How do you get rid of negative exponents in #(2y ^ { - 1} ) ( 3y ) ( 6y ^ { - 1} )#?

1 Answer
smendyka
Apr 11, 2017

See the entire solution process below:

Explanation:

Use this rule of exponents to eliminate the negative exponents:

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

#(2y^color(red)(-1))(3y)(6y^color(red)(-1)) = (2/y^color(red)(- -1))(3y)(6/y^color(red)(- -1)) =#

#(2/y^1)(3y)(6/y^1)#

If you want to also simplify this expression, first rewrite it as:

#(2 * 3 * 6)(1/y^1 * y * 1/y^1) = 36(1/y^1 * y * 1/y^1)#

Now, use these rules of exponents to complete the simplification:

#a^color(red)(1) = a# or #a = a^color(red)(1)#

#36(1/y^1 * y * 1/y^1) = 36(1/y^1 * y^color(red)(1) * 1/y^1) = 36(1/color(blue)(cancel(color(black)(y^1))) * color(blue)(cancel(color(black)(y^color(red)(1)))) * 1/y^1) =#

#36 * 1/y^color(red)(1) = 36/y#

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