Applied Calculus for the Managerial, Life, and Social Sciences
Applied Calculus for the Managerial, Life, and Social Sciences
10th Edition
ISBN: 9781305887831
Author: Tan
Publisher: CENGAGE CO
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Textbook Question
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Chapter DT, Problem 1E
  1. a. Evaluate the expression:

    (i) ( 16 9 ) 3 / 2

    (ii) 27 125 3

  2. b. Rewrite the expression using positive exponents only: ( x 2 y 1 ) 3

a.

Expert Solution
Check Mark
To determine

To evaluate: The expression (169)32.

Answer to Problem 1E

The value of the expression (169)32 is 6427_.

Explanation of Solution

Result used:

Laws of Exponents:

For any real numbers a, b and n(ab)n=anbn,(b0).

Calculation:

The given expression is (169)32.

Use Exponent law to evaluate the expression.

(169)32=1632932=161699=16493=6427

Thus, the value of the expression (169)32 is 6427_.

Expert Solution
Check Mark
To determine

To evaluate: The expression 271253.

Answer to Problem 1E

The value of the expression 271253 is 35_.

Explanation of Solution

The given expression is 271253.

Use Exponent law to evaluate the expression.

271253=(27125)13=271312513=35

Thus, the value of the expression 271253 is 35_.

b.

Expert Solution
Check Mark
To determine

To rewrite: The expression (x2y1)3 by using exponents.

Answer to Problem 1E

The expression (x2y1)3 can be rewritten as 1x6y3_.

Explanation of Solution

Negative exponent:

If n is a positive integer, then an=1an,a0.

Calculation:

The given expression is (x2y1)3.

Use negative exponent law to evaluate the expression.

(x2y1)3=(x2)3(y1)3=x(2)(3)y(1)(3)=x6y3=1x61y3         (Use Negative exponent law)

That is, (x2y1)3=1x6y3.

Thus, the value of the expression (x2y1)3 is 1x6y3_.

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