31.
If ÃÂÃÂ ÃÂÃÂ then x is equal to = ?
- (A) 0
- (B) 12
- (C) 128
- (D) 512
Solution:
$$\eqalign{ & {{\log }_2}\left[ {{\text{ }}{{\log }_3}\left( {{\text{ }}{{\log }_2}x} \right)} \right] = 1 \cr & \Rightarrow {\log _3}\left( {{{\log }_2}x} \right) = {2^1} = 2 \cr & \Rightarrow {\log _2}x = {3^2} = 9 \cr & \Rightarrow x = {2^9} = 512 \cr} $$
32. If ÃÂÃÂ ÃÂÃÂ the number of digits in is -
- (A) 30
- (B) 31
- (C) 100
- (D) 200
Solution:
$$\eqalign{ & \log {4^{50}} \cr & = 50\log 4 \cr & = 50\log {2^2} \cr & = \left( {50 \times 2} \right)\log 2 \cr & = 100 \times \log 2 \cr & = \left( {100 \times 0.30103} \right) \cr & = 30.103 \cr & \therefore {\text{characteristic}} = 30, \cr} $$ Hence, the number of digits in $${{\text{4}}^{50}} = 31$$
33.
If log 2 = 0.30103, the number of digits in 264 is:
- (A) 18
- (B) 19
- (C) 20
- (D) 21
Solution:
$$\eqalign{ & \log \left( {{2^{64}}} \right) \cr & = 64 \times \log 2 \cr & = \left( {64 \times 0.30103} \right) \cr & = 19.26592 \cr} $$ Its characteristic is 19. Hence, then number of digits in 264 is 20.
34.
If log10 2 = 0.3010, the value of log10 80 is:
- (A) 1.6020
- (B) 1.9030
- (C) 3.9030
- (D) None of these
Solution:
$$\eqalign{ & {\log _{10}}80 \cr & = {\log _{10}}\left( {8 \times 10} \right) \cr & = {\log _{10}}8 + {\log _{10}}10 \cr & = {\log _{10}}\left( {{2^3}} \right) + 1 \cr & = 3{\log _{10}}2 + 1 \cr & = \left( {3 \times 0.3010} \right) + 1 \cr & = 1.9030 \cr} $$
35.
If ÃÂÃÂ ÃÂÃÂ ÃÂÃÂ then x equals to -
- (A) 3
- (B) 9
- (C) 27
- (D) None of these
Solution:
$$\eqalign{ & \Rightarrow {\log _3}x + {\log _9}{x^2} + {\log _{27}}{x^3} = 9 \cr & \Rightarrow {\log _3}x + {\log _{{3^2}}}{x^2} + {\log _{{3^3}}}{x^3} = 9 \cr & \Rightarrow {\log _3}x + \frac{2}{2}{\log _3}x + \frac{3}{3}{\log _3}x = 9 \cr & \Rightarrow 3{\log _3}x = 9 \cr & \Rightarrow {\log _3}x = 3 \cr & \Rightarrow x = {3^3} = 27 \cr} $$
36. The number of digits in ÃÂÃÂ when expressed in usual form, is -
- (A) 16
- (B) 17
- (C) 18
- (D) 19
Solution:
$$\eqalign{ & \log \left( {{4^9} \times {5^{17}}} \right) \cr & = \log \left( {{4^9}} \right) + \log \left( {{5^{17}}} \right) \cr & = \log \left( {{2^2}} \right)^9 + \log \left( {{5^{17}}} \right) \cr & = \log \left( {{2^{18}}} \right) + \log \left( {{5^{17}}} \right) \cr & = 18\log 2 + 17\log 5 \cr & = 18\log 2 + 17\left( {\log 10 - \log 2} \right) \cr & = 18\log 2 + 17\log 10 - 17\log 2 \cr & = \log 2 + 17\log 10 \cr & = 0.3010 + 17 \times 1 = 17.3010 \cr & \therefore {\text{Characteristic}} = 17 \cr & {\text{Hence, the number of digits in }}\left( {{4^9} \times {5^{17}}} \right) = 18 \cr} $$
37.
The value of ÃÂÃÂ is = ?
Solution:
$$\eqalign{ & {\text{lo}}{{\text{g}}_{10}}\left( {0.0001} \right)\, \cr & = {\text{lo}}{{\text{g}}_{10}}\left( {\frac{1}{{10000}}} \right) \cr & \, = {\text{lo}}{{\text{g}}_{10}}\left( {\frac{1}{{{{10}^4}}}} \right) \cr & \, = {\text{lo}}{{\text{g}}_{10}}{10^{ - 4}} \cr & \, = - 4\,{\text{lo}}{{\text{g}}_{10}}10 \cr & = - 4{\text{ }} \cr} $$
38.
ÃÂÃÂ is equal to:
Solution:
$$\eqalign{ & \frac{{\log \sqrt 8 }}{{\log 8}} = {\frac{{\log \left( 8 \right)}}{{\log 8}}^{\frac{1}{2}}} \cr & = \frac{{\frac{1}{2}\log 8}}{{\log 8}} \cr & = \frac{1}{2} \cr} $$
39.
If ax = by, then:
- (A)
- (B)
- (C)
- (D) None of these
Solution:
$$\eqalign{ & {a^x} = {b^y} \cr & \Rightarrow \log {a^x} = \log {b^y} \cr & \Rightarrow x\log a = y\log b \cr & \Rightarrow {{\log a} \over {\log b}} = {y \over x} \cr} $$
40. If ÃÂÃÂ and ÃÂÃÂ then a in terms of b is -
Solution:
$$\eqalign{ & a = {\log _8}225 \cr & = {\log _{{2^3}}}\left( {{{15}^2}} \right) \cr & = \frac{2}{3}{\log _2}15 \cr & = \frac{{2b}}{3} \cr} $$