How do you simplify (y ^ { 9} ) ^ { \frac { 1} { 2} } ( p ^ { 9} ) ^ { 0}?

Jan 30, 2018

$y$^(9÷2)

Explanation:

Consider

y^((9)^((1/2))#

By the law of indices

${\left({a}^{m}\right)}^{n} = {a}^{m n}$

${y}^{{\left(9\right)}^{\frac{1}{2}}} = {y}^{9 \times \frac{1}{2}}$

$= {y}^{\frac{9}{2}}$

Consider

${\left({p}^{9}\right)}^{0}$

Anything raised to the power zero is $1$. Hence

${y}^{{\left(9\right)}^{\frac{1}{2}}} \times {\left({p}^{9}\right)}^{0} = {y}^{\frac{9}{2}} \times 1$

${y}^{\frac{9}{2}}$

Jan 30, 2018

${y}^{3}$

Explanation:

${\left({y}^{9}\right)}^{\setminus \frac{1}{2}} {\left({p}^{9}\right)}^{0}$

According to rules of exponents, any number with an exponent zero gives 1, so ${\left({p}^{9}\right)}^{0} = 1$ .

$\implies {\left({y}^{9}\right)}^{\setminus \frac{1}{2}} \cdot \left(1\right)$

$\implies {\left({y}^{{\left(3\right)}^{2}}\right)}^{\setminus \frac{1}{2}}$

$\implies {\left({y}^{3}\right)}^{\setminus \frac{2}{2}}$

$\implies {y}^{3}$