What is \sqrt[5]{(a+b)^{4^'}} ?

The solution to \sqrt[5]{(a+b)^{4^'}} is (a+b)^{(4^')/5}

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simplify \sqrt[5]{(a+b)^{4^'}}

simplify

(a + b)^{\frac{4 ^{'}}{5}}

Solution steps

Apply radical rule:

= ((a + b)^{4^{^{^{'}}}})^{\frac{1}{5}}

Apply exponent rule:

(a^{b})^{c} = a^{b c},

assuming

a \geq 0

= (a + b)^{4^{^{^{'}}} \cdot \frac{1}{5}}

Simplify

4^{^{'}} \cdot \frac{1}{5} : \frac{4 ^{^{'}}}{5}

= (a + b)^{\frac{4 ^{^{'}}}{5}}

(e^{- t})^{6}

6 X^{2} Y^{2} O

6 (1.3)^{- (- 6)}

150 \cdot e^{- \frac{135}{80}}

7.03 \cdot 1 0^{- 5}

What is \sqrt[5]{(a+b)^{4^'}} ?
The solution to \sqrt[5]{(a+b)^{4^'}} is (a+b)^{(4^')/5}