# To prove: the given Laws of exponents for the case in which m and n are positive integers and m &gt; n .

### 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 94E

(a)

To determine

## To prove: the given Laws of exponents for the case in which m and n are positive integers and m>n .

Expert Solution

### Explanation of Solution

Given information:

Law 2

Calculation:

Consider two positive numbers m and n such that m is greater than n .

Then, prove:

aman=amn

First, rewrite aman as a product and apply the law of negative exponents to obtain:

aman=am1an=aman

Since, am and an have a common base a , apply the additive law of exponents and use the fact that n is a positive number to simplify and obtain the desired result:

aman=am+(n)=amn

(b)

To determine

Expert Solution

### Explanation of Solution

Given information:

Law 5

Calculation:

Consider, n is a positive number,

Then prove:

(ab)n=anbn

First rewrite ab as a product within the expression (ab)n

(ab)n=(ab1)n

The law for powers of a product states that for ant two numbers x and y .

(xy)n=xnyn

Now, evaluate the expression (ab1)n using the law for powers of a product and apply the multiplicative law for exponents to (b1)n .

(ab1)=an(b1)n=anbn

Finally, rewrite bn using the law for negative exponents to obtain the desired result.

anbn=an1bn=anbn

(c)

To determine

Expert Solution

### Explanation of Solution

Given information:

Law 6

Calculation:

Consider, n is a positive number,

Then prove:

(ab)n=(ba)n

First rewrite the expression (ab)n as a product of two exponents.

(ab)n=((ab)1)n

Therefore, for any two numbers x,y, and c ,

(xy)c=xcyc

Apply the law to the expression (ab)1 to obtain.

((ab)1)n=(a1b1)n

Rewrite as a product and apply the law of negative exponents to a1 and the multiplicative law of exponents to (b1)1 , giving us the desired result.

(a1b1)n=(a1(b1)1)n=(a1b)n=(ba)n

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