PreCalculus
posted by Ann .
How is 16/(4x+6)^5 and 1/(2(2x+3)^5) equivalent?
I think 1/2(2x+3)^5 is the more simplified version of the two, but I just can't figure out how 16/(4x+6)^5 simplifies to 1/(2(2x+3)^5). please explain

When you factor the denominator you get
(4x+6)^5
= 2^5 (2x + 3)^5
= 32(2x+3)^5
so 16/(4x+6)^5
= 16/(32(2x+3)^5)
= 1/(2(2x+3)^5)
further explanation:
(4x+6)^5
= (4x+6)(4x+6)(4x+6)(4x+6)(4x+6)
= (2)(2x+3)(2)(2x+3)(2)(2x+3)(2)(2x+3)(2)(2x+3)
= 32(2x+3)^5 
thanks
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