_{3} 4 = 2\log_{3} 2$

$= 4\log_{2} 3 \cdot \log_{3} 2 = 4(1) = 4$
$9^{\log_{3} 4} =$
A $4$ B $2$ C $16$ D $8$

Answer and solution

Answer: (C) $16$

$9^{\log_{3} 4} = (3^{\log_{3} 4})^2$

$= 4^2 = 16$
If $x^2 + y^2 = 34xy$ , then $2\log(x + y) =$
A $\log x + \log y + 2\log 6$ B $\log x + \log y + \log 34$ C $\log x + \log y + 2\log 17$ D $\log x + \log y + 2\log 4$

Answer and solution

Answer: (A) $\log x + \log y + 2\log 6$

$(x + y)^2 = x^2 + y^2 + 2xy = 36xy$

$2\log(x + y) = \log 36xy = 2\log 6 + \log x + \log y$
If $\displaystyle \log_{\frac{1}{x}} 25 = -2$ , then
A $\displaystyle x = \frac{1}{5}$ B $x = 5$ C $x = -5$ D $x = 5$ (or) $x = -5$

Answer and solution

Answer: (B) $x = 5$

$\displaystyle 25 = \left(\frac{1}{x}\right)^{-2} = x^{2}$

$\Rightarrow x = 5$ (the negative value is rejected: the base must be positive)
The graph of $\displaystyle y = \log_{\frac{2}{3}} x$
A is a straight line B rises from left to right C passes through $(0, 1)$ D falls from left to right

Answer and solution

Answer: (D) falls from left to right

The base $\displaystyle \frac{2}{3}$ is between $0$ and $1$ .

For $0 < a < 1$ the graph of $\log_{a} x$ falls from left to right.

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MCQ test, Set 3