Logarithms: MCQ test, Set 5 | 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 $10^x = 10000$ , then $x =$
A $1000$ B $\displaystyle \frac{1}{4}$ C $4$ D $-4$

Answer and solution

Answer: (C) $4$

$10^{4} = 10000$

$\therefore \log_{10} 10000 = 4$
$\log_{12} 144 =$
A $-2$ B $12$ C $\displaystyle \frac{1}{2}$ D $2$

Answer and solution

Answer: (D) $2$

$12^{2} = 144$

$\therefore \log_{12} 144 = 2$
The base of a logarithm cannot be $1$ because
A $1$ is not a prime number B $\log_{1} 1 = 1$ always C $1^2 = 1$ and $1^6 = 1$ would give $2 = 6$ D $1^x$ is negative for some $x$

Answer and solution

Answer: (C) $1^2 = 1$ and $1^6 = 1$ would give $2 = 6$

$1^2 = 1 \Rightarrow 2 = \log_{1} 1$ , $1^6 = 1 \Rightarrow 6 = \log_{1} 1$

So $2 = 6$ , which has no meaning.
The logarithmic form of $\displaystyle 25^{\frac{3}{2}} = 125$ is
A $\displaystyle \frac{3}{2} = \log_{125} 25$ B $\displaystyle 25 = \log_{\frac{3}{2}} 125$ C $\displaystyle \frac{3}{2} = \log_{25} 125$ D $\displaystyle 125 = \log_{25} \frac{3}{2}$

Answer and solution

Answer: (C) $\displaystyle \frac{3}{2} = \log_{25} 125$

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

The power $\displaystyle \frac{3}{2}$ is the logarithm of $125$ to the base $25$ .
The exponential form of $\displaystyle \log_{81} 27 = \frac{3}{4}$ is
A $\displaystyle 27 = 81^{\frac{3}{4}}$ B $\displaystyle 27 = (\frac{3}{4})^{81}$ C $\displaystyle 81 = 27^{\frac{3}{4}}$ D $\displaystyle 27 = 81 \times \frac{3}{4}$

Answer and solution

Answer: (A) $\displaystyle 27 = 81^{\frac{3}{4}}$

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

So $\displaystyle 27 = 81^{\frac{3}{4}}$ .
$\log_{0.3} 0.3 - \log_{0.3} 1 =$
A $1$ B $0.3$ C $0$ D $-1$

Answer and solution

Answer: (A) $1$

$\log_{0.3} 0.3 = 1,\ \log_{0.3} 1 = 0$

$= 1 - 0 = 1$
$\log_{3} 8 + \log_{3} 5 =$
A $\log_{3} 13$ B $\log_{3} 3$ C $\log_{3} 8 \times \log_{3} 5$ D $\log_{3} 40$

Answer and solution

Answer: (D) $\log_{3} 40$

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

$= \log_{3} 40$
$\log_{8} 16 + \log_{8} 32 =$
A $\log_{8} 48$ B $\log_{8} 16 \times \log_{8} 32$ C $512$ D $3$

Answer and solution

Answer: (D) $3$

$= \log_{8}(16 \times 32) = \log_{8} 512$

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

Answer and solution

Answer: (A) $\displaystyle \frac{3}{5}$

$\displaystyle = \log_{2^{5}} 2^{3} = \frac{3}{5} \log_{2} 2$

$\displaystyle = \frac{3}{5}$
$\log_{2} 9^{5} =$
A $(\log_{2} 9)^{5}$ B $5\log_{2} 9$ C $\log_{2} 45$ D $9\log_{2} 5$

Answer and solution

Answer: (B) $5\log_{2} 9$

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

$\therefore \log_{2} 9^{5} = 5\log_{2} 9$
$\log_{13} 6 \cdot \log_{6} 13 =$
A $\displaystyle \frac{6}{13}$ B $78$ C $1$ D $0$

Answer and solution

Answer: (C) $1$

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

So the product is $1$ .
$\log_{q} p \cdot \log_{r} q \cdot \log_{t} r =$
A $\log_{t} pqr$ B $\log_{p} t$ C $\log_{t} p$ D $\log_{pqr} t$

Answer and solution

Answer: (C) $\log_{t} p$

$\log_{q} p \cdot \log_{r} q = \log_{r} p$

$\log_{r} p \cdot \log_{t} r = \log_{t} p$
$\displaystyle \frac{\log_{10} 1000}{\log_{10} 100} =$
A $\displaystyle \frac{3}{2}$ B $\displaystyle \frac{2}{3}$ C $10$ D $\log_{10} 10$

Answer and solution

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

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

$\displaystyle = \log_{100} 1000 = \frac{3}{2}$
$\displaystyle 15^{\log_{15} \frac{4}{7}} =$
A $\displaystyle \log_{15} \frac{4}{7}$ B $\displaystyle \frac{4}{7}$ C $\displaystyle \frac{7}{4}$ D $15$

Answer and solution

Answer: (B) $\displaystyle \frac{4}{7}$

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

So the value is $\displaystyle \frac{4}{7}$ .
$3^{\log_{10} 8} =$
A $8^{\log_{3} 10}$ B $8^{\log_{10} 3}$ C $10^{\log_{8} 3}$ D $3^{\log_{8} 10}$

Answer and solution

Answer: (B) $8^{\log_{10} 3}$

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

$\therefore 3^{\log_{10} 8} = 8^{\log_{10} 3}$
If $\log_{11} \sqrt[3]{x} = \log_{11} 3$ , then $x =$
A $27$ B $3$ C $9$ D $81$

Answer and solution

Answer: (A) $27$

$\Rightarrow \sqrt[3]{x} = 3$

$\Rightarrow x = 3^{3} = 27$
If $\log_{6}(4x - 2) = 1$ , then $x =$
A $1$ B $\displaystyle \frac{1}{2}$ C $\displaystyle \frac{3}{4}$ D $2$

Answer and solution

Answer: (D) $2$

$\log_{a} a = 1$

$4x - 2 = 6$

$\Rightarrow x = 2$
If $\log_{10} x + \log_{10}(x - 21) = 2$ , then
A $x = 25$ (or) $x = -4$ B $x = -4$ C $x = 25$ D $x = 4$

Answer and solution

Answer: (C) $x = 25$

$\Rightarrow \log_{10} x(x - 21) = 2$

$\Rightarrow x^2 - 21x = 10^{2} = 100$

$\Rightarrow x^2 - 21x - 100 = 0$

$\Rightarrow (x + 4)(x - 25) = 0$

$\Rightarrow x = -4$ (or) $x = 25$

$x = -4$ is rejected: the log of a negative number is not defined.

$\therefore x = 25$
$\log_{16} 64 - \log_{64} 16 =$
A $\displaystyle \frac{5}{6}$ B $\displaystyle \frac{3}{2}$ C $\displaystyle \frac{13}{6}$ D $0$

Answer and solution

Answer: (A) $\displaystyle \frac{5}{6}$

$\displaystyle = \frac{6}{4} - \frac{4}{6}$ (writing both numbers as powers of $2$ )

$\displaystyle = \frac{5}{6}$
If $\log_{m} n = q$ , then $\log_{n^2} m =$
A $\displaystyle \frac{q}{2}$ B $2q$ C $\displaystyle \frac{2}{q}$ D $\displaystyle \frac{1}{2q}$

Answer and solution

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

$\displaystyle \log_{n^2} m = \frac{1}{2}\log_{n} m$

$\displaystyle = \frac{1}{2} \cdot \frac{1}{\log_{m} n} = \frac{1}{2q}$
If $\log_{\sqrt[3]{p}} q \cdot \log_{\sqrt{r}} p = k\log_{r} q$ , then $k =$
A $\displaystyle \frac{3}{2}$ B $6$ C $5$ D $\displaystyle \frac{1}{6}$

Answer and solution

Answer: (B) $6$

$\displaystyle \log_{p^{\frac{1}{3}}} q = 3\log_{p} q,\ \log_{r^{\frac{1}{2}}} p = 2\log_{r} p$

$= 6\log_{p} q \log_{r} p = 6\log_{r} q \Rightarrow k = 6$
$100^{\log_{10} 3} =$
A $3$ B $9$ C $300$ D $30$

Answer and solution

Answer: (B) $9$

$100^{\log_{10} 3} = (10^{\log_{10} 3})^2$

$= 3^2 = 9$
If $c^3 - b^3 = 64$ , then $\log_{4}(c - b) + \log_{4}(c^2 + cb + b^2) =$
A $2$ B $64$ C $16$ D $3$

Answer and solution

Answer: (D) $3$

$= \log_{4}(c^3 - b^3) = \log_{4} 64$

$= \log_{4} 4^3 = 3$
If $\displaystyle \log_{x} 125 = \frac{3}{2}$ , then $x =$
A $25$ B $\displaystyle \frac{1}{25}$ C $-25$ D $\displaystyle \frac{1}{125}$

Answer and solution

Answer: (A) $25$

$\displaystyle 125 = x^{\frac{3}{2}}$

$\displaystyle \Rightarrow x = 125^{\frac{2}{3}} = 25$
The graph of $y = \log_{7} x$ cuts the x-axis where $x =$
A $7$ B $1$ C $\displaystyle \frac{1}{7}$ D $0$

Answer and solution

Answer: (B) $1$

$y = 0 \Rightarrow \log_{7} x = 0$

$\Rightarrow x = 7^0 = 1$

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MCQ test, Set 5