How do you simplify #1/(b^-7xxb^5)#?

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Answer 1 · 2015-09-25T18:20:32

The answer is #b^2#.

When multiplying variables with exponents, add the exponents.

#x^m*x^n=x^(m+n)#

#1/(b^(-7)xxb^5)=#

#1/(b^(-7+5))=#

#1/(b^(-2))#

Apply exponent rule #x^(-a)=1/x^a#

#b^(-2)=1/b^2#

#1/(1/(b^2)#

Invert and multiply.

#cancel1xxb^2/cancel1=b^2#