# given expression ( s − 2 t 2 ) 2 ( s 2 t ) 3 .

### Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

### Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

#### Solutions

Chapter 1.2, Problem 44E

(a)

To determine

Expert Solution

## Answer to Problem 44E

(s2t2)2(s2t)3=s2t7

### Explanation of Solution

Given information:

Given expression

(s2t2)2(s2t)3

Calculation:

Consider the given expression.

(s2t2)2(s2t)3

Now. Simplify using law of exponents (ab)n=anbn and (am)n=amn .

(s2t2)2(s2t)3={(s2)2(t2)2}{(s2)3(t)3}=(s4t4)(s6t3)

To simplify this, rewrite the above expression as a single product.

(s4t4)(s6t3)=s4s6t4t3

Now, use the first law of exponents, which states

aman=am+n

And combine like terms

s4s6t4t3=s4+6t4+3=s2t7

Hence,

(s2t2)2(s2t)3=s2t7

(b)

To determine

Expert Solution

## Answer to Problem 44E

(2u2v3)3(3u3v)3=216v12u3

### Explanation of Solution

Given information:

Given expression

(2u2v3)3(3u3v)3

Calculation:

Consider the given expression.

(2u2v3)3(3u3v)3

Now. Simplify using law of exponents (ab)n=anbn and (am)n=amn .

(2u2v3)3(3u3v)3={23(u2)3(v3)3}{33(u3)3(v)3}=(23u6v9)(33u9v3)

To simplify this, first rewrite the above expression as a single product.

(23u6v9)(33u9v3)=2333u6u9v9v3=216u6u9v9v3

Now, use the first law of exponents, which states

aman=am+n

And combine like terms

216u6u9v9v3=216u69v9+3=216u3v12

Hence,

(2u2v3)3(3u3v)3=216v12u3

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