Simplify and write the answers in exponential forms.
(a) \( 3 \times 3^{10} \)
(b) \( 5^{2} \times 5^{7} \)
(c) \( (-7)^{2} \times(-7)^{7} \)
(d) \( 27^{3} \p 3^{9} \)
(e) \( \left(25^{0}+5^{0}\right) \times 5^{0} \)
(f) \( \left(2^{0} \p 3^{0}\right) \times 4^{0} \)

Given:

(a) \( 3 \times 3^{10} \)
(b) \( 5^{2} \times 5^{7} \)
(c) \( (-7)^{2} \times(-7)^{7} \)
(d) \( 27^{3} \div 3^{9} \)
(e) \( \left(25^{0}+5^{0}\right) \times 5^{0} \)
(f) \( \left(2^{0} \div 3^{0}\right) \times 4^{0} \)

To do:

We have to simplify and write the answers in exponential forms.

Solution:

We know that,

$(a^{m})^{n}=a^{m n}$

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

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

$a^{0}=1$  

Therefore,

(a) $3 \times 3^{10}=3^{1+10}$

$=3^{11}$

(b) $ 5^{2} \times 5^{7} = 5^{2+7}$

$=5^9$

(c) $(-7)^{2} \times(-7)^{7}=(-7)^{2+7}$

$=(-7)^{9}$

(d) $27^{3} \div 3^{9}=(3^3)^3 \div 3^9$

$=3^{3\times3} \div 3^9$

$=3^9 \div 3^9$

$=3^{9-9}$

$=3^0$

$=1$

(e) $(25^{0}+5^{0}) \times 5^{0}=(1+1) \times 1$

$=2\times1$

$=2$

(f) $(2^{0} \div 3^{0}) \times 4^{0}=(1 \div 1) \times 1$

$=1\times1$

$=1$

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Updated on: 10-Oct-2022

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