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simplify this (image attached)
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how do i get the 125 down?
well \[\huge 125 = 5^3\] does that help?
do i use that with the 6?
yes
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actually...it depends on what you meant...
im a bit lost
(5^3)^(1/6) = 5 ^(3/6) = 5^(1/2) and (a^15)^(1/6) = ?
recall that 6th root = ^ (1/6)
so a^5/2 ?
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yes
dont forget to simplify the whole thing
how do i simplify 5^1/2 a^5/2
@lgbasallote
\[\huge x^{1/2} \times y^{1/2} \implies (xy)^{1/2} \implies \sqrt{xy}\]
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so i should put a square root over the five? err...
or is it liike 5a^?/2
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