SOLUTION: ^4 sqrt27^3sqrt3

Question 606547: ^4 sqrt27^3sqrt3
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
Radicals are cumbersome; fractional exponents are easier to work with.
The only problem is writing those fractional exponents so that they are clearly readable.
If your problem was I would solve it like this

I can write that final expressions as (3^(3/4))*(3^(1/3)) and hope it is more readable.
That is the product of two powers of base 3, which equals a power of base 3 whose exponent is the sum of the exponents.
(3^(3/4))*(3^(1/3))=3^(3/4+1/3)
I can add fractions:
so
(3^(3/4))*(3^(1/3))=3^(3/4+1/3)=3^(13/12)=
Or even better,
=

We could have kept writing fractional exponents to the end because
so
(3^(3/4))*(3^(1/3))=3^(3/4+1/3)=3^(1+1/12)=(3^1)*(3^(1/12))=

(I would like to write that front 3 a little lower, as I would with pen and paper, but I can't. I hope you understand what I mean, anyway).
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