# The expression x 2 n ⋅ x 3 n − 1 x n + 2 using the law of exponents.

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter 1.5, Problem 4E

(a)

To determine

## To simplify: The expression x2n⋅x3n−1xn+2 using the law of exponents.

Expert Solution

The simplified form of the given expression is x2nx3n1xn+2=x4n3.

### Explanation of Solution

Law of exponents:

If x and y are positive numbers and m and n are real numbers, then

a.xm+n=xmxnb.xmn=xmxnc.(xm)n=xmnd.(xy)m=xmyn

Calculation:

Use the law (xy)m=xmyn, and rewrite the given expression as,

x2nx3n1xn+2=x2n+3n1xn+2=x5n1xn+2

Then, use the law xmn=xmxn, and simplify further as follows.

x5n1xn+2=x5n1(n+2)=x5n1n2=x4n3

Thus, the given expression can be simplified as x2nx3n1xn+2=x4n3.

(b)

To determine

### To simplify: The expression abab3 using the law of exponents.

Expert Solution

The expression abab3=a16b112.

### Explanation of Solution

Use the rule xa=x1a, and rewrite the expression as (ab12)12(ab)13.

Then, apply the law (xy)m=xmyn, and rewrite the expression as follows.

(ab12)12(ab)13=(a)12(b12)12(a)13(b)13=a12b14a13b13

Apply the law xmn=xmxn, and simplify further as follows.

a12b14a13b13=a1213b1314=a16b112

Thus, the given expression can be simplified as abab3=a16b112.

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