How do you simplify #(81x^12)^1.25#?

1 Answer
Feb 9, 2017

#(81x^12)^1.25 = 243 abs(x)^15#

Explanation:

The identity:

#(x^a)^b = x^(ab)#

holds under any of the following conditions:

  • #x > 0# with #a, b# any real numbers.

  • #x = 0# with #a, b >= 0#.

  • #x < 0# with #a, b# any integers.

In other circumstances it can fail.

#color(white)()#
Given:

#(81x^12)^1.25#

FIrst note that:

#(81x^12)^1.25 = (3^4 (x^3)^4)^(5/4) = ((3x^3)^4)^(5/4)#

If #x >= 0# then #3x^3 >= 0#, so since both #4 >= 0# and #5/4 >= 0# we can assert:

#((3x^3)^4)^(5/4) = (3x^3)^(4*5/4) = (3x^3)^5 = 3^5 x^15 = 243 x^15#

If #x < 0# then #3x^3 < 0#, but #x^12 = (-x)^12#, so:

#(81x^12)^1.25 = (81(-x)^12)^1.25 = 243 (-x)^15 = -243 x^15#

To cover both cases, we can write:

#(81x^12)^1.25 = 243 abs(x)^15#