What is sqrt(\sqrt{=x^{{-1\)3e^1-3/2 e^{1^2}}}} ?

The solution to sqrt(\sqrt{=x^{{-1\)3e^1-3/2 e^{1^2}}}} is \sqrt[4]{=x^{3e^1{-1}-e^{1^2} 3/2}}

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simplify sqrt(\sqrt{=x^{{-1\)3e^1-3/2 e^{1^2}}}}

simplify

= x^{{- 1} 3 e^{1} - \frac{3}{2} e^{1^{2}}}

Solution steps

= x^{{- 1} \cdot 3 e^{1} - \frac{3}{2} e^{1^{2}}}

Apply radical rule:

a = a^{\frac{1}{2}}

= ((= x^{{- 1} \cdot 3 e^{1} - \frac{3}{2} e^{1^{2}}})^{\frac{1}{2}})^{\frac{1}{2}}

Apply exponent rule:

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

assuming

a \geq 0

= (= x^{{- 1} \cdot 3 e^{1} - \frac{3}{2} e^{1^{2}}})^{\frac{1}{2} \cdot \frac{1}{2}}

\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}

= (= x^{{- 1} \cdot 3 e^{1} - \frac{3}{2} e^{1^{2}}})^{\frac{1}{4}}

1.8 \cdot 1 0^{- 5} \cdot 9

1.8 \cdot 1 0^{- 5} \cdot 3

1.81 \cdot 1 0^{- 3} \cdot 48

\frac{2}{3} \cdot \frac{x}{6}

What is sqrt(\sqrt{=x^{{-1\)3e^1-3/2 e^{1^2}}}} ?
The solution to sqrt(\sqrt{=x^{{-1\)3e^1-3/2 e^{1^2}}}} is \sqrt[4]{=x^{3e^1{-1}-e^{1^2} 3/2}}