Logarithm

1.Which of the following statements is not correct?

1. log₁₀ 10 = 1
2. log (2 + 3) = log (2 x 3)
3. log₁₀ 1 = 0
4. log (1 + 2 + 3) = log 1 + log 2 + log 3

Answer:Option 1

Explanation:

Solution:

\begin{aligned} A  =  Since lo g_{a} a = 1, so lo g_{10} 10 = 1 \\ B = log (2 + 3) = 5 \\ and log (2 \times 3) = log 6 \\ = log 2 + log 3 \\ ∴ log (2 + 3) \neq log (2 \times 3) . \\ C = Since lo g_{a} 1 = 0, s o lo g_{10} 1 = 0. \\ D  =  log (1 + 2 + 3) = log 6 \\ = log (1 \times 2 \times 3) \\ = log 1 + log 2 + log 3. \\ So (2) is incorrect \end{aligned}

$\eqalign{ & {\text{A = Since lo}}{{\text{g}}_a}a = 1,{\text{ so lo}}{{\text{g}}_{10}}10 = 1 \cr & \cr & B = \,{\text{log}}\left( {2 + 3} \right) = 5\, \cr & \,\,\,{\text{and log}}\left( {2 \times 3} \right) = {\text{log}}6 \cr & = {\text{log}}2 + {\text{log}}3 \cr & \therefore \,{\text{log}}\left( {2 + 3} \right) \ne {\text{log}}\left( {2 \times 3} \right). \cr & \cr & C = \,{\text{ Since lo}}{{\text{g}}_a}1 = 0,so{\text{ lo}}{{\text{g}}_{10}}1 = 0. \cr & \cr & {\text{D = log}}\left( {1 + 2 + 3} \right) = {\text{ log}}6 \cr & = {\text{ log}}\left( {1 \times 2 \times 3} \right) \cr & = {\text{ log}}1 + {\text{ log}}2 + {\text{ log}}3. \cr & {\text{So (2) is incorrect}} \cr}$

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2.If log 2 = 0.3010 and log 3 = 0.4771, the value of log₅ 512 is:

1. 2.870
2. 2.967
3. 3.876
4. 3.912

Answer:Option 1

Explanation:

Solution:

\begin{aligned} \log_{5} 512 \\ = \frac{\log 512}{\log 5} \\ = \frac{\log 2^{9}}{\log (\frac{10}{2})} \\ = \frac{9 \log 2}{\log 10 - \log 2} \\ = \frac{(9 \times 0.3010)}{1 - 0.3010} \\ = \frac{2.709}{0.699} \\ = \frac{2709}{699} \\ = 3.876 \end{aligned}

$\eqalign{ & {\log _5}512 \cr & = {{\log 512} \over {\log 5}} \cr & = {{\log {2^9}} \over {\log \left( {{{10} \over 2}} \right)}} \cr & = {{9\log 2} \over {\log 10 - \log 2}} \cr & = {{\left( {9 \times 0.3010} \right)} \over {1 - 0.3010}} \cr & = {{2.709} \over {0.699}} \cr & = {{2709} \over {699}} \cr & = 3.876 \cr}$

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3.

\frac{\log \sqrt{8}}{\log 8}

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

1. ¹/₆
2. ¹/₄
3. ¹/₂
4. ¹/₈

Answer:Option 1

Explanation:

Solution:

\begin{aligned} \frac{\log \sqrt{8}}{\log 8} = {\frac{\log (8)}{\log 8}}^{\frac{1}{2}} \\ = \frac{\frac{1}{2} \log 8}{\log 8} \\ = \frac{1}{2} \end{aligned}

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

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4.

I f \log \frac{a}{b} + \log \frac{b}{a} = \log (a + b), t h e n :

${\rm{If}}\log {a \over b} + \log {b \over a} = \log \left( {a + b} \right),{\rm{then:}}$

1. a + b = 1
2. a - b = 1
3. a = b
4. a² - b² = 1

Answer:Option 1

Explanation:

Solution:

\begin{aligned} \log \frac{a}{b} + \log \frac{b}{a} = \log (a + b) \\ \Rightarrow \log (a + b) = \log (\frac{a}{b} \times \frac{b}{a}) = \log 1 \\ S o, a + b = 1 \end{aligned}

$\eqalign{ & \log {a \over b} + \log {b \over a} = \log \left( {a + b} \right) \cr & \Rightarrow \log \left( {a + b} \right) = \log \left( {{a \over b} \times {b \over a}} \right) = \log 1 \cr & So,a + b = 1 \cr}$

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5.

I f \log_{10} 7 = a, t h e n \log_{10} (\frac{1}{70}) i s e q u a l t o

${\rm{If}}{\kern 1pt} \,{\kern 1pt} {\log _{10}}7 = a,{\kern 1pt} {\kern 1pt} {\rm{then}}\,{\kern 1pt} {\kern 1pt} {\log _{10}}\left( {{1 \over {70}}} \right){\rm{is}}\,{\kern 1pt} {\rm{equal}}\,{\kern 1pt} {\rm{to}}$

1. - (1 + a)
2. (1 + a)^-1
3. ^a/₁₀
4. ¹/_10a

Answer:Option 1

Explanation:

Solution:

\begin{aligned} \log_{10} (\frac{1}{70}) \\ = \log_{10} 1 - \log_{10} 70 \\ = - \log_{10} (7 \times 10) \\ = - (\log_{10} 7 + \log_{10} 10) \\ = - (a + 1) \end{aligned}

$\eqalign{ & {\log _{10}}\left( {{1 \over {70}}} \right) \cr & = {\log _{10}}1 - {\log _{10}}70 \cr & = - {\log _{10}}\left( {7 \times 10} \right) \cr & = - \left( {{{\log }_{10}}7 + {{\log }_{10}}10} \right) \cr & = - \left( {a + 1} \right) \cr}$

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

1. ⁶⁹⁹/₃₀₁
2. ¹⁰⁰⁰/₃₀₁
3. 0.3010
4. 0.6990

Answer:Option 1

Explanation:

Solution:

\begin{aligned} \log_{2} 10 \\ = \frac{1}{\log_{10} 2} \\ = \frac{1}{0.3010} \\ = \frac{10000}{3010} \\ = \frac{1000}{301} \end{aligned}

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

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

1. 1.6020
2. 1.9030
3. 3.9030
4. None of these

Answer:Option 1

Explanation:

Solution:

\begin{aligned} \log_{10} 80 \\ = \log_{10} (8 \times 10) \\ = \log_{10} 8 + \log_{10} 10 \\ = \log_{10} (2^{3}) + 1 \\ = 3 \log_{10} 2 + 1 \\ = (3 \times 0.3010) + 1 \\ 1.9030 \end{aligned}

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

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8.

\begin{aligned} The value of \\ (\frac{1}{\log_{3} 60} + \frac{1}{\log_{4} 60} + \frac{1}{\log_{5} 60}) is : \end{aligned}

$\eqalign{ & {\text{The}}{\kern 1pt} {\text{value}}{\kern 1pt} {\text{of}} \cr & \left( {\frac{1}{{{{\log }_3}60}} + \frac{1}{{{{\log }_4}60}} + \frac{1}{{{{\log }_5}60}}} \right){\text{is}}: \cr}$

1. 0
2. 1
3. 5
4. 60

Answer:Option 1

Explanation:

Solution:

Given expression

\begin{aligned} = \log_{60} 3 + \log_{60} 4 + \log_{60} 5 \\ = \log_{60} (3 \times 4 \times 5) \\ = \log_{60} 60 \\ = 1 \end{aligned}

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

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9.If log 2 = 0.30103, the number of digits in 2⁶⁴ is:

1. 18
2. 19
3. 20
4. 21

Answer:Option 1

Explanation:

Solution:

\begin{aligned} \log (2^{64}) \\ = 64 \times \log 2 \\ = (64 \times 0.30103) \\ = 19.26592 \end{aligned}

$\eqalign{ & \log \left( {{2^{64}}} \right) \cr & = 64 \times \log 2 \cr & = \left( {64 \times 0.30103} \right) \cr & = 19.26592 \cr}$ Its characteristic is 19.
Hence, then number of digits in 2⁶⁴ is 20.

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10.

I f \log_{x} (\frac{9}{16}) = - \frac{1}{2}, t h e n x i s e q u a l t o :

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

1. - ³/₄
2. ³/₄
3. ⁸¹/₂₅₆
4. ²⁵⁶/₈₁

Answer:Option 1

Explanation:

Solution:

\begin{aligned} \log_{x} (\frac{9}{16}) = - \frac{1}{2} \\ \Rightarrow x^{- \frac{1}{2}} = \frac{9}{16} \\ \Rightarrow \frac{1}{\sqrt{x}} = \frac{9}{16} \\ \Rightarrow \sqrt{x} = \frac{16}{9} \\ \Rightarrow x = {(\frac{16}{9})}^{2} \\ \Rightarrow x = \frac{256}{81} \end{aligned}

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

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