cragajjar2 cragajjar2
  • 10-12-2018
  • Mathematics
contestada

What is -16 to the 3/4 power?

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ren963
ren963 ren963
  • 10-12-2018

First this I would like to explain is that ;
[tex] \sqrt{x} = \sqrt[2]{ {x}^{1} } = {x}^{ \frac{1}{2} } [/tex]
This means that when;
[tex] {x}^{ \frac{3}{4} } = \sqrt[4]{ {x}^{3} } [/tex]

either >>

[tex] ({ - 16})^{ \frac{3}{4} } = \sqrt[4]{ { - 16}^{3} } [/tex]
(-16)^3 = -4096

However the 4th root of anything negative is undefined, the answer to your question is undefined.

or >>

[tex] - ( {16}^{ \frac{3}{4} } ) = - \sqrt[4]{ {16}^{ 3 } } = - \sqrt[4]{4096} = - 8[/tex]

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