EVALUATING EXPRESSIONS WITH EXPONENTS

Example 1 :

Evaluate :

(-1)4

Solution :

Order of operations (PEMDAS) dictates that parentheses take precedence.

Here, the exponent 4 is an even number. So, the negative sign inside the parentheses will become positive.

When 1 is multiplied by itself any number of times, the result will be 1.

More clearly,

(-1)= (-1) ⋅ (-1) (-1) ⋅ (-1)

(-1)= 1

So, the value of (-1)4 is 1.

Example 2 :

Evaluate :

(-1)5

Solution :

Order of operations (PEMDAS) dictates that parentheses take precedence.

Here, the exponent 5 is an odd number. So, the negative sign inside the parentheses will remain same.

When 1 is multiplied by itself any number of times, the result will be 1.

More clearly,

(-1)= (-1) ⋅ (-1) ⋅ (-1) ⋅ (-1) ⋅ (-1)

(-1)= -1

So, the value of (-1)5 is -1.

Example 3 :

Evaluate :

-(-3)3

Solution :

Order of operations (PEMDAS) dictates that parentheses take precedence.

So, we have

-[(-3)3] = -[(-3) ⋅ (-3) ⋅ (-3)]

-[(-3)3] = -[-27]

-[(-3)3] = 27

So, the value of (-3)3 is 27.

Example 4 :

Evaluate :

50

Solution :

Anything to the power zero is equal to 1.

So, we have

50 = 1

So, the value of 50 is 1.

Example 5 :

Evaluate :

-30

Solution :

Anything to the power zero is equal to 1.

So, we have

-30 = -1

So, the value of -3is -1.

Example 6 :

Evaluate :

(-7)0

Solution :

Order of operations (PEMDAS) dictates that parentheses take precedence.

Anything to the power zero is equal to 1.

So, we have

(-7)0 = 1

So, the value of (-7)is -1.

Example 7 :

Evaluate :

5-3

Solution :

Using laws of exponents, we have

5-3 = 1/53

5-3 = 1/125

So, the value of 5-3 is 1/125.

Example 8 :

Evaluate :

23 ⋅ 3⋅ (-1)5

Solution :

In the above expression, first evaluate each term separately.

23 = 2 ⋅ 2 ⋅ 2 = 8

32 = 3 ⋅ 3 = 9

(-1)= (-1) ⋅ (-1) ⋅ (-1) ⋅ (-1) ⋅ (-1) = -1

Now, we have

23 ⋅ 3⋅ (-1)5 = 8 ⋅ 9 ⋅ (-1)

23 ⋅ 3⋅ (-1)5 = - 72

So, the value 23 ⋅ 3⋅ (-1)5 is -72.

Example 9 :

Evaluate :

(-1)4 ⋅ 3⋅ 22

Solution :

In the above expression, first evaluate each term separately.

(-1)= (-1) ⋅ (-1) ⋅ (-1) ⋅ (-1) = 1

33 = 3 ⋅ 3 ⋅ 3 = 27

22 = 2 ⋅ 2 = 4

Now, we have

(-1)4 ⋅ 3⋅ 2= 1 ⋅ 27 ⋅ 4

(-1)4 ⋅ 3⋅ 2= 108

So, the value (-1)4 ⋅ 3⋅ 22 is 108.

Example 10 :

If x2y3  =  10 and x3y2  =  8, then find the value of x5y5.

Solution :

x2y3 = 10 ----(1)

x3y2 = 8 ----(2)

Multiply (1) and (2) :

(1) ⋅ (2) ----> (x2y3) ⋅ (x3y2) = 10 ⋅ 8

x5y5 = 80

So, the value x5y5 is 80.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. SAT Math Resources (Videos, Concepts, Worksheets and More)

    Dec 23, 24 03:47 AM

    SAT Math Resources (Videos, Concepts, Worksheets and More)

    Read More

  2. Digital SAT Math Problems and Solutions (Part - 91)

    Dec 23, 24 03:40 AM

    Digital SAT Math Problems and Solutions (Part - 91)

    Read More

  3. Digital SAT Math Problems and Solutions (Part - 90)

    Dec 21, 24 02:19 AM

    Digital SAT Math Problems and Solutions (Part - 90)

    Read More