41.
The value of $${\frac{1}{{{{\log }_3}60}} + }$$ ÃÂÃÂ $${\frac{1}{{{{\log }_4}60}} + }$$ ÃÂÃÂ $${\frac{1}{{{{\log }_5}60}}}$$ ÃÂÃÂ is :
Solution:
Given expression $$\eqalign{ & = {\log _{60}}3 + {\log _{60}}4 + {\log _{60}}5 \cr & = {\log _{60}}\left( {3 \times 4 \times 5} \right) \cr & = {\log _{60}}60 \cr & = 1 \cr} $$
42.
If $${\log _{10000}}x = - \frac{1}{4}{\text{,}}$$ ÃÂÃÂ ÃÂÃÂ then the value of x is = ?
- (A) $$\frac{1}{{10}}$$
- (B) $$\frac{1}{{100}}$$
- (C) $$\frac{1}{{1000}}$$
- (D) $$\frac{1}{{10000}}$$
Solution:
$$\eqalign{ & {\text{let }}{\log _{10000}}x = - \frac{1}{4} \cr & \Rightarrow x = {\left( {10000} \right)^{ - \frac{1}{4}}} \cr & = {\left( {{{10}^4}} \right)^{ - \frac{1}{4}}} \cr & = {10^{ - 1}} \cr & = \frac{1}{{10}} \cr} $$
43.
$${\frac{1}{{\left( {{{\log }_a}bc} \right) + 1}} + }$$ ÃÂÃÂ $${\frac{1}{{\left( {{{\log }_b}ca} \right) + 1}} + }$$ ÃÂÃÂ $${\frac{1}{{\left( {{{\log }_c}ab} \right) + 1}}}$$ ÃÂÃÂ is equal to -
- (A) 1
- (B) $$\frac{3}{2}$$
- (C) 2
- (D) 3
Solution:
$$\eqalign{ & {\text{Given}}\,\,\,{\text{Expression}} \cr & = \frac{1}{{{{\log }_a}bc + {{\log }_a}a}} + \frac{1}{{{{\log }_b}ca + {{\log }_b}b}} + \frac{1}{{{{\log }_c}ab + {{\log }_c}c}} \cr & = \frac{1}{{{{\log }_a}\left( {abc} \right)}} + \frac{1}{{{{\log }_b}\left( {abc} \right)}} + \frac{1}{{{{\log }_c}\left( {abc} \right)}} \cr & = {\log _{abc}}a + {\log _{abc}}b + {\log _{abc}}c \cr & = {\log _{abc}}\left( {abc} \right) \cr & = 1 \cr} $$
44.
$$\frac{1}{2}\left( {\log x + \log y} \right)$$ ÃÂÃÂ ÃÂÃÂ will equal to $$\log \left( {\frac{{x + y}}{2}} \right)$$ ÃÂÃÂ if -
- (A) y = 0
- (B) x = $$\sqrt {\text{y}} $$
- (C) x = y
- (D) x = $$\frac{{\text{y}}}{2}$$
Solution:
$$\eqalign{ & \frac{1}{2}\left( {\log x + \log y} \right) = \log \left( {\frac{{x + y}}{2}} \right) \cr & \Rightarrow \frac{1}{2}\log \left( {xy} \right) = \log \left( {\frac{{x + y}}{2}} \right) \cr & \Rightarrow \log {\left( {xy} \right)^{\frac{1}{2}}} = \log \left( {\frac{{x + y}}{2}} \right) \cr & \Rightarrow {\left( {xy} \right)^{\frac{1}{2}}} = \left( {\frac{{x + y}}{2}} \right) \cr & \Rightarrow xy = {\left( {\frac{{x + y}}{2}} \right)^2} \cr & \Rightarrow 4xy = {x^2} + {y^2} + 2xy \cr & \Rightarrow {x^2} + {y^2} - 2xy = 0 \cr & \Rightarrow {\left( {x - y} \right)^2} = 0 \cr & \Rightarrow x - y = 0 \cr & \Rightarrow x = y\, \cr} $$
45.
$${\text{If}}\,{\text{log}}\frac{a}{b} + {\text{log}}\frac{b}{a} = {\text{log}}(a + b),$$ ÃÂÃÂ ÃÂÃÂ ÃÂÃÂ then:
- (A) a + b = 1
- (B) a - b = 1
- (C) a = b
- (D) a2 - b2 = 1
Solution:
$$\eqalign{ & \log {a \over b} + \log {b \over a} = \log \left( {a + b} \right) \cr & \Rightarrow \log \left( {a + b} \right) = \log \left( {{a \over b} \times {b \over a}} \right) = \log 1 \cr & So,a + b = 1 \cr} $$
46.
What is the value of $${\left[ {{\text{lo}}{{\text{g}}_{10}}\left( {{\text{5lo}}{{\text{g}}_{10}}100} \right)\,} \right]^2}$$ ÃÂÃÂ ÃÂÃÂ = ?
- (A) 1
- (B) 2
- (C) 10
- (D) 25
Solution:
$$\eqalign{ & {\left[ {{\text{lo}}{{\text{g}}_{10}}\left( {{\text{5lo}}{{\text{g}}_{10}}100} \right)\,} \right]^2} \cr & = {\left[ {{\text{lo}}{{\text{g}}_{10}}\left\{ {{\text{5lo}}{{\text{g}}_{10}}{{\left( {10} \right)}^2}} \right\}\,} \right]^2} \cr & = \,{\left[ {{\text{lo}}{{\text{g}}_{10}}\left( {5 \times 2} \right)} \right]^2} \cr & = {\left( {{\text{lo}}{{\text{g}}_{10}}10} \right)^2} \cr & = 1 \cr} $$
47.
If $${\log _{10}}125 + {\log _{10}}8 = x,$$ ÃÂÃÂ ÃÂÃÂ then x is equal to -
- (A) $$\frac{1}{3}$$
- (B) 0.064
- (C) -3
- (D) 3
Solution:
$$\eqalign{ & {\log _{10}}125 + {\log _{10}}8 = x \cr & \Rightarrow {\log _{10}}\left( {125 \times 8} \right) = x \cr & \Rightarrow x = {\log _{10}}\left( {1000} \right) \cr & \Rightarrow x = {\log _{10}}{\left( {10} \right)^3} \cr & \Rightarrow x = 3{\log _{10}}10 \cr & \Rightarrow x = 3 \cr} $$