How do you simplify $\frac { ( 9s ^ { - 5} t ^ { 4} ) ^ { 3/ 2} } { ( 27s ^ { 6} t ^ { - 2} ) ^ { 2/ 3} }$ ?

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Answer 1 · 2018-01-09T02:49:20

Assuming $s$ and $t$ are both positive;

$((9s^(-5)t^4)^(3/2))/((27s^6t^(-2))^(2/3))=(3t^(22/3))/s^(23/2)$

$((9s^(-5)t^4)^(3/2))/((27s^6t^(-2))^(2/3))=((3^2s^(-5)t^4)^(3/2))/((3^3s^6t^(-2))^(2/3)$

$((3^2s^(-5)t^4)^(3/2))/((3^3s^6t^(-2))^(2/3))=(3^3s^(-15/2)t^6)/(3^2s^4t^(-4/3))$

$(3^3s^(-15/2)t^6)/(3^2s^4t^(-4/3))=3^(3-2)s^((-15-8)/2)t^((18-(-4))/3$

$3^(3-2)s^((-15-8)/2)t^((18-(-4))/3)=3s^(-23/2)t^(22/3)$

$3s^(-23/2)t^(22/3)=(3t^(22/3))/s^(23/2)$

Hope it helps :)

How do you simplify \frac { ( 9s ^ { - 5} t ^ { 4} ) ^ { 3/ 2} } { ( 27s ^ { 6} t ^ { - 2} ) ^ { 2/ 3} }\frac { ( 9s ^ { - 5} t ^ { 4} ) ^ { 3/ 2} } { ( 27s ^ { 6} t ^ { - 2} ) ^ { 2/ 3} }?