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# How do i simpifiy 81x^4y^2

i need help with this question

^ Assuming that is the actual question then that is the correct answer. Moreover, that likely is the question depending on what level math Devanee is taking. However, if one assumes the question is to simplify exactly what is given verbatim then the actual term that is to be simplified is 81X^4y^2, which is a different question entirely. If there are no parenthesis to dictate the order of operations we work from the top down and can use the identity b^p^q=b^(p^q)=b^(p*q) (* is multiplication). So, for 81X^4y^2 b=81x, p=4y, and q=2. Thus we multiply p by q or 4y*2 and get 8y so b^(p*q)= 81x^8y. This can be further simplified to 81^8y*x^8y by the identity (b*c)^n= b^n*c^n.

I'm going to assume that the question is to simplify

81x4 • y2

If so, ask yourself "what number to the 4th power (what number times itself 4 times) equals 81"

34  or 3 • 3 • 3 • 3 = 81

You can rewrite 81x4 as: (3x)4 because both the 3 and the x need to go to the 4th power to give you 81x4

You can't make y2 any simpler