What is # 3^(3/2)# in radical form?

2 Answers
Apr 10, 2018

The answer is #sqrt(3^3)#, or to simplify it even further, #sqrt27#.

Explanation:

Say we have a whole number c, raised to the power of a fraction n over d, with n being the numerator and d being the denominator (#c^(n/d)#).

You can rewrite #c^(n/d)# as #c^(n * 1/d)#. When a number is raised to a fractional exponent, it is equivalent to taking the #d^(th)# root of that number.

For example, say we have #16^(1/4)#. This is the same as taking the fourth root of 16, which can be written as #root4(16)#, whose answer is equal to 2 (#2^4# = 16).

Apr 10, 2018

#3sqrt3=sqrt27#

Explanation:

When using indicies, #x^(1/2)=sqrtx#.

I think the easiest way to solve this is probably to split it up using our Index Laws:

#a^n + a^m=a^(n+m)#

Hence:

#3^(3/2)=3^1 times 3^(1/2)#

And this is obviously the same as:

#3 times sqrt3= 3sqrt3#

Then if you want it entirely as a radical, you must "insert" the 3 under the radical sign which is done by taking the square of the number in front of the radical (as this is the inverse operation of taking the square root of something), and placing it under the root sign:

#3sqrt3=sqrt(9 times 3)=sqrt27#