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:

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

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