Logarithm and Antilogarithm Test 5

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Logarithm and Antilogarithm Test 5

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Weekly Quiz Competition

Question 1
1 / -0

$$\displaystyle \left \{ \left ( \frac{5}{3} \right )^{15} \right \}^0$$ is equal to

A
$$\displaystyle \frac{5}{3}$$

B
$$\displaystyle \frac{3}{5}$$

C
1

D
None of these

Solution

Anything raised to the power of 0 equals 1
Question 2
1 / -0

State true or false:
$$\displaystyle \left ( \frac{2}{3} \right )^{4} \div \left ( \frac{2}{3} \right )^{6} = \left ( \frac{2}{3} \right )^{2}$$

A
True

B
False

C
Ambiuous

D
Data insufficient

Solution

$$\left(\cfrac { 2 }{ 3 } \right)^{ 4 }\div \left(\cfrac { 2 }{ 3 } \right)^{ 6 }=\left(\cfrac { 2 }{ 3 } \right)^{ 4 } \times \left(\cfrac { 2 }{ 3 } \right)^{ -6 }=\left(\cfrac { 2 }{ 3 } \right)^{ 4-6 }=\left(\cfrac { 2 }{ 3 } \right)^{ -2 }$$
Question 3
1 / -0

State true or false: $$\displaystyle \left [ \left ( \frac{3}{7} \right )^{2}\right ]^3 = \left ( \frac{3}{7} \right )^{5}$$

A
True

B
False

C
Ambiguous

D
Data insufficient

Solution

$$\left[\left(\cfrac { 3 }{ 7 } \right)^{ 2 }\right]^{ 3 }=\left(\cfrac { 3 }{ 7 } \right)^{ 2 \times 3 }=\left(\cfrac { 3 }{ 7 } \right)^{ 6 }$$
Question 4
1 / -0

State true or false: $$\displaystyle \left ( \frac{-7}{9} \right )^{2} \div \frac{49}{81} = - 1$$

A
True

B
False

C
Ambiguous

D
Data insufficient

Solution

$$\left(\cfrac { -7 }{ 9 } \right)^{ 2 }\div \cfrac { 49 }{ 81 } =\left(\cfrac { -1 \times 7 }{ 9 } \right)^{ 2 }\times \left(\cfrac { 7 }{ 9 } \right)^{ -2 }=\left(\cfrac { 7 }{ 9 } \right)^{ 2-2 }=1$$
Here, $$\left(\cfrac { -7 }{ 9 } \right)^{ 2 }=\left(\cfrac { 7 }{ 9 } \right)^{ 2 }$$
(As the square of $$-ve$$ sign number, will give a positive number)
Question 5
1 / -0

Which expression is equivalent to 81?

A
$$2^9$$

B
$$\displaystyle \left ( \frac{1}{3} \right )^{-4}$$

C
$$3^{-4}$$

D
$$\displaystyle \left ( \frac{1}{3} \right )^{4}$$

Solution

$$({ \frac { 1 }{ 3 } ) }^{ -4 }\quad =\quad 3^{ 4 }\quad =\quad 81$$
Question 6
1 / -0

$$\displaystyle \left ( \frac{1}{3} \right )^{7-7} = 2^0$$

A
True

B
False

C
Ambiguous

D
Data insufficient

Solution

7-7 = 0,
anything raised to the power of 0 equals 1
Question 7
1 / -0

The value of $$\displaystyle \left ( \frac{5}{3} \right )^{-8} \div \left ( \frac{5}{3} \right )^{-8}$$ is equal to

A
$$\displaystyle \frac{5}{3}$$

B
$$\displaystyle \frac{3}{5}$$

C
$$1$$

D
None of these

Solution

According to the properties of exponents,
$$\dfrac{a^m}{a^n}=a^{m-n}$$
According to the given condition,

$$\left(\cfrac { 5 }{ 3 } \right)^{ -8 }\div \left(\cfrac { 5 }{ 3 } \right)^{ -8 }=\left(\cfrac { 5 }{ 3 } \right)^{ -8-(-8) } \\= \left(\cfrac { 5 }{ 3 } \right)^{ 0 } \\=1$$
Question 8
1 / -0

Which of the following expresses zero law of exponents?

A
$$0^{x} = 0$$

B
$$x^{0} = 1$$

C
$$x^{1} = x$$

D
None of the above

Solution

According to the zero law of exponents,
any non - zero number raised to the power of zero is equal to $$1$$.
$$\therefore x^{0} = 1$$, where $$x\neq 0$$
So, option $$B$$ is correct.
Question 9
1 / -0

The value of $$(2^0 - 3^0) \times 4^2$$ is---

A
$$16$$

B
$$0$$

C
$$-16$$

D
None of these

Solution

$$(2^0 - 3^0) \times 4^2\\ = (1 - 1) \times 4^2\\ = 0\times 16\\=0$$
Question 10
1 / -0

If $$\displaystyle \log x=n$$ then 2n is equal to

A
$$\displaystyle \log\left ( x^{2} \right )$$

B
$$\displaystyle \left ( \log x \right )^{2}$$

C
$$\displaystyle \log \left ( x+2 \right )$$

D
$$\displaystyle \log 2x$$

Solution

Given , $$\log x = n $$
We know that $$\log x^{2} = 2 \log x$$
$$\Rightarrow 2n = 2 \log x = \log x^{2}$$
Or $$ \log x^{2} = 2 \log x = 2n$$.

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Re-Attempt Weekly Quiz Competition

Logarithm and Antilogarithm Test 5

Result

Logarithm and Antilogarithm Test 5

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SHARING IS CARING

Question 1

$$\displaystyle \frac{5}{3}$$

$$\displaystyle \frac{3}{5}$$

1

None of these

Solution

Question 2

True

False

Ambiuous

Data insufficient

Solution

Question 3

True

False

Ambiguous

Data insufficient

Solution

Question 4

True

False

Ambiguous

Data insufficient

Solution

Question 5

$$2^9$$

$$\displaystyle \left ( \frac{1}{3} \right )^{-4}$$

$$3^{-4}$$

$$\displaystyle \left ( \frac{1}{3} \right )^{4}$$

Solution

Question 6

True

False

Ambiguous

Data insufficient

Solution

Question 7

$$\displaystyle \frac{5}{3}$$

$$\displaystyle \frac{3}{5}$$

$$1$$

None of these

Solution

Question 8

$$0^{x} = 0$$

$$x^{0} = 1$$

$$x^{1} = x$$

None of the above

Solution

Question 9

$$16$$

$$0$$

$$-16$$

None of these

Solution

Question 10

$$\displaystyle \log\left ( x^{2} \right )$$

$$\displaystyle \left ( \log x \right )^{2}$$

$$\displaystyle \log \left ( x+2 \right )$$

$$\displaystyle \log 2x$$

Solution

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