Logarithm and Antilogarithm Test 23

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Logarithm and Antilogarithm Test 23

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Weekly Quiz Competition

Question 1
1 / -0

Approximate of $$\log_{11}21$$ is

A
1.27

B
1.21

C
1.18

D
1.15

E
1.02

Solution

Approximate value of $$\log _{ 11 }{ 21 } $$
$$=\log _{ 11 }{ (7\times 3) } $$
$$=\log _{ 11 }{ 7 } +\log _{ 11 }{ 3 }$$
$$ =0.8115+0.4581$$
$$=1.27$$
Question 2
1 / -0

$$2^{1/4}4^{1/8}8^{1/16}16^{1/32}$$....... is equal to

A
1

B
2

C
$$\frac{3}{2}$$

D
$$\frac{5}{2}$$
Question 3
1 / -0

If $$\log_{10} 2 = 0.3010$$, then the number of digits in $$2^{64}$$ is

A
$$18$$

B
$$24$$

C
$$22$$

D
$$20$$

Solution

Given $$\log _{ 10 }{ 2 } =0.301$$
$$\log _{ 10 }{ 2^{64} } =64 \times \log _{ 10 }{ 2 } =64 \times 0.3010=19.264$$
$$\Rightarrow 2^{64}=10^{19.264}$$
The number of digits in $$10^{19}$$ is $$20$$ , there will be $$21$$ digits from $$10^{21}$$
The number $$10^{19.264}$$ lies between them
Therefore the number of digits in $$10^{19.264}$$ is $$20$$
Therefore the correct option is $$D$$
Question 4
1 / -0

Evaluate using logarithm table: $$\dfrac {28.45 \times \sqrt [3] {0.3254}}{32.43 \times \sqrt [5] {0.3046}}$$

A
$$0.7666$$

B
$$0.7656$$

C
$$0.5686$$

D
$$0.2936$$

Solution

Let $$y=\dfrac { 28.45\times \sqrt [ 3 ]{ .3254 } }{ 32.43\times \sqrt [ 3 ]{ .3046 } } $$
$$ \ln { y } =\ln { 28.45 } +\ln { \sqrt [ 3 ]{ .3254 } } -(\ln { 32.43 } +\ln { \sqrt [ 5 ]{ .3046 } } )\\ \ln { y } =\ln { 25.45 } +\dfrac { 1 }{ 3 } \ln { .3245- } \ln { 32.43 } -\dfrac { 1 }{ 5 } \ln { .4046 } \\ \ln { y } =3.236+(-.375)-3.479-(-.237)\\ \ln { y } =-.381$$
$$ y=$$ anti $$\ln { (-.381) } $$
$$ y=.7656$$
So, option B is correct.
Question 5
1 / -0

If $$\dfrac{\log_{2}a}{4} = \dfrac{\log_{2}b}{6} = \dfrac{\log_{2}c}{3p}$$ and also $$a^{3}b^{2}c = 1$$, then the value of $$p$$ is equal to

A
$$-6$$

B
$$-7$$

C
$$-8$$

D
$$-9$$

Solution

$$\dfrac { \log _{ 2 }{ a } }{ 4 } =\dfrac { \log _{ 2 }{ b } }{ 6 } =\dfrac { \log _{ 2 }{ c } }{ 3p } = x\\ \log _{ 2 }{ a } = 4x $$

$$\Rightarrow a = { 2 }^{ 4x }$$

$$ \log _{ 2 }{ b } = 6x$$

$$ \Rightarrow b = { 2 }^{ 6x }$$

$$ \log _{ 2 }{ c } = 3px$$

$$ \Rightarrow c = { 2 }^{ 3px }$$

$$ \\ { a }^{ 3 }{ b }^{ 2 }c = 1$$

$${ 2 }^{ 12x }{ \cdot 2 }^{ 12x }\cdot { 2 }^{ 3px } ={ 2 }^{ 0 }$$

$$ { 2 }^{ 24x+3px } ={ 2 }^{ 0 }$$

$$ \therefore 24x +3px = 0$$

$$ \therefore 3px = -24x$$

$$ \therefore p = -8 $$
Question 6
1 / -0

Given $$log_3(a) = c$$ and $$log_3(b)=2c, a =$$

A
$$3c$$

B
$$c + 3$$

C
$$b^2$$

D
$$\sqrt{b}$$

E
$$\dfrac{b}{2}$$

Solution

Given, $$\log _{ 3 }{ (a) } =c$$ and $$\log _{ 3 }{ (b) } =2c$$
$$\therefore a={ 3 }^{ c }$$ and $$b={ 3 }^{ 2c }$$
Now consider $$a^2$$ as follows:
$$a^{ 2 }={ (3 }^{ c })^{ 2 }$$
$$ \Rightarrow a^{ 2 }={ 3 }^{ 2c }$$
$$ \Rightarrow a^{ 2 }=b$$
$$\Rightarrow a=\sqrt { b }$$
Question 7
1 / -0

The number $$ N=6 \log_{10}2+\log_{10}31$$ lies between two successive integers, whose sum is equal to

A
$$5$$

B
$$7$$

C
$$9$$

D
$$10$$

Solution

$$N=6\log_{10}2+\log_{10}31$$

$$=\log_{10}2^{6}+log_{10}(31)$$

$$=\log_{10}(2^{6}.31)$$

$$=\log_{10}(64\times31)$$
$$=\log_{10}(1984)$$

Now
$$\log_{10}(10^{3})=\log_{10}(1000)=3$$
Similarly
$$\log_{10}(10^{4})=4$$

Since
$$10^{3}<1984<10^{4}$$

$$3<\log_{10}(1984)<4$$
Hence it lies between $$3$$ and $$4$$.
The sum of these two integers is $$7$$.
Question 8
1 / -0

Let $$a = \log_3\log_32$$. An integer k satisfying $$1< 2^{(-k+3^{-a})} < 2,$$ must be less than _____.

A
$$1.25766$$

B
$$2.256$$

C
$$3$$

D
$$1$$

Solution
Question 9
1 / -0

Directions For Questions

Given that $$N= 7^{log_{49} 900}, A= 2^{log_24}+ 3 ^{log_24}+ 4^{log_2^2}- 4^{log_23}, D= (log_549)(log_7125)$$ Then answer the following questions: ( Using the value of N, A, D)

...view full instructions

If $$log_AD= a, $$ then value of $$log_612$$ is (in terms of a)

A
$$\frac{1+3a}{3a}$$

B
$$\frac{1+2a}{3a}$$

C
$$\frac{1+2a}{2a}$$

D
$$\frac{1+3a}{2a}$$

Solution
Question 10
1 / -0

If $$x=198!$$ then value of the expression $$\dfrac {1}{\log_{2}x}+\dfrac {3}{\log_{2}x}+...\dfrac {198}{\log_{2}x}$$ equals ?

A
$$-1$$

B
$$0$$

C
$$1$$

D
$$198$$

Solution

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Logarithm and Antilogarithm Test 23

Result

Logarithm and Antilogarithm Test 23

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SHARING IS CARING

Question 1

1.27

1.21

1.18

1.15

1.02

Solution

Question 2

1

2

$$\frac{3}{2}$$

$$\frac{5}{2}$$

Question 3

$$18$$

$$24$$

$$22$$

$$20$$

Solution

Question 4

$$0.7666$$

$$0.7656$$

$$0.5686$$

$$0.2936$$

Solution

Question 5

$$-6$$

$$-7$$

$$-8$$

$$-9$$

Solution

Question 6

$$3c$$

$$c + 3$$

$$b^2$$

$$\sqrt{b}$$

$$\dfrac{b}{2}$$

Solution

Question 7

$$5$$

$$7$$

$$9$$

$$10$$

Solution

Question 8

$$1.25766$$

$$2.256$$

$$3$$

$$1$$

Solution

Question 9

$$\frac{1+3a}{3a}$$

$$\frac{1+2a}{3a}$$

$$\frac{1+2a}{2a}$$

$$\frac{1+3a}{2a}$$

Solution

Question 10

$$-1$$

$$0$$

$$1$$

$$198$$

Solution

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