What is ((e^{1n(2x)}e^{1n(3x)})^4)^{1/2} ?

The solution to ((e^{1n(2x)}e^{1n(3x)})^4)^{1/2} is (e^{1*2nx}e^{1*3nx})^2

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simplify ((e^{1n(2x)}e^{1n(3x)})^4)^{1/2}

simplify

((e^{1 n (2 x)} e^{1 n (3 x)})^{4})^{\frac{1}{2}}

(e^{1 \cdot 2 n x} e^{1 \cdot 3 n x})^{2}

Solution steps

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

Apply exponent rule:

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

assuming

a \geq 0

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

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

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

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

25 ma s^{60} ce n t a v os

e^{- 6573}

e^{2 x^{2}} \cdot e^{x^{2}}

13 y^{4} (x + y)^{4} (x + z)^{4} z^{7} (y + 1)^{3}

(6.28 x 1 0^{- 6}) (sin (9 0^{\circ})) (0.2)

What is ((e^{1n(2x)}e^{1n(3x)})^4)^{1/2} ?
The solution to ((e^{1n(2x)}e^{1n(3x)})^4)^{1/2} is (e^{1*2nx}e^{1*3nx})^2