How do you simplify #[a^(1/5) b^½ c^¼]÷[a^-½ b^2.5 c^2] #?

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Answer 1 · 2017-04-01T22:41:22

#a^(7/10)/(b^2c^(7/4))#

You're asking:

#[a^(1/5)b^(1/2)c^(1/4)]-:[a^(-1/2)b^2.5c^2]#

Which is equal to:

#=(a^(1/5)b^(1/2)c^(1/4))/(a^(-1/2)b^2.5c^2)=(a^(1/5)b^(1/2)c^(1/4))/(a^(-1/2)b^(5/2)c^2)#

Which we can sort individually by base:

#=a^(1/5)/a^(-1/2)(b^(1/2)/b^(5/2))c^(1/4)/c^2#

We can simplify each of these using the rule #x^m/x^n=x^(m-n)#:

#=a^(1/5-(-1/2))(b^(1/2-5/2))c^(1/4-2)#

Finding common denominators:

#=(a^(2/10+5/10))(b^(1/2-5/2))(c^(1/4-8/4))#

#=a^(7/10)b^(-2)c^(-7/4)#

Simplify this using the rule #x^-m=1/x^m#:

#=a^(7/10)(1/b^2)1/c^(7/4)#

Which can be combined:

#=a^(7/10)/(b^2c^(7/4))#