How do you write $\sqrt{d^{11}}$ $sqrt(d^11)$ as a exponential form?

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Answer 1 · 2016-02-01T16:12:49

$d^{\frac{11}{2}}$ $d^(11/2$

Answer 2 · 2016-02-01T16:14:27

exponential form $= d^{\frac{11}{2}}$ $=d^(11/2)$

Recall that the square root of a term can be written as $\sqrt{x}$ $sqrt(x)$ or $x^{\frac{1}{2}}$ $x^(1/2)$ .

Thus:

$\sqrt{d^{11}}$ $sqrt(d^11)$

$= {(d^{11})}^{\frac{1}{2}}$ $=(d^11)^(1/2)$

Multiply the exponents 11 and 1/2 together.

$= d^{\frac{11}{2}}$ $=d^(11/2)$

$:.$ , $sqrt(d^11)$ in exponential form is $d^(11/2)$ .

Answer 3 · 2016-02-02T00:35:20

If $d$ is Real and non-negative then $sqrt(d^11) = d^(11/2)$

Otherwise about the best you can say is $sqrt(d^11) = (d^11)^(1/2)$

Suppose $d = -1$

Then $d^11 = -1$ and $sqrt(d^11) = sqrt(-1) = i$

However, $d = cos(pi) + i sin(pi)$

So, using De Moivre's formula:

$d^(11/2) = cos((11pi)/2) + i sin((11pi)/2)$

$= cos((3pi)/2) + i sin((3pi)/2)$

$= 0 + i (-1)$

$=-i$

So we find $d^(11/2) != (d^11)^(1/2)$

How do you write sqrt(d^11)√d11sqrt(d^11) as a exponential form?