Exponents and Powers Test

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Exponents and Powers Test - 19

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Weekly Quiz Competition

Question 1
1 / -0

$$3^{-2}$$ can be written as

A
$$3^2$$

B
$$\dfrac {1}{3^2}$$

C
$$\dfrac {1}{3^{-2}}$$

D
$$-\dfrac {2}{3}$$

Solution

We will use, law of exponent $$a^{-m}=\dfrac {1}{a^m}$$, where $$a=$$ non zero integer
Therefore, $$3^{-2}=\dfrac {1}{3^2}$$
Question 2
1 / -0

The value of $$\left(\dfrac 25 \right)^{-2}$$ is

A
$$\dfrac 45$$

B
$$\dfrac {4}{25}$$

C
$$\dfrac {25}{4}$$

D
$$\dfrac {5}{2}$$

Solution

We will use, law of exponent $$a^{-m}=\dfrac {1}{a^m}$$ where $$a=$$ non-zero integer

$$\left(\dfrac {2}{5}\right)^{-2}=\dfrac {1}{\left(\dfrac {2}{5}\right)^2} \\=\dfrac {1}{\dfrac {4}{25}}\\=\dfrac {25}{4}$$
Question 3
1 / -0

Mark tick against the correct answer in each of the following:
$$[\left\{(-1/2) ^{2}\right\}^{-2}]^{-1}=$$?

A
$$(1/16)$$

B
$$16$$

C
$$(-1/16)$$

D
$$-16$$

Solution

$$[\left\{(-1/2) ^{2}\right\}^{-2}]^{-1}=[\left\{(-1^2/2) ^{2}\right\}^{-2}]^{-1}=$$
$$=[\left\{1/4\right\}^{-2}]^{-1}$$
$$=[\left\{4\right\}^{2}]^{-1}$$
$$=[16^{-1}$$
$$=[1/16]$$
Question 4
1 / -0

The value of $$\left(\dfrac 25 \right)^{-1}$$ is

A
$$\dfrac {2}{5}$$

B
$$\dfrac {5}{2}$$

C
$$-\dfrac {5}{2}$$

D
$$-\dfrac {2}{5}$$

Solution

We will use, law of exponent $$a^{-m}=\dfrac {1}{a^m}$$ where $$a=$$ non-zero integer
$$\left(\dfrac {2}{5}\right)^{-1}=\dfrac {1}{\left(\dfrac {2}{5}\right)^1} =\dfrac {5}{2}$$
Question 5
1 / -0

If $$x$$ be any integer different from zero and $$m$$ be any integers then $$x^{-m}$$ is equal to

A
$$x^m$$

B
$$-x^m$$

C
$$\dfrac {1}{x^m}$$

D
$$\dfrac {-1}{x^m}$$

Solution

Using law of exponents,
$$a^{-m} =\dfrac {1}{a^m}$$
[Where,$$a$$ is non-zero integer]
Similarly, $$x^{-m}=\dfrac {1}{x^m}$$
Question 6
1 / -0

The value of $$(-2)^{2\times 3-1}$$ is

A
$$32$$

B
$$64$$

C
$$-32$$

D
$$-64$$

Solution

To find the value of $$(-2)^{2\times 3-1}$$

$$(-2)^{2\times 3-1}$$
$$=(-2)^{6-1}$$
$$=(-2)^5$$
$$=(-2)\times (-2)\times (-2)\times (-2)\times (-2)$$
$$=(4)\times (4)\times (-2)$$
$$=(16)\times (-2)$$
$$=-32$$

Hence, the result $$=-32$$
Question 7
1 / -0

On multiplying ________ by $$2^{-5}$$ we get $$2^5$$

A
$$1$$

B
$$2$$

C
$$-1$$

D
$$0$$

Solution

$$2^{-5} \times 2^{5}= \dfrac{1}{2^5} \times 2^{5} = 1$$
Question 8
1 / -0

The value of $$(7^{-1}-8^{-1})^{-1} -(3^{-1}-4^{-1})$$ is

A
$$44$$

B
$$56$$

C
$$68$$

D
$$12$$

Solution

Using law of exponents, $$a^{-m}=\dfrac {1}{a^m}$$

$$\Rightarrow \ (7^{-1}-8^{-1})^{-1}-(3^{-1} -4^{-1})^{-1}=\left(\dfrac {1}{7}-\dfrac {1}{8}\right)^{-1}-\left(\dfrac {1}{3}-\dfrac {1}{4}\right)^{-1}$$

$$\Rightarrow \ \left(\dfrac {1\times 8-1\times 7}{56}\right)^{-1}- \left(\dfrac {1\times 4-1\times 3}{12}\right)^{-1}$$

$$=\left(\dfrac {8-7}{56}\right)^{-1} -\left(\dfrac {4-3}{12}\right)^{-1}$$

$$=\left(\dfrac {1}{56}\right)^{-1} -\left(\dfrac {1}{12}\right)^{-1}$$

Again we will use $$a^{-m} =\dfrac {1}{a^m}$$

$$\Rightarrow (7^{-1} -8^{-1})^{-1} -(3^{-1} -4^{-1})^{-1}=\left(\dfrac {1}{56}\right)^{-1}-\left(\dfrac {1}{12}\right)^{-1}$$

$$=56-12=44$$.
Question 9
1 / -0

$$\left(-\dfrac 57 \right)^{-5}$$ is equal to

A
$$\left(\dfrac 57 \right)^{-5}$$

B
$$\left(\dfrac 57 \right)^{5}$$

C
$$\left(\dfrac 75 \right)^{5}$$

D
$$\left(-\dfrac 75 \right)^{5}$$

Solution

Using law of exponent $$a^{-m}=\dfrac {1}{a^m}$$
[Where, $$a$$ is non-zero integer]
$$\left(\dfrac {-5}{7}\right)^{-5}=\dfrac {1}{\left(\dfrac {-5}{7}\right)^{5}}=\left(-\dfrac {7}{5}\right)^{5}$$
Question 10
1 / -0

$$\left(\dfrac {-7}{5} \right)^{-1}$$ is equal

A
$$\dfrac 57$$

B
$$-\dfrac 57$$

C
$$\dfrac 75$$

D
$$\dfrac {-7}{5}$$

Solution

Using law of exponent $$a^{-m}=\dfrac {1}{a^m}$$
[Where, $$a$$ is non-zero integer]
$$\left(\dfrac {-7}{5}\right)^{-1}=\dfrac {1}{\left(\dfrac {-7}{5}\right)}=\left(-\dfrac {5}{7}\right)$$

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Re-Attempt Weekly Quiz Competition

Exponents and Powers Test - 19

Result

Exponents and Powers Test - 19

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SHARING IS CARING

Question 1

$$3^2$$

$$\dfrac {1}{3^2}$$

$$\dfrac {1}{3^{-2}}$$

$$-\dfrac {2}{3}$$

Solution

Question 2

$$\dfrac 45$$

$$\dfrac {4}{25}$$

$$\dfrac {25}{4}$$

$$\dfrac {5}{2}$$

Solution

Question 3

$$(1/16)$$

$$16$$

$$(-1/16)$$

$$-16$$

Solution

Question 4

$$\dfrac {2}{5}$$

$$\dfrac {5}{2}$$

$$-\dfrac {5}{2}$$

$$-\dfrac {2}{5}$$

Solution

Question 5

$$x^m$$

$$-x^m$$

$$\dfrac {1}{x^m}$$

$$\dfrac {-1}{x^m}$$

Solution

Question 6

$$32$$

$$64$$

$$-32$$

$$-64$$

Solution

Question 7

$$1$$

$$2$$

$$-1$$

$$0$$

Solution

Question 8

$$44$$

$$56$$

$$68$$

$$12$$

Solution

Question 9

$$\left(\dfrac 57 \right)^{-5}$$

$$\left(\dfrac 57 \right)^{5}$$

$$\left(\dfrac 75 \right)^{5}$$

$$\left(-\dfrac 75 \right)^{5}$$

Solution

Question 10

$$\dfrac 57$$

$$-\dfrac 57$$

$$\dfrac 75$$

$$\dfrac {-7}{5}$$

Solution

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