Logarithm and Antilogarithm Test 14

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Logarithm and Antilogarithm Test 14

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Weekly Quiz Competition

Question 1
1 / -0

The value of $$\displaystyle \log_{3} 27$$ is equal to

A
$$3$$

B
$$9$$

C
$$16$$

D
$$25$$

Solution

$$ \log _{ 3 }{ 27 } = \log _{ 3 }{ { 3 }^{ 3 } } \\ =3\log _{ 3 }{ 3 } \\ = 3$$
Question 2
1 / -0

If $$\displaystyle 2 \log y - \log x - 3 = 0$$ and $$x$$ = $$\dfrac {y^2}{m}$$, then the value of $$m$$ is equal to

A
$$4000$$

B
$$3000$$

C
$$2000$$

D
$$1000$$

Solution

Since, $$\displaystyle 2 \log y - \log x - 3 = 0$$

$$\therefore \log y^2 - \log x = \log 1000$$ ($$\log a^b = b \log a$$)

$$\Rightarrow \log\dfrac{y^2}{x} = \log 1000$$ ($$\log\dfrac{a}{b} = \log a - \log b$$)

$$\Rightarrow \dfrac{y^2}{x} = 1000$$

$$\Rightarrow x = \dfrac{y^2}{1000}$$

$$ \Rightarrow m = 1000$$
Question 3
1 / -0

If $$\displaystyle \log_{10} 8 = 0.90$$ and $$\displaystyle \log \sqrt {32}$$ = $$\cfrac{m}{4}$$, then the value of $$m$$ is equal to

A
$$43$$

B
$$8$$

C
$$3$$

D
$$6$$

Solution

Goven, $$ \log_{10}8 = 0.9$$
$$\Rightarrow \log_{10}2^3 = 0.9$$
$$\Rightarrow 3\log_{10} = 0.9$$ (Using $$\log a^b = b \log a$$)
$$\Rightarrow \log_{10}2 = 0.3$$
Now, $$\log_{10}\sqrt{32} = \dfrac{5}{2}\log_{10}2 = 2.5\times 0.3 = 0.75= \dfrac{3}{4}$$
$$\Rightarrow m=3$$.
Question 4
1 / -0

The value of $$\displaystyle \log_{2} 0.125$$ is equal to

A
$$-3$$

B
$$3$$

C
$$-2$$

D
$$2$$

Solution

$$ \log _{ 2 }{ 0.125 } = \log _{ 2 }{ { \dfrac { 1 }{ 8 } } } \\ =\log _{ 2 }{ { 2 }^{ -3 } } \\ = -3$$
Question 5
1 / -0

The value of $$\log_{0.5}16$$ is equal to

A
$$-4$$

B
$$-1$$

C
$$-2$$

D
$$0$$

Solution

Given, $$ \log _{ 0.5 }{ 16 } = x $$

$$\implies 16 = {0.5 }^{ x }$$

$$\implies { 2 }^{ 4 } = { 2 }^{ -x }$$

$$\implies x = -4$$
Question 6
1 / -0

If $$\log_{16} 8$$ = $$\displaystyle \frac{m}{4}$$, then value of $$m$$ is equal to

A
$$1$$

B
$$3$$

C
$$4$$

D
$$2$$

Solution

Since, $$\log _{16}{8}=\frac {m}{4}$$ ...(1)
Let $$ \log _{ 16 }{ 8 } =x$$
Convert in exponential form,
$$ 8 = { 16 }^{ x }$$
$${ 2 }^{ 3 } = { 2 }^{ 4x }$$
$$ \therefore x = \displaystyle \frac { 3 }{ 4 } $$ ...(2)
From (1) and (2),
$$m=3$$
Question 7
1 / -0

If the value of $$(\log_{10} 2 )+ 1$$ is in the form of $$\log_{10}m$$, then $$m$$ is equal to

A
$$11$$

B
$$14$$

C
$$20$$

D
$$17$$

Solution

$$ \log _{ 10 }{ 2 } +1 = \log _{ 10 }{ 2 } +\log _{ 10 }{ 10 } \\ \\ \\ =\log _{ 10 }{ (2\times 10 } )\\ = \log _{ 10 }{ 20 } $$
Question 8
1 / -0

The value of $$ \log_{5} 0.2$$ is equal to

A
$$-1$$

B
$$1$$

C
$$10$$

D
$$-10$$

Solution

Let $$\log _{ 5 }{ 0.2 } = x$$
$$ \therefore 0.2 = { 5 }^{ x } \Rightarrow \dfrac { 2 }{ 10 } = { 5 }^{ x } $$
$$\therefore \dfrac{1}{5} = {5}^{x} \Rightarrow 5^{-1} = 5^{x}$$
$$ \therefore x = -1$$
Question 9
1 / -0

The value of $$\displaystyle \log_{10}0.001 $$ is equal to

A
$$-3$$

B
$$3$$

C
$$-2$$

D
$$2$$

Solution

Consider, $$ \log _{ 10 }{ 0.001 } =x$$
$$\implies 0.001 = { 10 }^{ x }$$
$$\implies \dfrac { 1 }{ 1000 } = 10^x$$
$$\implies { 10 }^{ -3 }= 10^x$$
$$\implies x=-3 $$
Question 10
1 / -0

The value of $$\displaystyle \log_{5} 625$$ is equal to

A
$$2$$

B
$$6$$

C
$$4$$

D
$$1$$

Solution

$$ \log _{ 5 }{ 625 } = \log _{ 5 }{ { 5 }^{ 4 } } \\ = 4\log _{ 5 }{ 5 } \\ = 4$$

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Re-Attempt Weekly Quiz Competition

Logarithm and Antilogarithm Test 14

Result

Logarithm and Antilogarithm Test 14

Score

-

Rank

-

SHARING IS CARING

Question 1

$$3$$

$$9$$

$$16$$

$$25$$

Solution

Question 2

$$4000$$

$$3000$$

$$2000$$

$$1000$$

Solution

Question 3

$$43$$

$$8$$

$$3$$

$$6$$

Solution

Question 4

$$-3$$

$$3$$

$$-2$$

$$2$$

Solution

Question 5

$$-4$$

$$-1$$

$$-2$$

$$0$$

Solution

Question 6

$$1$$

$$3$$

$$4$$

$$2$$

Solution

Question 7

$$11$$

$$14$$

$$20$$

$$17$$

Solution

Question 8

$$-1$$

$$1$$

$$10$$

$$-10$$

Solution

Question 9

$$-3$$

$$3$$

$$-2$$

$$2$$

Solution

Question 10

$$2$$

$$6$$

$$4$$

$$1$$

Solution

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Question Analysis

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Rank

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