Logarithm and Antilogarithm Test 17

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Logarithm and Antilogarithm Test 17

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Weekly Quiz Competition

Question 1
1 / -0

The value of $$\displaystyle \log_{2}\left [ \log_{2}\left \{ \log_{3}\left ( \log_{3}27^{3} \right ) \right \} \right ]$$ is equal to

A
2

B
3

C
0

D
1

Solution

Required value of $$\displaystyle \log _{2}\left [ \log _{2}\left \{ \log _{3}\left ( \log _{3}27^{3} \right ) \right \} \right ]$$
$$=$$ $$\displaystyle \log _{2}\left [ \log _{2}\left \{ \log _{3}\left ( \log _{3}3^{9} \right ) \right \} \right ]$$
$$=$$ $$\displaystyle \log _{2}\left [ \log _{2}\left \{ \log _{3}\left (9\cdot \log _{3}3 \right ) \right \} \right ]$$
$$=$$ $$\displaystyle \log _{2}\left [ \log _{2} \left \{\log _{3}\cdot 9 \right \}\right ]$$ ....$$\displaystyle \left [ \because \log _{3}3=1 \right ]$$
$$=$$ $$\displaystyle \log _{2} \left [ \log _{2}\left \{ \log _{3}3^{2} \right \} \right ]$$
$$=$$ $$\displaystyle \log _{2} \left [ \log _{2} 2\cdot \log _{3} 3\right ]$$
$$=$$ $$\displaystyle \log _{2}\left [ \log _{2}2\right ]$$ ....$$\displaystyle \left [ \because \log _{2}2=1 \right ]$$
$$=$$ $$\displaystyle \log _{2}$$ $$1 = 0$$
Question 2
1 / -0

If there are $$n$$ zeros after the decimal point, then the characteristic of that number will be

A
$$n+1$$

B
$$-n+1$$

C
$$-(n+1)$$

D
$$n-1$$

Solution

For number less than $$1,$$ if there are $$n$$ zeroes after the decimal point, then the characteristics of that number will be $$-(n+1).$$
and For number greater than $$1,$$ if there are $$n$$ number of zeroes are on the left sides of the digits then characteristics will be $$(n+1).$$
Hence, C is the correct option.
Question 3
1 / -0

Simplify : $$(6^{-1} - 8^{-1})^{-1} + (2^{-1} - 3^{-1})^{-1}$$ is-

A
$$20$$

B
$$30$$

C
$$10$$

D
$$40$$

Solution

Law of exponents,
Given - $$(6^{-1}-8^{-1})^{-1}+(2^{-1}-3^{-1})^{-1}$$
$$=\left(\dfrac{1}{6}-\dfrac{1}{8}\right)^{-1}+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)^{-1}$$
$$=\left(\dfrac{8-6}{6\times 8}\right)^{-1}+\left(\dfrac{3-2}{2\times 3}\right)^{-1}$$
$$=\left(\dfrac{2}{36\times 8}\right)^{-1}+\left(\dfrac{1}{2\times 3}\right)^{-1}$$
$$=\left(\dfrac{1}{24}\right)^{-1}+\left(\dfrac{1}{6}\right)^{-1}$$
$$=24+6$$
$$=30$$
Hence, option B is correct.
Question 4
1 / -0

The value of $$\log_{10} 0.0006024$$ is equal to

A
$$\overline {3}.7979$$

B
$$\overline {1}.9779$$

C
$$\overline {4}.7799$$

D
$$0.7279$$

Solution

$$\log_{10}0.0006024$$
Characteristics$$=-4$$
For mantissa ,read from the table $$6024.$$
Mantissa$$=7799$$
Thus, $$\log_{10}0.0006024=$$ characteristics of $$0.0006024+$$ mantissa of $$0.0006024$$
$$=-4+0.7799$$
$$=\bar4.7799$$
Hence, C is the correct option.
Question 5
1 / -0

The logarithm of number lying between $$0$$ and $$1$$ is

A
positive

B
negative

C
zero

D
none of the above

Solution

From the graph, we can see that logarithmic value of number lying between $$0$$ and $$1$$ is negative.
Hence, B is the correct option.
Question 6
1 / -0

If $$3^{n-2}=\dfrac{1}{81},n=$$

A
$$0$$

B
$$1$$

C
$$2$$

D
$$-2$$

E
$$-4$$

Solution

Given, $${3}^{n-2}=\dfrac {1}{81} $$
$$\Rightarrow 3^{n-2}=\dfrac {1}{3^4}$$
$$\Rightarrow 3^{n-2}=3^{-4}$$
$$\Rightarrow n-2=-4$$
$$\Rightarrow n=-2$$
Question 7
1 / -0

Using logarithm table, determine the value of $$\log_{10}0.5432$$.

A
$$\overline {1}.7350$$

B
$$\overline {2}.7350$$

C
$$0.7350$$

D
$$0.07350$$

Solution

$$\log_{10}0.5432$$
Characteristics$$=-1$$
For mantissa ,read from the table $$5432.$$
Mantissa$$=7350$$
Thus, $$\log_{10}0.5432=$$ characteristics of $$0.5432+$$ mantissa of $$0.5432$$
$$=-1+0.7350$$
$$=\bar1.7350$$
Hence, A is the correct option.
Question 8
1 / -0

If $$x=\log _{ a }{ bc } ,y=\log _{ b }{ ca } ,z=\log _{ c }{ ab } $$, then the value of $$\dfrac { 1 }{ 1+x } +\dfrac { 1 }{ 1+y } +\dfrac { 1 }{ 1+z } $$ will be

A
$$x+y+z$$

B
$$1$$

C
$$ab+bc+ca$$

D
$$abc$$

Solution

Now, $$1+x=\log _{ a }{ a } +\log _{ a }{ bc } =\log _{ a }{ abc } $$
$$\Rightarrow \dfrac { 1 }{ 1+x } =\log _{ abc }{ a } $$
Similarly, $$\dfrac { 1 }{ 1+y } =\log _{ abc }{ b } $$ and $$\dfrac { 1 }{ 1+z } =\log _{ abc }{ c } $$
$$\therefore \dfrac { 1 }{ 1+x } +\dfrac { 1 }{ 1+y } +\dfrac { 1 }{ 1+z } $$
$$=\log _{ abc }{ a } +\log _{ abc }{ b } +\log _{ abc }{ c } $$
$$=\log _{ abc }{ abc } =1$$
Question 9
1 / -0

If $$\displaystyle { log }_{ 5 }{ log }_{ 5 }{ log }_{ 2 }x=0$$, then the value of $$x$$ is

A
$$32$$

B
$$125$$

C
$$625$$

D
$$25$$

Solution

Given, $$\displaystyle { log }_{ 5 }{ log }_{ 5 }{ log }_{ 2 }x=0$$
$$\displaystyle \Rightarrow \quad { log }_{ 5 }{ log }_{ 2 }x=50$$
$$\displaystyle \Rightarrow \quad { log }_{ 5 }{ log }_{ 2 }x=1\Rightarrow { log }_{ 2 }x=5$$
$$\displaystyle \Rightarrow x={ 2 }^{ 5 }=32$$
Question 10
1 / -0

Find the value of $$\log_{10} {\dfrac{64^{2.1}\times 81^{4.2}}{49^{3.4}}}$$ using log table

A
$$2.1 \times 6 \times .303+ 4.2 \times 2 \times .854- 3.4 \times 2 \times .745$$

B
$$2.1 \times .303+ 4.2 \times .954- 3.4 \times .845$$

C
$$2.1 \times 6 \times .303- 4.2 \times 2 \times .954+3.4 \times 2 \times .845$$

D
$$2.1 \times 6 \times .303+ 4.2 \times 2 \times .954- 3.4 \times 2 \times .845$$

Solution

$$\log_{10}{64^{2.1}}+\log_{10}{81^{4.2}}-\log_{10}{49^{3.4}}$$
$$= 2.1\log_{10}{64}+4.2\log_{10}{81}-3.4\log_{10}{49}$$
$$=2.1\log_{10}{2^{6}}+4.2\log_{10}{3^{4}}-3.4\log_{10}{7^{2}}$$
$$=2.1\times6\times(0.303)+4.2\times4\times(0.477)-3.4\times2\times(0.845)$$
$$=2.1\times6\times(0.303)+4.2\times2\times(0.954)-3.4\times2\times(0.845)$$

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Logarithm and Antilogarithm Test 17

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Logarithm and Antilogarithm Test 17

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SHARING IS CARING

Question 1

2

3

0

1

Solution

Question 2

$$n+1$$

$$-n+1$$

$$-(n+1)$$

$$n-1$$

Solution

Question 3

$$20$$

$$30$$

$$10$$

$$40$$

Solution

Question 4

$$\overline {3}.7979$$

$$\overline {1}.9779$$

$$\overline {4}.7799$$

$$0.7279$$

Solution

Question 5

positive

negative

zero

none of the above

Solution

Question 6

$$0$$

$$1$$

$$2$$

$$-2$$

$$-4$$

Solution

Question 7

$$\overline {1}.7350$$

$$\overline {2}.7350$$

$$0.7350$$

$$0.07350$$

Solution

Question 8

$$x+y+z$$

$$1$$

$$ab+bc+ca$$

$$abc$$

Solution

Question 9

$$32$$

$$125$$

$$625$$

$$25$$

Solution

Question 10

$$2.1 \times 6 \times .303+ 4.2 \times 2 \times .854- 3.4 \times 2 \times .745$$

$$2.1 \times .303+ 4.2 \times .954- 3.4 \times .845$$

$$2.1 \times 6 \times .303- 4.2 \times 2 \times .954+3.4 \times 2 \times .845$$

$$2.1 \times 6 \times .303+ 4.2 \times 2 \times .954- 3.4 \times 2 \times .845$$

Solution

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