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

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Answer 1 · 2016-01-11T01:16:53

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

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)#