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

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Answer 1 · 2016-02-01T16:12:49

Feb 1, 2016

#d^(11/2#

Answer 2 · 2016-02-01T16:14:27

exponential form#=d^(11/2)#

Recall that the square root of a term can be written as #sqrt(x)# or #x^(1/2)#.

Thus:

#sqrt(d^11)#

#=(d^11)^(1/2)#

Multiply the exponents 11 and 1/2 together.

#=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)#