Exponents and Powers Test

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Exponents and Powers Test - 20

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Weekly Quiz Competition

Question 1
1 / -0

Find the value of $$x$$ in: $$\left (\dfrac {12}{13}\right )^{3x} \div \left (\dfrac {13}{12}\right )^{4} = \left (\dfrac {12}{13}\right )^{10}$$

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

We can rewrite the given equation as,

$$\bigg(\dfrac{12}{13}\bigg)^{3x}\times \bigg(\dfrac{12}{13}\bigg)^{4} = \bigg(\dfrac{12}{13}\bigg)^{10}$$

Now sum of the powers on the right hand side must be equal to the left hand side since the base is same.
$$3x+4=10$$
$$3x=6$$
$$x=2$$
Question 2
1 / -0

Which of the following expressions is equal to $$x^{mn}$$?

A
$$(X^m)^n$$

B
$$X^{m+n}$$

C
$$X^mX^n$$

D
$$nX^{m}$$

Solution
Question 3
1 / -0

Express $$\displaystyle \left ( 125 \right )^{5}$$ as a power with the base $$5$$.

A
$$\displaystyle 5^{-2}$$

B
$$\displaystyle 5^{-15}$$

C
$$\displaystyle 5^{-8}$$

D
$$\displaystyle 5^{15}$$

Solution

Given, $$\displaystyle \left ( 125 \right )^{5}$$
$$=\left ( 5^{3} \right )^{5}$$
$$={5}^{3 \times 5}$$
$$=5^{15}$$
Hence, option $$B$$ is correct.
Question 4
1 / -0

Convert $$62000+39000$$ to scietific form.

A
$$1.01\times 10^5$$

B
$$1.1\times 10^5$$

C
$$1.01\times 10^4$$

D
$$1.1\times 10^4$$

Solution

On adding, we get
$$62000+39000=101000$$
Therefore, $$ 101000=1.01\times 10^5$$
Hence, option A is correct.
Question 5
1 / -0

Simplify the following using law of exponents.
$$9^2 \times 9^{18}\times 9^{10}$$

A
$$9^{30}$$

B
$$9^{25}$$

C
$$9^{15}$$

D
$$9^{10}$$

Solution

We know that,

$$a^{m}\times a^{n}\times a^{p}=a^{m+n+p}$$

Therefore,

$$9^{2}\times 9^{18}\times 9^{10}=9^{2+18+10}$$

$$=9^{30}$$
Question 6
1 / -0

Simplify the following using law of exponents.
$$(-3)^3\times (-3)^{10}\times (-3)^7$$

A
$$3^{20}$$

B
$$3^{40}$$

C
$$3^{60}$$

D
$$3^{10}$$

Solution

we know,

$$a^{m}*a^{n}*a^{p}=a^{m+n+p}$$

so,

$$(-3)^{3}*(-3)^{10}*(-3)^{7}=(-3)^{3+10+7}$$

$$=(-3)^{20}$$

$$=(3)^{20}$$
Question 7
1 / -0

Find: $$7^{5} \div 7^{3}=$$

A
$$7^{2}$$

B
$$7^{-2}$$

C
$$7^{8}$$

D
$$7^{-8}$$

Solution

We have the quotient law of exponents,

$$\dfrac{a^m}{a^n}=a^{m-n}$$

Applying this,

$$\dfrac{7^5}{7^3}=7^{5-3}=7^2=49$$
Question 8
1 / -0

$$(2^{5} \div 2^{2})=?$$

A
$$\dfrac{1}{128}$$

B
$$\dfrac{-1}{128}$$

C
$$\dfrac{-1}{8}$$

D
$${8}$$

Solution

Using law of exponent $$\dfrac {a^m}{a^n}=(a)^{m-n}$$ Where, $$a$$ is non-zero integer
Therefore, $$\dfrac {(2)^5}{(2)^2}=(2)^{5-2}=(2)^{3}$$
Question 9
1 / -0

$$a\times a\times b\times b\times b$$ can be written as

A
$$a^2b^3$$

B
$$a^3b^2$$

C
$$a^3b^3$$

D
$$a^5b^5$$

Solution

we have, $$a\times a\times b\times b\times b$$
$$=a^2 \times b^3 =a^2b^3$$

Hence, $$a\times a\times b\times b\times b$$ can be written as $$a^2b^3$$
Question 10
1 / -0

The value of $$\left(-\dfrac 23 \right)^4$$ is equal to

A
$$\left(\dfrac {16}{81} \right)$$

B
$$\left(-\dfrac {81}{16} \right)$$

C
$$\left(\dfrac {-16}{81} \right)$$

D
$$\left(-\dfrac {81}{-16} \right)$$

Solution

Given, $$\left(\dfrac {-2}{3}\right)^4=\left(\dfrac {-2}{3}\right)\times \left(\dfrac {-2}{3}\right)\times \left(\dfrac {-2}{3}\right)\times \left(\dfrac {-2}{3}\right)=\dfrac {16}{81}$$
[for $$(-a)^m$$, if $$m$$ is even, then $$(-a)^m$$ is positive.]

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Exponents and Powers Test - 20

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Exponents and Powers Test - 20

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SHARING IS CARING

Question 1

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 2

$$(X^m)^n$$

$$X^{m+n}$$

$$X^mX^n$$

$$nX^{m}$$

Solution

Question 3

$$\displaystyle 5^{-2}$$

$$\displaystyle 5^{-15}$$

$$\displaystyle 5^{-8}$$

$$\displaystyle 5^{15}$$

Solution

Question 4

$$1.01\times 10^5$$

$$1.1\times 10^5$$

$$1.01\times 10^4$$

$$1.1\times 10^4$$

Solution

Question 5

$$9^{30}$$

$$9^{25}$$

$$9^{15}$$

$$9^{10}$$

Solution

Question 6

$$3^{20}$$

$$3^{40}$$

$$3^{60}$$

$$3^{10}$$

Solution

Question 7

$$7^{2}$$

$$7^{-2}$$

$$7^{8}$$

$$7^{-8}$$

Solution

Question 8

$$\dfrac{1}{128}$$

$$\dfrac{-1}{128}$$

$$\dfrac{-1}{8}$$

$${8}$$

Solution

Question 9

$$a^2b^3$$

$$a^3b^2$$

$$a^3b^3$$

$$a^5b^5$$

Solution

Question 10

$$\left(\dfrac {16}{81} \right)$$

$$\left(-\dfrac {81}{16} \right)$$

$$\left(\dfrac {-16}{81} \right)$$

$$\left(-\dfrac {81}{-16} \right)$$

Solution

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