Practice MCQ Questions and Answer on Logarithm

1.

If $${\log }_{10000}x=-\frac{1}{4}\text{,}$$    then the value of x is = ?

• (A) $$\frac{1}{10}$$
• (B) $$\frac{1}{100}$$
• (C) $$\frac{1}{1000}$$
• (D) $$\frac{1}{10000}$$

Show Answer

 $$\eqalign{ & {\text{let }}{\log _{10000}}x = - \frac{1}{4} \cr & \Rightarrow x = {\left( {10000} \right)^{ - \frac{1}{4}}} \cr & = {\left( {{{10}^4}} \right)^{ - \frac{1}{4}}} \cr & = {10^{ - 1}} \cr & = \frac{1}{{10}} \cr} $$

2.

The value of $$\frac{1}{{\log }_{3}60}+$$  $$\frac{1}{{\log }_{4}60}+$$  $$\frac{1}{{\log }_{5}60}$$   is :

• (A) 0
• (B) 1
• (C) 5
• (D) 60

Show Answer

 Given expression $$\eqalign{ & = {\log _{60}}3 + {\log _{60}}4 + {\log _{60}}5 \cr & = {\log _{60}}\left( {3 \times 4 \times 5} \right) \cr & = {\log _{60}}60 \cr & = 1 \cr} $$

3.

If $${\log }_x\left(\frac{9}{16}\right)=-\frac{1}{2},$$    then x is equal to:

• (A) $$-\frac{3}{4}$$
• (B) $$\frac{3}{4}$$
• (C) $$\frac{81}{256}$$
• (D) $$\frac{256}{81}$$

Show Answer

 $$\eqalign{ & {\log _x}\left( {{9 \over {16}}} \right) = - {1 \over 2} \cr & \Rightarrow {x^{ - {1 \over 2}}} = {9 \over {16}} \cr & \Rightarrow {1 \over {\sqrt x }} = {9 \over {16}} \cr & \Rightarrow \sqrt x = {{16} \over 9} \cr & \Rightarrow x = {\left( {{{16} \over 9}} \right)^2} \cr & \Rightarrow x = {{256} \over {81}} \cr} $$

4.

$$2{{\log }_{10}}^5+$$  $${\log }_{10}8-$$  $$\frac{1}{2}{\log }_{10}4$$   = ?

• (A) 2
• (B) 4
• (C) $$\text{2 - 2 lo}{{\text{g}}_{10}}^2$$
• (D) $$\text{4 - 4 lo}{{\text{g}}_{10}}^2$$

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 $$\eqalign{ & 2{\log _{10}}5 + {\log _{10}}8 - \frac{1}{2}{\log _{10}}4 \cr & = {\log _{10}}\left( {{5^2}} \right) + {\log _{10}}8 - {\log _{10}}\left( {{4^{\frac{1}{2}}}} \right) \cr & = {\log _{10}}25 + {\log _{10}}8 - {\log _{10}}2 \cr & = {\log _{10}}\left( {\frac{{25 \times 8}}{2}} \right) \cr & = {\log _{10}}100 \cr & = 2 \cr} $$

5.

If log₁₀ 2 = 0.3010, then log₂ 10 is equal to:

• (A) $$\frac{699}{301}$$
• (B) $$\frac{1000}{301}$$
• (C) 0.3010
• (D) 0.6990

Show Answer

 $$\eqalign{ & {\log _2}10 \cr & = \frac{1}{{{{\log }_{10}}2}} \cr & = \frac{1}{{0.3010}} \cr & = \frac{{10000}}{{3010}} \cr & = \frac{{1000}}{{301}} \cr} $$

6.

If $${\log }_{10}2=a$$  and $${\log }_{10}3=b,$$   then $${\log }_{5}12$$ = ?

• (A) $$\frac{a+b}{1+a}$$
• (B) $$\frac{2a+b}{1+a}$$
• (C) $$\frac{a+2b}{1+a}$$
• (D) $$\frac{2a+b}{1-a}$$

Show Answer

 $$\eqalign{ & {\log _5}12 = {\log _5}\left( {3 \times 4} \right) \cr & = {\log _5}3 + {\log _5}4 \cr & = {\log _5}3 + 2{\log _5}2 \cr & = \frac{{{{\log }_{10}}3}}{{{{\log }_{10}}5}} + \frac{{2{{\log }_{10}}2}}{{{{\log }_{10}}5}} \cr} $$ $$ = \frac{{{{\log }_{10}}3}}{{{{\log }_{10}}10 - {{\log }_{10}}2}}$$   $$ + \frac{{2{{\log }_{10}}2}}{{{{\log }_{10}}10 - {{\log }_{10}}2}}$$ $$\eqalign{ & = \frac{b}{{1 - a}} + \frac{{2a}}{{1 - a}} \cr & = \frac{{2a + b}}{{1 - a}} \cr} $$

7.

If log₁₀ 2 = 0.3010, the value of log₁₀ 80 is:

• (A) 1.6020
• (B) 1.9030
• (C) 3.9030
• (D) None of these

Show Answer

 $$\eqalign{ & {\log _{10}}80 \cr & = {\log _{10}}\left( {8 \times 10} \right) \cr & = {\log _{10}}8 + {\log _{10}}10 \cr & = {\log _{10}}\left( {{2^3}} \right) + 1 \cr & = 3{\log _{10}}2 + 1 \cr & = \left( {3 \times 0.3010} \right) + 1 \cr & = 1.9030 \cr} $$

8.

If log 27 = 1.431, then the value of log 9 is:

• (A) 0.934
• (B) 0.945
• (C) 0.954
• (D) 0.958

Show Answer

 $$\eqalign{ & \log 27 = 1.431 \cr & \Rightarrow \log \left( {{3^3}} \right) = 1.431 \cr & \Rightarrow 3\,\log \,3 = 1.431 \cr & \Rightarrow \log \,3 = 0.477 \cr & \therefore \log \,9 = \log \left( {{3^2}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2\,\log \,3 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {2 \times 0.477} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.954 \cr} $$

9.

$$\frac{\log \sqrt{8}}{\log 8}$$  is equal to:

• (A) $$\frac{1}{6}$$
• (B) $$\frac{1}{4}$$
• (C) $$\frac{1}{2}$$
• (D) $$\frac{1}{8}$$

Show Answer

 $$\eqalign{ & \frac{{\log \sqrt 8 }}{{\log 8}} = {\frac{{\log \left( 8 \right)}}{{\log 8}}^{\frac{1}{2}}} \cr & = \frac{{\frac{1}{2}\log 8}}{{\log 8}} \cr & = \frac{1}{2} \cr} $$

10.

The value of $${\log }_5\frac{\left(125\right)\left(625\right)}{25}$$    is equal to -

• (A) 725
• (B) 5
• (C) 3125
• (D) 6

Show Answer

 $$\eqalign{ & {\text{lo}}{{\text{g}}_5}\frac{{\left( {125} \right)\left( {625} \right)}}{{25}} \cr & = {\text{lo}}{{\text{g}}_5}\left( {\frac{{{5^3} \times {5^4}}}{{{5^2}}}} \right) \cr & = {\text{lo}}{{\text{g}}_5}{5^5} \cr & {\text{ = 5lo}}{{\text{g}}_5}5 \cr & = 5 \cr} $$