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3(b^5c)^-2 Cannot remember how to simplify this..Do i distribute? Help:)
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\[3(b ^{5c})^{-2}\] ______________________________ first do \[(b ^{5c})^{-2}\] -2 * 5c = -10c so \[(b ^{-10c})\] ______________________________ which is 1/\[(b ^{10c})\] because a negative exponent is the same as the 1 over the monomial Now multiply 1/ \[(b ^{10c})\] by 3, so the answer is \[3/(b ^{10c})\]
Awesome thank you...Actually the "c" is not an exponent though..Does the same action apply?
If it is just -C you don't put it under 1 if that's what you're asking? but if its X^-c then you put it under one without the negative.
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