({x^5})^{1/3} (16{x^3})^{2/3} (frac{1}{4} x^{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given expression: $E = (x^5)^{1/3} (16x^3)^{2/3} (\frac{1}{4} x^{4/9})^{-3/2}$Step $1$: Simplify each term:$(x^5)^{1/3} = x^{5/3}$$(16x^3)^{2/3} =

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) Given expression: $E = (x^5)^{1/3} (16x^3)^{2/3} (\frac{1}{4} x^{4/9})^{-3/2}$Step $1$: Simplify each term:$(x^5)^{1/3} = x^{5/3}$$(16x^3)^{2/3} = (2^4)^{2/3} (x^3)^{2/3} = 2^{8/3} x^2$$(\frac{1}{4} x^{4/9})^{-3/2} = (2^{-2})^{-3/2} (x^{4/9})^{-3/2} = 2^3 x^{-6/9} = 2^3 x^{-2/3}$Step $2$: Combine the terms:$E = x^{5/3} 	imes 2^{8/3} x^2 	imes 2^3 x^{-2/3}$$E = 2^{(8/3 + 3)} 	imes x^{(5/3 + 2 - 2/3)}$$E = 2^{17/3} 	imes x^{(3/3 + 2)} = 2^{17/3} x^3$Since $2^{17/3} x^3 
eq 8x^3$ or other options,the correct answer is None of these.

$({x^5})^{1/3} (16{x^3})^{2/3} (\frac{1}{4} x^{4/9})^{-3/2} = $

Similar Questions

If $3^x - 3^{x - 1} = 6$,then $x^x$ is equal to

If $a^{1/x} = b^{1/y} = c^{1/z}$ and $b^2 = ac$,then $x + z = $

The rationalising factor of $a^{1/3} + a^{-1/3}$ is

If ${x^{x\sqrt[3]{x}}} = {(x \cdot \sqrt[3]{x})^x}$,then $x =$

If $4^{x} - 3^{x - \frac{1}{2}} = 3^{x + \frac{1}{2}} - 2^{2x - 1}$,then the value of $x$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module