How do you simplify (s^-2t^-7u^98)/(s^23t^79u^-77 * s^0t^83u^-1?

1 Answer
Jan 11, 2016

=u^176/(s^25*t^169)

Explanation:

Remembering that:

a^0=1

a^(-n)=1/a^n

a^n*a^m=a^(n+m)

a^n/a^m=a^(n-m)

you can write the expression as:

(1/s^2*1/t^7u^98)/(s^23*t^79*1/(u^77)*1*t^83*1/(u))=

(u^98*u^77*u)/(s^23*t^79*t^83*s^2*t^7)=

ordering the therms

=(u^98*u^77*u)/(s^23*s^2*t^79*t^83*t^7)=

applying the rules

=u^(98+77+1)/(s^(23+2)*t^(79+83+7))=

=u^176/(s^25*t^169)

alternatively

(s^-2*t^-7u^98)/(s^23*t^79*u^-77*s^0*t^83*u^-1=

(s^-2/(s^23*s^0))*(t^-7/(t^79*t^83))*(u^98/(u^(-77)*u^-1))=

s^(-2-23+0)*t^(-7-79-83)*u^(98+77+1)=

s^-25*t^-169*u^176=u^176/(s^25*t^169)