Question

9^-1/2 and 27^-2/3 what is this

Answer 1

Answer:

9^(-1/2) = 1/3

27^(-2/3) = 1/9

Step-by-step explanation:

We know that:

$\displaystyle \large {{a}^{ - n} = \frac{1}{ {a}^{n} } } \\ \displaystyle \large { {a}^{ \frac{m}{n} } = \sqrt[n]{ {a}^{m} } }$

If we add negative sign in m/n.

$\displaystyle \large{ {a}^{ - \frac{m}{n} } = \frac{1}{ {a}^{ \frac{m}{n} } } }$

Therefore,

$\displaystyle \large{ {a}^{ - \frac{m}{n} } = \frac{1}{ \sqrt[n]{ {a}^{m} } } }$

From the expression:

1. 9^(-1/2)

$\displaystyle \large{ {9}^{ - \frac{1}{2} } = \frac{1}{ \sqrt{9} } }$

From above, apply the exponent rules. Simplify the expression:

$\displaystyle \large{ {9}^{ - \frac{1}{2} } = \frac{1}{ 3 }}$

2. 27^(-2/3)

$\displaystyle \large{ {27}^{ - \frac{2}{3} } = \frac{1}{ \sqrt[3]{ {27}^{2} } } }$

27 can be factored as 3^3.

$\displaystyle \large{ {27}^{ - \frac{2}{3} } = \frac{1}{ \sqrt[3]{ { ({3}^{3}) }^{2} } } }$

From above:- Use the following exponent rules to multiply the exponents and leave the base as 3 so we can cancel the cube (3 out of the root) and the exponent.

$\displaystyle \large{ {27}^{ - \frac{2}{3} } = \frac{1}{ \sqrt[3]{ {3}^{6} } }}$

Divide 6 by 3 or the cube root.

$\displaystyle \large{ {27}^{ - \frac{2}{3} } = \frac{1}{ {3}^{2} }}$

Therefore:

$\displaystyle \large{ {27}^{ - \frac{2}{3} } = \frac{1}{ 9}}$

Answer 2

Answer:

Multiply by the exponents

Step-by-step explanation:

9^-1/2=1/3

27^-2/3=1/9