How do you condense #\frac { s ^ { - 1} t ^ { 22} \cdot s ^ { 22} t ^ { 0} } { s t ^ { 0} }# into one term?

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Answer 1 · 2017-06-16T14:02:30

See a solution process below:

First, we will use this rule of exponents to simplify some of the terms:

#a^color(red)(0) = 1#

#(s^-1t^22 * s^22t^color(red)(0))/(st^color(red)(0)) => (s^-1t^22 * s^22 * 1)/(s * 1) => #

#(s^-1t^22 * s^22)/s#

We can now use this rule of exponents to consolidate the #s# terms in the numerator:

#x^color(red)(a) xx x^color(blue)(b) = x^(color(red)(a) + color(blue)(b))#

#(s^-1t^22 * s^22)/s => (s^color(red)(-1) * s^color(blue)(22) * t^22)/s => (s^(color(red)(-1)+color(blue)(22)) * t^22)/s => #

#(s^21 * t^22)/s#

We can now use these two rules of exponents to complete the simplification:

#a = a^color(red)(1)# and #x^color(red)(a)/x^color(blue)(b) = x^(color(red)(a)-color(blue)(b))#

#(s^21 * t^22)/s => (s^color(red)(21) * t^22)/s^color(blue)(1) => (s^(color(red)(21)-color(blue)(1)) * t^22) => s^20 * t^22 =>#

#s^20t^22#

Answer 2 · 2017-06-16T15:52:42

#s^20 t^22#

The expression is already one term because there are no addition or subtraction signs between different parts.

It can only be written in its simplest form....

Use the laws of indices to simplify the different bases:

#(s^-1t^22xxs^22cancel(t^0))/(scancel(t^0))" "larr (t^0 =1)#

Add the indices of like bases in the numerator and subtract the index in the denominator.

#(color(blue)(s^-1)color(red)(t^22)xxcolor(blue)(s^22))/color(blue)(s^1)#

#color(blue)(s^((-1+22-1)))color(red)(t^22)#

#=s^20t^22#