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

Answer 1

Genesis Allen

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#

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Answer 2

Eva Abramson

#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#

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Answer 3

Ethan Adkins

#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#

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Answer 4

Eli Assouline

To find ( k ), we need to equate the exponents of both sides of the equation.

( 27\sqrt{3} = \sqrt{3} \times 3^k )

( 3^3 \times \sqrt{3} = \sqrt{3} \times 3^k )

Since the square root of ( 3 ) appears on both sides, we can equate the exponents of ( 3 ) to solve for ( k ):

( 3^3 = 3^k )

( 3 = k )

So, ( k = 3 ).

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Answer from HIX Tutor

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