How do you simplify #\frac { 3 t^ { 2 } m ^ { 3 } \cdot 2 f ^ { 4 } m ^ { 1 3 } } { ( \lm ^ { 4 } ) ^ { 3 } }#?

1 Answer
smendyka
Feb 14, 2017

See the entire simplification process below:

Explanation:

First, use these rules of exponents to simplify the denominator:

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

#(3t^2m^3 * 2f^4m^13)/(lm^4)^3 -> (3t^2m^3 * 2f^4m^13)/(l^color(red)(1)m^color(red)(4))^color(blue)(3) -> (3t^2m^3 * 2f^4m^13)/(l^(color(red)(1)xxcolor(blue)(3))m^(color(red)(4)xxcolor(blue)(3))) ->#

#(3t^2m^3 * 2f^4m^13)/(l^3m^12)#

Now, use this rule of exponents to simplify the numerator:

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

#(3t^2m^color(red)(3) * 2f^4m^color(blue)(13))/(l^3m^12) -> (6f^4m^(color(red)(3)+color(blue)(13))t^2)/(l^3m^12) -> (6f^4m^16t^2)/(l^3m^12)#

Now, we can use this rule for exponents to simplify the #m# terms in the numerator and denominator:

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

#(6f^4m^color(red)(16)t^2)/(l^3m^color(blue)(12)) -> (6f^4m^(color(red)(16)-color(blue)(12))t^2)/l^3 ->#

#(6f^4m^4t^2)/l^3#

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