SOLUTION: ^4 sqrt27^3sqrt3

Algebra ->  Exponents-negative-and-fractional -> SOLUTION: ^4 sqrt27^3sqrt3      Log On


   



Question 606547: ^4 sqrt27^3sqrt3
Answer by KMST(5328) About Me  (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 %28root%284%2C27%29%29%2A%28root%283%2C3%29%29 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:
3%2F4%2B1%2F3=9%2F12%2B4%2F12=%289%2B4%29%2F12=13%2F12 so
(3^(3/4))*(3^(1/3))=3^(3/4+1/3)=3^(13/12)=root%2812%2C3%5E13%29
Or even better,
=3root%2812%2C3%29
We could have kept writing fractional exponents to the end because
3%2F4%2B1%2F3=13%2F12=1%2B1%2F12 so
(3^(3/4))*(3^(1/3))=3^(3/4+1/3)=3^(1+1/12)=(3^1)*(3^(1/12))=3root%2812%2C3%29
(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).