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Intermediate Algebra: Week 3 Discussion

Question 1

10002/3

This problem will be solved using power exponent rule

The first step will be rewriting the number 1000 as a prime to a power.

1000 = 103

Thus the new expression will be (103)2/3

Using power rules the outside exponent will be multiplied with the inner exponent c as follows

3 * 2/3 = 2

Thus, the new exponent will be 2 and the new expression will be

10 = 10 *10 =100

Question 2

(54/46) ½

This problem will be solved using power exponent rule. The power rule will be used in handling the outside component. The inner exponents will be multiplied with outside exponents as follows:

4* ½ and 6 * ½ . The new exponents will be 2 and 3 respectively

The new expression will be 52/63

The numerator will be squared; 5 x 5 = 25 while the denominator will be cubed to get 6*6*6 =216

The new fraction will be:

52/63 = 25/216

 

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