How do you simplify #(16^(5/9) * 5 ^ (7/9)) ^-3#?

1 Answer
Mar 4, 2018

Answer:

See a solution process below:

Explanation:

First, use this rule for exponents to remove the outer exponent:

#(x^color(red)(a))^color(blue)(b) = x^(color(red)(a) xx color(blue)(b))#

#(16^color(red)(5/9)5^color(red)(7/9))^color(blue)(-3) =>#

#16^(color(red)(5/9) xx color(blue)(-3))5^(color(red)(7/9) xx color(blue)(-3)) =>#

#16^color(red)(-5/3)5^color(red)(-7/3)#

Next, we can use this rule of exponents to eliminate the negative exponents:

#x^color(red)(a) = 1/x^color(red)(-a)#

#1/(16^color(red)(- -5/3)5^color(red)(- -7/3)) =>#

#1/(16^(5/3)5^(7/3))#

Depending on how you are being requested to simplify the expression you can use this rule of exponents and radicals to write the expression in radical form:

#x^(1/color(red)(n)) = root(color(red)(n))(x)#

#1/(16^(5 xx 1/3)5^(7 xx 1/3)) =>#

#1/((16^5)^(1/3)(5^7)^(1/3)) =>#

#1/(root(3)(16^5)root(3)(5^7)) =>#

#1/(root(3)(16^3 * 16^2)root(3)(5^6 * 5)) =>#

#1/(root(3)(16^3)root(3)(16^2)root(3)(5^6)root(3)(5)) =>#

#1/(16root(3)(16^2)5^2root(3)(5)) =>#

#1/(16 xx 25root(3)(16^2)root(3)(5)) =>#

#1/(400root(3)(16^2)root(3)(5)) =>#

#1/(400root(3)(256)root(3)(5)) =>#

#1/(400root(3)(256 xx 5)) =>#

#1/(400root(3)(1280))#