How do you write the expression #root6(2)# in exponential form?

1 Answer
Reg H.
Dec 12, 2016

#root(6)2 = 2^(1/6)#

Explanation:

Well, 2 itself can be written as #2^1#, and the square root of two can be written as #sqrt2# and #sqrt2 * sqrt2# = 2, but alternatively #sqrt2# can be written as #2^(1/2)# and this works nicely because #2^(1/2) * 2^(1/2)# =2 and, as you recall, exponents add when multiplying the identical base.

Similarly #root(3)2# can be written as #2^(1/3)# and so #2^(1/3) * 2^(1/3) * 2^(1/3) = 2#

Carrying the same logic alone we see (I hope we do) #root(6)2# = #2^(1/6)#

I am doing these basic questions to refresh my rusty skills, so, thank you.
if anyone can write this more elegantly, I welcome the assistance.

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