The problem asks to evaluate the following expressions: i. $36^{\frac{1}{2}}$ ii. $625^{\frac{1}{4}}$ iii. $10000^{\frac{1}{4}}$ iv. $49^{-\frac{3}{2}}$ v. $(\frac{27}{125})^{-\frac{2}{3}}$

ArithmeticExponentsRootsSimplification
2025/6/19

1. Problem Description

The problem asks to evaluate the following expressions:
i. 361236^{\frac{1}{2}}
ii. 62514625^{\frac{1}{4}}
iii. 100001410000^{\frac{1}{4}}
iv. 493249^{-\frac{3}{2}}
v. (27125)23(\frac{27}{125})^{-\frac{2}{3}}

2. Solution Steps

i. 3612=36=636^{\frac{1}{2}} = \sqrt{36} = 6
ii. 62514=6254=544=5625^{\frac{1}{4}} = \sqrt[4]{625} = \sqrt[4]{5^4} = 5
iii. 1000014=100004=1044=1010000^{\frac{1}{4}} = \sqrt[4]{10000} = \sqrt[4]{10^4} = 10
iv. 4932=(4912)3=(49)3=73=173=134349^{-\frac{3}{2}} = (49^{\frac{1}{2}})^{-3} = (\sqrt{49})^{-3} = 7^{-3} = \frac{1}{7^3} = \frac{1}{343}
v. (27125)23=(12527)23=((12527)13)2=(1253273)2=(53)2=259(\frac{27}{125})^{-\frac{2}{3}} = (\frac{125}{27})^{\frac{2}{3}} = ((\frac{125}{27})^{\frac{1}{3}})^2 = (\frac{\sqrt[3]{125}}{\sqrt[3]{27}})^2 = (\frac{5}{3})^2 = \frac{25}{9}

3. Final Answer

i. 6
ii. 5
iii. 10
iv. 1343\frac{1}{343}
v. 259\frac{25}{9}

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