To calculate: To convert the product 3a4b25⋅7a2b66 into a single radical with lowest index.

Calculation: Here the radical are 5 and 6 The LCM is given by 5×6=30 Hence applying the 30-th power and then 30-th root. 3a4b25⋅7a2b66=(3a4b25⋅7a2b66)3030 According to product rule, (ab)n=an⋅bn Applying the above rule: 3a4b25⋅7a2b66=(3a4b25)30⋅(7a2b66)3030 According to exponential rules, an=a1n Applying the above rule: 3a4b25⋅7a2b66=((3a4b2)15)30⋅((7a2b6)16)3030 According to the power rule, (am)n=amn 3a4b25⋅7a2b66=(3a4b2)(15)(30)⋅(7a2b6)(16)(30)30 3a4b25⋅7a2b66=(3a4b2)6⋅(7a2b6)530 According to product rule and power rule: 3a4b25⋅7a2b66=(36a(4)(6)b(2)(6))⋅(75a(2)(5)b(6)(5))30 3a4b25⋅7a2b66=(36a24b12)⋅(75a10b30)30 3a4b25⋅7a2b66=3675a24b12a10b3030 According to the product rule, aman=am+n Applying the above rule:

Question: "To calculate: To convert the product 3a4b25⋅7a2b66 into..."

High School
Other

Globokim8

Answered question

2021-10-26

To calculate:
To convert the product $\sqrt[5]{3 a^{4} b^{2}} \cdot \sqrt[6]{7 a^{2} b^{6}}$ into a single radical with lowest index.

Answer & Explanation

funblogC

Skilled2021-10-27Added 91 answers

Calculation:
Here the radical are 5 and 6
The LCM is given by $5 \times 6 = 30$
Hence applying the 30-th power and then 30-th root.
$\sqrt[5]{3 a^{4} b^{2}} \cdot \sqrt[6]{7 a^{2} b^{6}} = \sqrt[30]{(\sqrt[5]{3 a^{4} b^{2}} \cdot \sqrt[6]{7 a^{2} b^{6}})^{30}}$
According to product rule, ${(a b)}^{n} = a^{n} \cdot b^{n}$
Applying the above rule:
$\sqrt[5]{3 a^{4} b^{2}} \cdot \sqrt[6]{7 a^{2} b^{6}} = \sqrt[30]{(\sqrt[5]{3 a^{4} b^{2}})^{30} \cdot (\sqrt[6]{7 a^{2} b^{6}})^{30}}$
According to exponential rules, $\sqrt[n]{a} = a^{\frac{1}{n}}$
Applying the above rule:
$\sqrt[5]{3 a^{4} b^{2}} \cdot \sqrt[6]{7 a^{2} b^{6}} = \sqrt[30]{((3 a^{4} b^{2})^{\frac{1}{5}})^{30} \cdot ((7 a^{2} b^{6})^{\frac{1}{6}})^{30}}$
According to the power rule, ${(a^{m})}^{n} = a^{m n}$
$\sqrt[5]{3 a^{4} b^{2}} \cdot \sqrt[6]{7 a^{2} b^{6}} = \sqrt[30]{(3 a^{4} b^{2})^{(\frac{1}{5}) (30)} \cdot (7 a^{2} b^{6})^{(\frac{1}{6}) (30)}}$
$\sqrt[5]{3 a^{4} b^{2}} \cdot \sqrt[6]{7 a^{2} b^{6}} = \sqrt[30]{(3 a^{4} b^{2})^{6} \cdot (7 a^{2} b^{6})^{5}}$
According to product rule and power rule:
$\sqrt[5]{3 a^{4} b^{2}} \cdot \sqrt[6]{7 a^{2} b^{6}} = \sqrt[30]{(3^{6} a^{(4) (6)} b^{(2) (6)}) \cdot (7^{5} a^{(2) (5)} b^{(6) (5)})}$
$\sqrt[5]{3 a^{4} b^{2}} \cdot \sqrt[6]{7 a^{2} b^{6}} = \sqrt[30]{(3^{6} a^{24} b^{12}) \cdot (7^{5} a^{10} b^{30})}$
$\sqrt[5]{3 a^{4} b^{2}} \cdot \sqrt[6]{7 a^{2} b^{6}} = \sqrt[30]{3^{6} 7^{5} a^{24} b^{12} a^{10} b^{30}}$
According to the product rule, $a^{m} a^{n} = a^{m + n}$
Applying the above rule:

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