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Evaluate. fourth root of 9 multiplied by square root of 9 over the fourth root of 9 to the power of 5


5 to the power of negative 1 over 6

5 to the power of 3 over 2

9

92

Verified Answer :

JcAlmighty

The answer is 1/3

To calculate this will use several rules:
\(\sqrt[n]{x^m } = x^{ \frac{m}{n} }\)
\(x^{-m}= \frac{1}{ x^{m} }\)
\(x^{a} * x^{b} =x ^{a+b}\)
\(\frac{ x^{a} }{ x^{b} } = x^{a-b}\)


The fourth root of 9 is \(\sqrt[4]{9}= \sqrt[4]{ 9^{1} }= 9^{ \frac{1}{4} }\)
Square root of 9 is \(\sqrt[2]{9}= \sqrt[2]{ 9^{1} } = 9^{ \frac{1}{2} }\)
The fourth root of 9 to the power of 5 is \(\sqrt[4]{9^{5} } = 9^{ \frac{5}{4} }\)

The fourth root of 9 multiplied by square root of 9 over the fourth root of 9 to the power of 5 is:
\(\frac{9^{ \frac{1}{4} }*9^{ \frac{1}{2} }}{9^{ \frac{5}{4} }} = \frac{ 9^{ \frac{1}{4}+ \frac{1}{2}} }{9 \frac{5}{4} } =9^{ \frac{1}{4}+ \frac{1}{2}- \frac{5}{4} }=9^{ \frac{1}{4}+ \frac{1*2}{2*2}- \frac{5}{4} }=9^{ \frac{1}{4}+ \frac{2}{4}- \frac{5}{4} }=9^{ \frac{1+2-5}{4} }= 9^{ \frac{-2}{4} }\) \(= 9^{- \frac{1}{2} }= \frac{1}{9^{ \frac{1}{2} } } =\frac{1}{ \sqrt[2]{9^{1} } } = \frac{1}{ \sqrt[2]{9} } = \frac{1}{3}\)

(1) Fourth root of 9 multiplied by square root of 9 over the fourth root of 9 to the power of 5.= 9^(1/4) * 9^(1/2) / (9^(1/4))^5, cancel 9^(1/4)= 9^(1/2) / (9^(1/4))^4= 9^(1/2) / 9= 3 / 9= 1/3

(2) Five to the power of negative 1 over 6= 5 ^ (-1/6)= (1/5)^(1/6)

(3) Five to the power of 3 over 2= 5 ^ (3/2)= sqrt(125)


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