How do you simplify #(8r^-5s^-4)/(6qr^6s^-7)#?

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Answer 1 · 2017-03-18T17:38:38

#(4s^3)/(3qr^11)#

Using the Index law #a^(-n)=1/a^n#, and conversely #1/a^(-n)=a^n#, swap around the values so that they all have positive indices:

#(8s^7)/(6qr^5r^6s^4)#

Using the index law #a^m+a^n=a^(m+n)#, you can simplify #r#:

#(8s^7)/(6qr^(5+6)s^4)=(8s^7)/(6qr^(11)s^4)#

Using the index law #a^m/a^n=a^(m-n)# gives

#(8s^(7-4))/(6qr^(11))=(8s^3)/(6qr^(11))#

Reduce:

#(cancel8^color(red)(4)s^3)/(cancel6^color(red)(3)qr^(11))=(4s^3)/(3qr^(11))#