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(y^9)(2y^2)^3

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2 Answers

Recall three of the laws of exponents I assume your teacher must have covered in your class:

i) xmxn = xm+n

ii) (xm)n = xmn

iii) (xy)n = xnyn

These two laws can be used here:

y9(2y2)3 = 23y9y2*3 = 8y9y6 = 8y9+6 = 8y15

 

Audy,

(y^9) the first term is y to the 9th power

(2y^2) is (y squared ) times 2.  Note that the exponent 2 does not apply to the factor 2.

(2y^2)^3 is the previous term raised to the 3rd power.  To see exactly how this calculates expand it:

   (2y^2)^3  =  (2y^2) *  (2y^2)  * (2y^2)

                   = 4*y^4  *  (2y^2)

                  =  8*y^6

  Go back and include the first term:

        (y^9)(2y^2)^3  =  (y^9) * 8*y^6

        (y^9)(2y^2)^3 = 8*y^15

There are two keys to this problem:

  Exponents add when their base numbers are multiplied.

  Observe which factors are raised to powers and which are not.

BruceS