What is \sqrt[5]{(x^2+cos(x)+2e^x)^2} ?

The solution to \sqrt[5]{(x^2+cos(x)+2e^x)^2} is (x^2+2e^x+cos(x))^{2/5}

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simplify \sqrt[5]{(x^2+cos(x)+2e^x)^2}

simplify

(x^{2} + 2 e^{x} + cos (x))^{\frac{2}{5}}

Solution steps

Apply radical rule:

= ((x^{2} + cos (x) + 2 e^{x})^{2})^{\frac{1}{5}}

Apply exponent rule:

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

assuming

a \geq 0

= (x^{2} + cos (x) + 2 e^{x})^{2 \cdot \frac{1}{5}}

2 \cdot \frac{1}{5} = \frac{2}{5}

= (x^{2} + cos (x) + 2 e^{x})^{\frac{2}{5}}

= (x^{2} + 2 e^{x} + cos (x))^{\frac{2}{5}}

7.4094 \cdot 1 0^{- 3}

140 e^{- 0.4 (6)}

\frac{72 y ^{2 n}}{3} (9^{0}) (2 y^{n + 2})^{- 1}

e^{- \frac{410}{400}}

What is \sqrt[5]{(x^2+cos(x)+2e^x)^2} ?
The solution to \sqrt[5]{(x^2+cos(x)+2e^x)^2} is (x^2+2e^x+cos(x))^{2/5}