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

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Answer 1 · 2018-01-30T17:03:28

#y#^(9÷2)#

Consider

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

By the law of indices

#(a^m)^n = a^(mn)#

#y^((9)^(1/2))=y^(9xx1/2)#

#=y^(9/2)#

Consider

#(p^9)^0#

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

#y^((9)^(1/2)) xx (p^9)^0 = y^(9/2) xx 1#

The answer is

#y^(9/2)#

Answer 2 · 2018-01-30T17:07:52

# y ^3#

#(y ^ { 9} ) ^ { \frac { 1} { 2} } ( p ^ { 9} ) ^ { 0}#

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

#=> (y ^ { 9} ) ^ { \frac { 1} { 2} } cdot( 1 ) #

#=> (y ^ ((3)^2))^ { \frac { 1} { 2} } #

#=> (y ^ (3))^{ \frac { 2} { 2} } #

#=> y ^3#