## Wednesday, May 30, 2012

When given the lenghts of all 3 sides of a triangle, how do we determine if the triangle is right, acute or obtuse?

Recall that for a right triangle with sides of length a,b and c (with c being the hypotenuse),
a^2 + b^2 = c^2

So a triangle with sides 7, 24 and 25 is a right triangle

7^2 + 24^2 = 25^2
49 + 576 = 625

If a^2 + b^2 < c^2, the triangle is obtuse.

A triangle with sides 4, 6 and 10 is an example of an obtuse triangle

4^2 + 6^2 < 10^2

16 + 36 < 100

52 < 100

If a^2 + b^2 > c^2, the triangle is acute.

A triangle with sides 7, 8, 9 is an acute triangle

7^2 + 8^2 > 9^2

49 + 64 > 81

113 > 81

## Saturday, May 26, 2012

Suppose you wish to find the asymptotes of the following:

2x/(3x + 2)

To find the vertical asymptote, find where the denominator is undefined for x. Set the denominator equal to zero.

3x + 2 = 0

x = -2/3

Therefore the vertical asymptote is the line x = -2/3

To find the horizontal asymptote, divide the coefficients of the x term (numerator/denominator). Notice the coefficients are 2 and 3.

The horizontal aymptote is the line y = 2/3.

The graph will approach but never touch the asymptotes.

## Thursday, May 24, 2012

Some basic trigonometric identities are as follows:

sin^2(x) + cos^2(x) = 1

1 + tan^2(x) = sec^2(x)

1 + cot^2(x) = csc^2(x)

Recall that

sin(x) = 1/csc(x)

cos(x) = 1/sec(x)

tan(x) = 1/cot(x)

The trigonometric functions and identities and can derived from the basic trigonometric functions sin(x) and cos(x).

For example:

therefore tan(x) = sin(x)/cos(x)

1 + tan^2(x) = sec^2(x) using sin(x) and cos(x)

1+ sin^2(x)/cos^2(x) = 1/cos^2(x)

[cos^2(x) + sin^2(x)]/cos^2(x) = 1/cos^2(x)

multiply both sides by cos^2(x) to get

cos^2(x) + sin^2(x) = 1, which confirms the first identity.

## Thursday, May 17, 2012

Suppose we wish to solve the following problem:

2x/(x + 1) + 3/x = 12

First we need to get a common denominator.  Notice the factors in the denominators are (x+1) for the first term and x for the second term.  We multiply the factors to get x(x+ 1) for the common denominator.

The numerator and denominator of the first term is multiplied by x and the numerator and denominator of the second term is multiplied by (x + 1).  The 12 is multiplied by x(x + 1)/x(x + 1).

This gives us

2x(x)/x(x+1) + 3(x+ 1)/x(x + 1) = 12x(x + 1)/x(x + 1)

When solving this, we can ignore the denominators. The resason for this is since all the denominators are the same, we can multiply the entire equation by x(x + 1), which cancels the denominators

This leaves us with

2x(x) + 3(x + 1) = 12x(x + 1)

2x^2 + 3x + 3 = 12x^2 + 12x

0 = 10x^2 + 9x - 3

x = [(-9 +/- sqrt(81- 4(10)(-3))]/2(10)

x = [-9 +/- sqrt(201)]/20

Solving for x, we get x is approximately 0.259 or -1.16

## Monday, May 14, 2012

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Here's part of the second book, chapter on sequences and series.

# Chapter 7

## Sequences and Geometric Applications

When we think of a sequence, we think of an ordered list. For example, a history teacher might discuss a sequence of battles during World War II. A baseball historian might discuss the sequence of events leading up to the first organized professional baseball league. In terms of mathematics, a sequence refers to a list of numbers written in a specific order.

Sequences can either be finite, which contains a set number of terms, or infinite, where the sequence continues on forever. A finite sequence is a function whose domain is the set of all natural numbers for a specified number n. An infinite sequence is a function whose domain is the set of all natural numbers. An easier way to denote terms of a sequence is to use an , which denotes the nth term of the sequence. Similarly a1 denotes the first term in the sequence, a2 denotes the second term of the sequence and so on.

Some examples of finite and infinite sequences are as follows:

Finite: 2, 5, 8, 11, 14, 17, 20, 23, 26

1, -1, -3, -5, -7, -9, -11, -13, -15

Infinite: 1, 2, 3, 4, 5, 6, 7, …....

4, 8, 12, 16, 20, 24, 28, …...

Consider the following infinite sequence:

2, 8, 32, 128, 512, …..

We can define the nth term of the sequence by finding the relationship between the terms. Notice a1 = 2 and a2 is 4 times a1. It is also 6 more than a1 but that is not the pattern throughout the sequence. If it was the pattern, a3 would be 8 + 6 = 14, but the third term is 32, which is 4 times 8. The formula holds true for the next term as well since 32 times 4 equals 128. Therefore the formula for the nth term in the sequence is an = 4an-1

When we apply the formula we just got for an , we obtain the first 5 numbers in the series and therefore we know our formula is correct.

a1 = 2

a2 = 4(a1) = 4(2) = 8

a3 = 4(a2) = 4(8) = 32

a4 = 4(a3) = 4(32) = 128

a5 = 4(a4) = 4(128) = 512

Example: Find the first 6 terms and the 46th term of the infinite sequence with an = 5n – 11.

To find the first 6 terms of the sequence, we need to substitute 1, 2, 3, 4, 5 and 6 for n in the formula. Recall that the first 6 terms of the sequence are denoted as a1, a2, a3, a4, a5 and a6.

Therefore,

a1 = 5(1) – 11 = 5 – 11 = -6

a2 = 5(2) – 11 = 10 – 11 = -1

a3 = 5(3) – 11 = 15 – 11 = 4

a4 = 5(4) – 11 = 20 – 11 = 9

a5 = 5(5) – 11 = 25 – 11 = 14

a6 = 5(6) – 11 = 30 – 11 = 19

Therefore, the first 6 terms of the sequence are -6, -1, 4, 9, 14, 19

Note the pattern is each term is 5 greater than the previous term. Once that pattern has been established, we can just add 5 to the previous term and don't have to perform each calculation. In the next part where we have to find the 46th term, it's easier to perform the calculation instead of adding 5 each time until we reach the 46th term.

To find the 46th term of the sequence, we need to substitute 46 for n in the formula.

Therefore, a46 = 5(46) – 11 = 230 -11 = 219.

### Arithmetic Sequence

In another example, suppose the sequence is 2, 8, 14, 20, 26, 32, 38, ….... Notice that each term is 6 more than the previous term. A sequence such as this is known as an arithmetic sequence because each term is found by adding the same number to the previous term. An arithmetic sequence is a sequence in the form

a1., a1 + d, a1 + 2d, a1 + 3d, a1 + 4d, …... a1 + (n – 1)d, where a1 is the first term of the sequence and d is the common difference between the terms.

The nth term of an arithmetic sequence is given by the formula an = a1 + (n – 1)d.

Consider the following arithmetic sequences below. Notice the common difference by subtracting the first term from the second, the second term from the third and so on.

Finite arithmetic sequence: 3, 10, 17, 24, 31, 38

The common difference d is 7 since 10 – 3 = 7, 17 – 10 = 7, 24 – 17 = 7, 31 – 24 = 7 and 38 – 31 = 7.

Infinite arithmetic sequence: 6, 3, 0, -3, -6, -9, ….....

The common difference d is -3 since 3 - 6 = -3, -3 - 0 = -3, -6 - (-3) = -3, -9 - (-6) = -3 and so on.

Example: In an arithmetic sequence, the first term is 8 and the common difference is -2. What are the first 6 terms of the sequence and what is the 29th term?

To solve this problem we start with 8 and add the common difference to each term thereafter until we get all 6 terms.

Therefore,

a1 = 8, a2 = 8 – 2 = 6, a3 = 6 – 2 = 4, a4 = 4 – 2 = 2, a5 = 2 – 2 = 0, a6 = 0 – 2 = -2

The first 6 terms in the series are 8, 6, 4, 2, 0, -2.

To find the 29th term in the sequence, we use the formula to find the nth term in a sequence. Substitute 29 for n in the formula an = a1 + (n – 1)d to get

a29 = 8 + (29 – 1)(-2)

a29 = 8 + (28)(-2)

a29 = 8 – 56

a29 = -48

Example: The first 4 terms of an arithmetic sequence are 4, 10, 16 and 22. What is the 95th term?

To solve this problem we have to find the common difference between each term. Since a1 = 4 and a2 = 10, the common difference is 10 – 4 = 6.

Now we use the formula an = a1 + (n – 1)d and substitute 95 for n and 6 for d and solve for an to get

a95 = 4 + (95 – 1)(6)

a95 = 4 + 94(6)

a95 = 4 + 564

a95 = 568

Sometimes we don't know the first several terms of a sequence. Maybe we know the first term and the 30th term and we need to find the 15th term. How do we approach such a problem?

First, we have to determine the common difference by substituting the value of the 30th term for an, the value of the first term for a1 and 30 for n into the formula an = a1 + (n – 1)d.

Example: The first term of an arithmetic sequence is 6 and the 45th term is 490. What are the first 5 terms of the sequence?

To solve this problem we need to find the common difference. We use the formula an = a1 + (n – 1)d. Since the 45th term is 490, we substitute 490 for an . Since the first term is 6, we substitute 6 for a1 and 45 for n to get

490 = 6 + (45 – 1)d

490 = 6 + 44d

484 = 44d

11 = d

Since the common difference is 11, the first 5 terms of the sequence are 6, 17, 28, 39, 50 and are determined as follows

a1 = 6

a2 = 6 + 11 = 17

a3 = 17 + 11 = 28

a4 = 28 + 11 = 39

a5 = 39 + 11 = 50

___________________________________________________________________________________________________

Suppose we know two terms of an arithmetic sequence and we wish to insert numbers between them. The numbers are known as arithmetic means. If a single number is inserted between the two numbers, it's known as the arithmetic mean between the two numbers.

For example, the numbers 4 and 7 are the arithmetic means between 1 and 10 in the sequence 1, 4, 7, 10. The common difference between the terms is 3. The number 6 is the arithmetic mean between 3 and 9 in the sequence 3, 6, 9. The common difference between the terms is 3.

How do we determine the arithmetic means? Consider the following examples.

Example: Find two arithmetic means between 8 and 23.

To solve this problem, we have to determine how many terms there will be and then find the common difference between the terms.

We know a1 = 8 but we don't know a2 or a3. We know that a4 = 23. There is a jump from a1 to a2, from a2 to a3 and from a3 to a4. Therefore there are 3 equal distances going from 8 to 23. The total distance from 8 to 23 is 15 since 23 – 8 = 15. Now we take 15 divided by 3 to get 5. Therefore the common difference is 5. The arithmetic means, a2 and a3 are then

a2 = a1 + 5 = 8 + 5 = 13, a3 = a2 + 5 = 13 + 5 = 18

## Friday, May 11, 2012

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## Wednesday, May 9, 2012

Here's part of chapter 2 from my second book, Algebra Simplified Intermediate & Advanced. If interested in the book, go to http://www.lulu.com/spotlight/KKauffman1969

# Chapter 2

## Square Roots, Cube Roots and Higher Roots

Recall when raising a number to a power n, where n is an integer greater than 1, we multiply the number by itself n times. For example, 43 = 4 ∙ 4 ∙ 4. Now suppose we want to know what number multiplied by itself 2 times equals 169. Problems of this kind can be represented using radicals.

A radical symbol √ is used to show the square root, or principal square root of a number or expression that appears under the radical symbol. Recall that the square root is defined as a number or expression multiplied by itself twice to equal the number or expression under the radical symbol, known as the radicand.

For example, if we want to know what number multiplied by itself 2 times equals 169, we can set this up with the radical symbol as follows:

169, read as “square root of 169”. The answer to this is 13.

- √169, read as “negative square root of 169”. The answer to this is -13.

0.09 = 0.3 and -0.3 since (0.3)2 and (-0.3)2 equals 0.09. The principal square root is 0.3. Another way to simplify this is to change √0.09 to √(9/100) and simplify to 3/10.

(25/49) = 5/7 and -5/7 since (5/7)2 and (-5/7)2. The principal square root is 5/7.

Note that a square root also has a negative value since a negative times a negative equals a positive, but we will deal with only the principal square root unless otherwise noted.

Note that you can also simplify the square root of a fraction by taking the square root of the numerator and then the square root of the denominator instead of the square root of the fraction as a whole. In the previous example, you can take the square root of 25 first, then the square root of 49.

Note that the square root of many positive integers are not whole numbers or rational numbers. For example, √19 can be found on a calculator or by leaving the answer as √19.

An easy way to solve many square root problems is to know the perfect squares from 1 to 25. They are as follows:

12 = 1 102 = 100 192 = 361

22 = 4 112 = 121 202 = 400

32 = 9 122 = 144 212 = 441

42 = 16 132 = 169 222 = 484

52 = 25 142 = 196 232 = 529

62 = 36 152 = 225 242 = 576

72 = 49 162 = 256 252 = 625.

82 = 64 172 = 289

92 = 81 182 = 324

Sometimes we have to find the square root of a number that is not a perfect square. In these cases, we break down the radicand into factors, one of which is a perfect square.

Examples: Find each square root.

1. √68

First, find factors of 68.

Since 68 is even, we can divide it by 2. Therefore, 68 = 2 ∙ 34. Notice 34 is also divisible by 2, therefore 34 = 2 ∙ 17.

So 68 is factored into 2 ∙ 2 ∙ 17. Notice that 17 is prime and cannot be factored further and 2 ∙ 2 = 4, which is a perfect square. Therefore √68 = √4 ∙ √17 = 2√17.

2. √108

First, find factors of 108.

Since 108 is even, we can divide it by 2. Therefore 108 = 2 ∙ 54. Notice 54 is also divisible by 2, therefore 54 = 2 ∙ 27. Next, we know that 27 = 3 ∙ 3 ∙ 3.

The factors of 108 are 2 ∙ 2 ∙ 3 ∙ 3 ∙ 3. Notice 2 ∙ 2 = 4, which is a perfect square and 3 ∙ 3 = 9, which is also a perfect square. Therefore, 108 = 4 ∙ 9 ∙ 3 and √108 = √4 ∙ √9 ∙ √3 = 2 ∙ 3 ∙ √3 = 6√3.

Note that 108 = 36 ∙ 3 and 36 is a perfect square. But if you can't see right away that 3 is a factor of 108, you can break down by dividing 108 by 2 first and then simplify further at that point. It's easy to determine that 108 is divisible by 3. If the sum of the digits of a number are divisible by 3, the number is divisible by 3.

Recall, for a variable x, where x ≠ 0, √x2 = x and -x. If we are dealing with just the principal square root, then the answer is |x|. Recall that |x| is the absolute value of x.

Examples: Find the square root of the following.

a. √(25x2)

Simplify the numerical part and the variable part separately.

25 = 5 since 52 = 25 and √x2 = x since (x)2 = x2.

Therefore, √(25x2) = 5x.

b. √(64x4y6)

64 = 8 since 82 = 64, √x4 = x2 since (x2)2 = x4 and √y6 = y3 since (y3)2 = y6.

Therefore, multiply the terms in bold to get √(64x4y6) = 8x2y3.

c. √(x + 4)2 = (x + 4) because (x + 4)(x + 4) = (x + 4)2

Note that the square root of any number or expression squared is just that number. For example, notice in the previous examples that √(x + 4)2 = (x + 4) and √25 = √(5)2 = 5.

d. √(x2 + 12x + 36) = √(x + 6)(x + 6)

Notice that we factored (x2 + 12x + 36) to get (x + 6)(x + 6). That is a perfect square which enables us to simplify the radical expression.

(x2 + 12x + 36) = √(x + 6)(x + 6) = √(x + 6)2 = (x + 6)

Note that you can check by squaring your answer. After squaring, you should get the expression under the √.

In the previous examples, we simplified radicals that were perfect squares. Many times we can simplify radicals that are not perfect squares. The idea is to break down the number or variable into factors, one of which is a perfect square. By the  multiplication property of radicals n√(ab) = na nb, where n√a and nb are real numbers.

Note that in a square root n is 2 but is not written. When solving any root higher than 2, it is noted in the upper left of the radical. For example, the cube root (n = 3) of a is noted as 3√a.

The cube root of a number n is the number when multiplied by itself 3 times equals n. For example, the cube root of -125 is -5 because (-5) ∙ (-5) ∙ (-5 ) = -125, or (-5)3 = -125. By definition, the cube root of a is defined as 3a = b, if b3 = a.

Examples:

3√27 = 3 since 33 = 27

3√(-27) = -3 since (-3)3 = -27

3√125 = 5 since 53 = 125

3√(-125) = -5 since (-5)3 = -125

Note that 27 and 125 have 1 real number cube root and 2 real number square roots. Any negative number will have a cube root that is negative and any positive number will have a cube root that is positive.

Examples: Simplify the following.

a. 3√-8

To solve this, think of what number multiplied by itself 3 times gives you -8. Notice that (-2) ∙ (-2) ∙ (-2) = -8. Therefore, 3√-8 = -2.

b. 3√343 = 7 since 7 ∙ 7 ∙ 7 = 343.

c. 3√729 = -9

Note that if the sum of the digits of a number is divisible by 9, the number is divisible by 9.

The sum of the digits of 729 is 18, which is divisible by 9. Therefore, 729 is divisible by 9 and -729 is divisible by 9.

729/9 = 81 and we know the square root of 81 is 9.

We reviewed how to take the cube root of positive and negative numbers. Now we will review how to take the cube root involving variables. Recall that by definition, if x is a variable representing a real number, 3x3 = x.

Examples: Simplify the following.

a. 3√(27x3) = 3√(27) ∙ 3√(x3) = 3 ∙ x = 3x

Check by taking the cube of 3x.

(3x)3 = 3x ∙ 3x ∙ 3x = 27x3.

b. 3√(5x + 2y)3 = (5x + 2y) because of the rule that 3x3 = x.

c. 3√(-216x6y12)

3√(-216) = -6, 3√(x6) = x2 since (x2)3 = x6 and 3√(y12) = y4 since (y4)3 = y12.

Multiply the terms in bold to get -6x2y4.

The previous examples were perfect cubes, but oftentimes we can find the cube root of expressions that are not perfect cubes. The idea is to break down the number or variable into factors, one of which is a perfect cube. For example, if you solve 3√81, break it down into 3√27 ∙ 3√3 because 3√27 = 3. If you solve 3√(y5), break it down into 3√(y3) ∙ 3√(y2) because 3√(y3) = y. Remember from the multiplication property of radicals that 3ab = 3a3b.

Example: Simplify the following.

3√(16x4)

3√16 = 3√8 ∙ 3√2 = 23√2 (3√8 = 2 because 23 = 8)

3√(x4) = 3√(x3) ∙ 3x = x3x (3√(x3) = x by the definition of cube root (x)3 = x3)

Multiply like terms to get 3√(16x4) = 2x3√(2x)

Note that you can check by cubing the answer, [2x3√(2x)]3 = 8x3 ∙ 2x = 16x4.

Example: Simply the following.

3√(-192x4y5)

3√(-192) = 3√(-64) ∙ 3√(3) = -43√(3) (note that -64 ∙ 3 = -192 and (-4)3 = -64)

3√(x4) = 3√(x3) ∙ 3√(x) = x3√(x) (note that x3x = x4 and (x)3 = x3)

3√(y5) = 3√(y3) ∙ 3√(y2) = y 3√(y2) (note that y3y2 = y5 and (y)3 = y3)

Now multiply the like terms to get -4xy3√(3xy2).

In some cases cube roots will be the form of a fraction of two numbers, a number and a expression or two expressions. In these cases, try to simplify the ratio first. By the division property of radicals 3√(x/y) = 3x / 3y.

Example: Simplify the following.

3√(8x3/64y6)

By the division property of radicals 3√(8x3/64y6) = 3√(8x3) / 3√(64y6)

3√(8x3) = 2x because (2x)3 = 8x3

3√(64y6) = 4y2 because (4y2)3 = 64y6

Therefore 3√(8x3/64y6) = (2x)/(4y2) = x/(2y2).

Example: Simplify the following.

3√(216y5/8y8)

By the division property of radicals 3√(216y5/8y8) = 3√(27/y3) because 216/8 = 27 and (y5)/(y8) = 1/(y3).

3√(27) = 3 and 3√(1/y3) = 1/y.

Therefore 3√(216y5/8y8) = 3/y.

Note that it's easier to solve cube roots if you know the first several perfect cubes. The first ten perfect cubes are 1, 8, 27, 64, 125, 216, 343, 512, 729 and 1000.

We reviewed how to take the square root and cube root of numbers and expressions. There are also fourth roots, fifth roots, sixth roots, seventh roots and so on. The procedure for solving these roots is the same as for square roots and cube roots. For example, a fourth root of a natural number x is a number multiplied by itself four times to equal x, and so on for higher roots. If n is an odd natural number greater than 1 (n > 1), then n√(x) is an odd root. When n is an even natural number greater than 1 (n > 1) and x > 0, then n√(x) is an even root.

Examples: Simplify the following radical expressions.

a. 4√10000

4√10000 = 10 since 104 = 10 ∙ 10 ∙ 10 ∙ 10 = 10000

b. 5√(32x15)

5√32 = 2 because 25 = 32

5√(x15) = x3 because (x3)5 = x15

Therefore 5√(32x15) = 2x3

c. 6√(729x6y18)

6√729 = 3 because 36 = 729

6√(x6) = x because (x)6 = x6

6√(y18) = y3 because (y3)6 = y18

Therefore 6√(729x6y18) = 3xy3

d. 7√(x + 10)14

7√(x + 10)14 = (x + 10)2 because [(x + 10)2]7 = (x + 10)14

### Review Problems: Set 1

Simplify each of the following square roots.

1. √121 2. √(4/25) 3. -√0.49

4. √60 5. √125 6. √252

Simplify each of the following radical expressions.

7. √(9y2) 8. √(64x4) 9. √(x2 + 14x + 49)

10. √(y2 – 8y + 16) 11. √(81x8y6) 12. √(144x4y2z6)

13. √(80x5) 14. √(75xy2z3) 15. √(98z3)

Simplify each of the following cube roots.

16. 3√-64 17. 3√(27/125) 18. -3√-216

19. 3√40 20. 3√320 21. 3√-54

Simplify each of the following radical expressions.

22. 3√(8x4y5z6) 23. 3√(-108z4) 24. 4√256

25. 5√(-243a5) 26. 6√(32x12) 27. 7√(107)

### Rational Exponents

Just as we can raise a number or expression to exponents that are integers, we can raise them to fractional exponents as well. If n is a positive integer greater then 1, then x1/n is the nth root of x and 1/n is a rational exponent. In other words x1/n = nx.

To make this more clear, consider √3 and 31/2. By squaring both of these, we can determine that they are equal.

(√3)2 = (√3)(√3) = √9 = 3

(31/2)2 = 3(1/2 ∙ 2) = 31 = 3

Notice that squaring both gives a result of 3. Therefore √3 = 31/2.

We can rewrite exponential expressions in radical form and radical expressions in exponential form. The radicand of a radical expression becomes the base of an exponential expression and one over the index of the radical expression becomes the exponent. We'll illustrate this in the following

36 = 6 1/3. Notice the index in red becoming the denominator of the exponent and the radicand in blue becomes the base.

Examples: Evaluate the following.

1. 161/2 = √16 = 4. Recall that the index of a square root is 2, although the 2 is not written.

2. (-27)1/3 = 3√(-27) = -3 because (-3)3 = -27.

3. (16/81)1/4 = 4√(16/81) = 2/3 because (2/3)4 = 16/81.

4. -321/5 = -2 because (-2)5 = -32.

Examples: Simplify the following rational expressions.

1. (25x4)1/2

We will break this down into the number part and the variable part.

251/2 = 5 and (x4)1/2 = √(x4) = x2 since (x2)2 = x4.

Therefore (25x4)1/2 = 5x2.

Note that (x4)1/2 can also be simplified using the rules for raising an exponent to an exponent. Therefore, the exponents 4 and ½ are multiplied together to get 2, which is the new exponent and (x4)1/2 = x2.

2. (-125x3)1/3

(-125)1/3 = 3√(-125) = -5 since (-5)3 = -125 (x3)1/3 = 3√(x3) = x

Therefore, (-125x3)1/3 = -5x.

3. [(2a + 7)4]1/4 = 4√(2a + 7)4 = |2a + 7|

Notice the answer is |2a + 7| and not 2a + 7. |2a + 7|4 = (2a + 7)4 and a can be any number, therefore 2a + 7 could be a negative number. But we can't have a real number solution if 2a + 7 is negative because no real number raised to the 4th power is negative. Therefore we must have absolute value to ensure a positive real number solution.

4. [(4b – 3)5]1/5 = 5√(4b – 3)5 = 4b – 3.

In this case, since n is odd, there is no need for the absolute value symbols in the solution. The value of 4b- 3 could still be negative but a negative quantity raised to an odd power will be negative.

In the previous examples all of the exponents were in the form 1/n. Now we will learn how to simplify exponential expressions where the exponent is in the form p/n where p and n are positive integers and n ≠ 1. To illustrate how to simplify , consider the following example.

272/3. We can rewrite this in the form x1/n raised to the 2nd power, (x1/n)2 = x2/n.

Therefore, 272/3 = (271/3)2

= (3√27)2 = 32 = 9

We can rewrite the exponential expression in radical form. The numerator of the rational exponent becomes the power of the radical expression. The denominator of the rational exponent becomes the index of the radical expression and the base of the exponential expression becomes the radicand. In the previous example,

272/3 = (327)2

We would solve this as follows:

Take the cube root of 27: 3√27 = 3

Square 3: 32 = 9

Or we can think of 272/3 as

272/3 = 3√(27)2

We would solve this as follows:

Square 27 first : 272 = 729

Take the cube root of 729 : 3√729 = 9 since 93 = 729.

Note that we know that 729 is divisible by 9 because the sum of the digits in 729 is divisible by 9. The sum of the digits of 729 is 18, which is divisible by 9.