If #27sqrt(3)=sqrt(3)xx3^k#, what is #k#?

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Answer 1 · 2017-01-19T22:10:58

Write each value in the expression as a power of three, and you will see the answer is #k=3#

Since #sqrt3=3^(1/2)# and #27=3^3# we can write the expression as

#(3^3*3^(1/2))/3^(1/2) = 3^k#

Cancel both #3^(1/2)# powers to get

#3^3 = 3^k#

Answer 2 · 2017-01-19T22:12:10

#k=3#

Note that #27 = 3^3#, so we find:

#3^k = (27color(red)(cancel(color(black)(sqrt(3)))))/color(red)(cancel(color(black)(sqrt(3)))) = 27 = 3^3#

Hence #k = 3#

Answer 3 · 2017-01-19T22:12:43

#k=3#

#(27sqrt3)/sqrt3=3^k#

i.e. #3^k=(27cancelsqrt3)/cancelsqrt3=27=3xx3xx3=3^3#

And thus #k=3#.

If this were not a perfect cube, we could take #"logs"# of both sides:

#klog3=3log3#