Question

Yeah uh I have no idea how to do this.....

Use exponent laws to simplify this :
9√(a^-5 b^2)

The answer is 9b / a^5/2 I just don't know how so please show steps! Thank you

Answer 1

`(9b)/(a^(2) )`
`\sqrt{(1)/(a) }` should be the answer of the problem that you posted

if you try to solve using
`√(x)` as as
`x^{(1)/(2) }`

then
`(a^(-5)) ^{(1)/(2) } \\` =
`(a^(-5/2))` =
`\frac{1}{a^{(5)/(2) } }`

`\frac{1}{a^{(5)/(2) } }` * 9b =
`\frac{9b}{a^{(5)/(2 ) } }`

Explanation:

`9 √((a^-5 b^2))`

`x^(-1) = (1)/(x)`

`a^(-5) = (1)/(a^(5) )`

`\sqrt{b^(2) } = b`

`a^(4)` a's can be removed from the radicle (one will be left in because you have a^5)

`(9b)/(a^(2) )`
`\sqrt{(1)/(a) }`

Answer 2

9b /a^5/2

Explanation:

9√(a^-5 b^2)

9 sqrt( a^ -5 b^2)

Rewriting sqrt as ^1/2

9 ( a^ -5 b^2)^1/2

9 ( a^ -5) ^1/2 ( b^2)^1/2

we know that an exponent to an exponent means multiply

9 a^ (-5*1/2) b^(2*1/2)

9 a^-5/2 b^1

9b a^ -5/2

We know that x^-y = 1/x^y

9b /a^5/2