What is \sqrt[ 3/2 ]{8^2}*\sqrt[ 5/3 ]{32^{2/3}} ?

The solution to \sqrt[ 3/2 ]{8^2}*\sqrt[ 5/3 ]{32^{2/3}} is 64

Popular >

simplify \sqrt[ 3/2 ]{8^2}*\sqrt[ 5/3 ]{32^{2/3}}

simplify

64

Solution steps

= 8^{\frac{4}{3}} \cdot 3 2^{\frac{2}{5}}

Factor integer

8 = 2^{3}

= (2^{3})^{\frac{4}{3}} \cdot 3 2^{\frac{2}{5}}

Apply exponent rule:

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

(2^{3})^{\frac{4}{3}} = 2^{3 \cdot \frac{4}{3}}

= 2^{3 \cdot \frac{4}{3}} \cdot 3 2^{\frac{2}{5}}

Refine

= 2^{4} \cdot 3 2^{\frac{2}{5}}

Factor integer

32 = 2^{5}

= 2^{4} (2^{5})^{\frac{2}{5}}

Apply exponent rule:

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

(2^{5})^{\frac{2}{5}} = 2^{5 \cdot \frac{2}{5}}

= 2^{4} \cdot 2^{5 \cdot \frac{2}{5}}

Refine

= 2^{4} \cdot 2^{2}

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

2^{4} \cdot 2^{2} = 2^{4 + 2}

= 2^{4 + 2}

Add the numbers:

4 + 2 = 6

= 2^{6}

2^{6} = 64

= 64

(\frac{2}{3} x^{4} - \frac{1}{5} y^{5})^{2} (\frac{2}{3} x^{4} + \frac{1}{5} y^{5})^{2}

(\frac{3 d e ^{- 1}}{12 n e ^{2} f})^{- 3}

e^{- 0.65} \cdot 0.6 5^{0} \cdot \frac{1}{0 !}

(2.3 x 1 0^{1}) (7 x 1 0^{6})

(e^{y} + 1)^{2} e^{- y} d x

What is \sqrt[ 3/2 ]{8^2}*\sqrt[ 5/3 ]{32^{2/3}} ?
The solution to \sqrt[ 3/2 ]{8^2}*\sqrt[ 5/3 ]{32^{2/3}} is 64