Logarithm and Antilogarithm Test 2

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Logarithm and Antilogarithm Test 2

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Weekly Quiz Competition

Question 1
1 / -0

The value of $$\log_2 \displaystyle \frac{1}{8}$$ is equal to

A
$$2$$

B
$$0$$

C
$$-3$$

D
$$-1$$

Solution

Given, $$ \log _{ 2 }{ \frac { 1 }{ 8 } } =x$$
Convert in exponential form,
$$ \frac { 1 }{ 8 } = { 2 }^{ x }$$
$$2^{-3}=2^{x}$$
$$ \therefore x = -3$$
Question 2
1 / -0

The exponential form of $$\log_8 0.125 = -1$$ is $$8 ^{-m} = 0.125$$. Then value of $$m$$ is

A
$$1$$

B
$$2$$

C
$$4$$

D
$$3$$

Solution

$$ \log _{ 8 }{ 0.125 } = -1\\ \implies 0.125 = { 8 }^{ -1 }$$
Also, by given we have
$$8^{-m} = 0.125$$
Therefore $$m\ =\ 1$$
Question 3
1 / -0

The value of $$\log_5\ 125$$ is equal to

A
$$0$$

B
$$1$$

C
$$2$$

D
$$3$$

Solution

$$\text {Consider}, \log _{ 5 }{ 125 } =x\\ \implies 125 = { 5 }^{ x }\\ \implies 5^3 =5^x \\\implies x = 3 $$
Question 4
1 / -0

The logarithm of $$27$$ to the base $$9$$ is $$\displaystyle \frac{3}{m}$$. Then the value of $$m$$ is equal to

A
$$0$$

B
$$2$$

C
$$3$$

D
$$6$$

Solution

Since, $$\log _9 {27}= \displaystyle \frac {3}{m}$$ ...(1)
Let $$ \log _{ 9 }{ 27 } = x$$
Converting in exponential form,
$$ 27 = { 9 }^{ x }$$
$${ 3 }^{ 3 } = { 3 }^{ 2x }$$
$$ x=\displaystyle \frac { 3 }{ 2 } $$ ...(2)
Comparing equations (1) and (2), we get
$$m=2$$
Question 5
1 / -0

The value of $$\log_{10}0.01$$ is equal to

A
$$0$$

B
$$-2$$

C
$$-1$$

D
$$4$$

Solution

$$\log _{ 10 }{ 0.01= } \log _{ 10 }{ { 10 }^{ -2 } } $$
$$\implies \log _{ 10 }{ 0.01= } -2$$.
Question 6
1 / -0

If exponential form of $$\log_{10} 0.01 = -2$$ is $$10^{m} = 0.01$$, then value of $$m$$ is equal to

A
$$-1$$

B
$$3$$

C
$$-2$$

D
$$4$$

Solution

$$ \log _{ 10 }{ 0.01 } = -2 \Rightarrow 0.01 = { 10 }^{ -2 } $$
$$\therefore m=-2$$
Question 7
1 / -0

If $$\log_{x} 2 = -1$$ then value of $$2x$$ is equal to

A
$$2$$

B
$$1$$

C
$$0$$

D
$$5$$

Solution

Given, $$\log _{ x }{ 2 } = -1$$

Converting in exponential form,

$$ 2 = { x }^{ -1 }$$

$$ 2 = \displaystyle \dfrac { 1 }{ x } $$

$$\therefore x=\displaystyle \dfrac {1}{2}$$

$$2 x=1$$
Question 8
1 / -0

Value of $$\sqrt [4]{(81)^{-2}}$$ is

A
$$\dfrac {1}{9}$$

B
$$\dfrac {1}{3}$$

C
$$9$$

D
$$\dfrac {1}{81}$$

Solution

As given $$\sqrt[4]{{(81)}^{-2}}$$

$$=\sqrt[4]{\dfrac{1}{\left ( 81 \right )^{2}}}$$

$$=\sqrt[4]{\dfrac{1}{\left ( \left (9 \right )^{2} \right )^{2}}}$$

$$=\dfrac{\left ( 1 \right )^{\frac{1}{4}}}{\left ( \left ( 9 \right )^{4} \right )^{\frac{1}{4}}}$$

$$=\dfrac{1}{9}$$
Question 9
1 / -0

Value of $$\displaystyle\frac{{2}^{100}}{2}$$ is-

A
$$1$$

B
$${50}^{100}$$

C
$${2}^{50}$$

D
$${2}^{99}$$

Solution

$$\Rightarrow \dfrac{{2}^{n}}{2}={2}^{n-1}$$
$$\Rightarrow \dfrac{{2}^{100}}{2}={2}^{100-1}={2}^{99}$$
Hence option 'D' is correct.
Question 10
1 / -0

What is the value of $$2^{0.64}*2^{0.36}$$ ?

A
2

B
1

C
16

D
32

Solution

$$2^{0.64}*2^{0.36} =2^{0.64+0.36} = 2^{1.00} = 2^1=2\\ (\because a^m*a^n=a^{m+n})$$

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Re-Attempt Weekly Quiz Competition

Logarithm and Antilogarithm Test 2

Result

Logarithm and Antilogarithm Test 2

Score

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SHARING IS CARING

Question 1

$$2$$

$$0$$

$$-3$$

$$-1$$

Solution

Question 2

$$1$$

$$2$$

$$4$$

$$3$$

Solution

Question 3

$$0$$

$$1$$

$$2$$

$$3$$

Solution

Question 4

$$0$$

$$2$$

$$3$$

$$6$$

Solution

Question 5

$$0$$

$$-2$$

$$-1$$

$$4$$

Solution

Question 6

$$-1$$

$$3$$

$$-2$$

$$4$$

Solution

Question 7

$$2$$

$$1$$

$$0$$

$$5$$

Solution

Question 8

$$\dfrac {1}{9}$$

$$\dfrac {1}{3}$$

$$9$$

$$\dfrac {1}{81}$$

Solution

Question 9

$$1$$

$${50}^{100}$$

$${2}^{50}$$

$${2}^{99}$$

Solution

Question 10

2

1

16

32

Solution

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