In this article, we will discuss roots and powers or exponents of numbers.

Contents

## Powers and Exponents

Exponents are a way of doing a lot of multiplications all at once.

Fundamentally, A^{n} means that we multiply n factors of A together. 1 to any power is one. 0 to any power is 0. A negative to an even power is positive. A negative to an odd power is negative.

Any square of an equation that equals a positive number has two solutions e.g x^{2 }= 4 has two solutions x = 2 and x = -2.

An equation with an expression to an even power equal to a negative has no solution, but an odd power can equal a negative e.g.

(x-1)^{2} = -4 has no solution but an equation of the form (x-4)^{3 }= -1 has a solution x-4 = -1 -> x = 3.

### Exponential Growth

Under this subtopic, we will explore some of the important patterns with powers of different kinds of numbers. For different cases, we will look at what happens to the powers when the exponents increase through the integers.

#### Positive base greater than 1

For a positive base greater than one, the values continually get larger at a fast rate.

7^{1 }= 1 ……… 7^{3} = 343 …….. 7^{6} = 117649.

#### Positive base less than 1

For a positive base less than one, the values continually get smaller.

(1/2)^{1}=1/2 ….. (1/2)^{4}=1/16 …… (1/2)^{8} = 1/256

#### Negative base less than -1

The absolute values here get larger but the +/- signs alternate.

(-3)^{1}=-3

(-3)^{2}=9

(-3)^{3}=-27

(-3)^{4}=81

(-3)^{5}=-243

(-3)^{6}=729

#### Negative base between -1 and 0

The absolute values get smaller approaching zero but the +/- signs alternate.

(-1/2)^{1 }= -1/2

(-1/2)^{2 }= 1/4

(-1/2)^{3}^{ }= -1/8

(-1/2)^{4}^{ }= 1/16

Let’s look at an example:

If x < 1 and x != 0, is x^{7 }> x^{6} ?

If x is any negative number, then x^{7 }is negative and x^{6} is positive so x^{7 }< x^{6} so the answer is no.

Also if o < x < 1, values get smaller as power increases therefore x^{7 }< x^{6}, answer is no.

So in all cases, the answer is no.

In summary, we have discussed the patterns of exponential growth, how increasing the exponent changes the size of the powers of different kinds of bases.

### Laws of Exponents

- a
^{m}* a^{n}= a^{m+n} - a
^{m}/a^{n}= a^{m-n} - a
^{0}= 1 (prove a^{m}/a^{m}= 1 = a^{m-m}= a^{0 }= 1) for all a != 0 - (a
^{m})^{n}= a^{m*n} - a
^{-n}= 1/a^{n}(prove a^{0-n}) - (a/b)
^{-n}= (b/a)^{n} - (ab)
^{n}= (a^{n})(b^{n}) - (a/b)
^{n}= a^{n}/b^{n}

Let’s look at an example:

Rank the following from smallest to biggest.

(a) (1/3)^{-8} (b) 3^{-3} (c) (1/3)^{5}

(1/3)^{5 } = 1/3^{5} < 3^{-3} = 1/3^{3 } < (1/3)^{-8 }= 3^{8}

#### The Units Digit

The units digit of any product will be influenced only by the units digits of the factors. Therefore we only need to consider **single digit products **when trying to determine the units digit of a product.

For example, what is the units digit of 57^{123}?

The idea is to find a pattern for the last digit of the value of each power as follow.

7^{1} = …7

7^{2} = …9

7^{3} = …3

7^{4} = …1

7^{5} = …7

7^{6} = …9

7^{7} = …3

7^{8} = …1

We begin to see a pattern at this point, 7,9,3,1 repeat as last digits.

And we also notice for every power that’s a multiple of 4, the value ends with 1, we can jump to 120 and the pattern continues to 123.

7^{120} = …1

7^{121} = …7

7^{122} = …9

7^{123} = …3

And so the units digit of 57^{123 }is 3.

##### Strategy for finding units digit

- Focus on the single-digit multiplication only.
- Look for the repeating pattern and determine the period of the pattern as in 4 (7,9,3,1) in the example above.
- Extend the pattern using multiples of the period.

## Roots

The root of a number A is another number X, which when multiplied by itself a given number of times, equals A e.g the square root of A is X in X*X = A. The following symbols are used:

√ for square root, ^{3}√ for cube root and generally ^{n}√ for all other roots.

### Square Root

In many contexts in math, we know the square of a number and we have to find the original number that was squared. If the square of the desired number is a perfect square, then it is easy to solve for the number from its square.

Below are some basic ideas about square root.

##### Idea 1

It is not possible to take the square root of a negative number e.g √-1 because positives squared re positive and negatives squared are positive so it is not possible to find the square root of a negative. For A ≥ 0, √A ≥ 0.

##### Idea 2

For positive numbers, the square root preserves the order of inequality.

In other words, if A < B < C, then √A < √B < √C

We know that 36 < 41 <49, therefore √36 < √41 < √49 → 6 < √41 < 7. From this, without knowing the exact value of √41, we can know that it is between 6 and 7.

##### Idea 3

There are a few simple square roots for which we can memorize the rough approximations:

√2 ∼= 1.4

√3 ∼= 1.7

√5 ∼= 2.2

If we know just these three to the nearest tenth, we can do a great deal of approximating.

##### Idea 4

The √ sign is an operator just like +,-,×,÷ and like those, it can operate on numbers or variables.

This operator undoes squaring but not necessarily exact opposite. For example, suppose x is negative, then y^{2} would be positive and √y^{2 }also would be positive. We don’t wind back where we started which is negative y.

So technically, because the result of √ is always positive regardless of whether y is positive or negative, we can say √y^{2} = |y| (absolute value of y).

When we see the √ operation, the result is always positive. This is in contrast to when we a variable is squared and equated to something and we have to solve for the variable like x^{2 }= A.

In this case, the solution has 2 parts as we already discussed above under exponents, x = ±√A

##### Idea 5

In the discussion about exponential growth above, we discussed that if A > 1, then positive powers of A gets bigger i.e A^{2 }> A.

If squaring makes these numbers bigger, then taking a square root must make them smaller

√A < A

But as we saw also in the exponential growth lesson, everything is opposite for positive numbers less than 1.

If 0 < A < 1, then positive powers of A get smaller i.e A^{2 }< A.

If squaring make these numbers smaller, then taking a square root must make them bigger.

√A > A

√(4/9) = 2/3 > 4/9

Let’s consider the following example:

If 7K is a positive integer and if √K > K, is K an integer?

If √K > K, ten K must be a number between 0 and 1. So it can not be an integer. No is the answer.

### Cube Root and Other Roots

Just as the square root undoes the act of squaring, so there are higher-order roots that undo other powers as well.

For example, the **cube roots** undo the act of cubing a number.

Just as we use the notation √ for square root, the notation for all other roots is similar. We use the same radical sign √ but for all other roots, we put a small number in front of the radical to denote the order of the root.

Thus, the cube root of x is written as ^{3}√x.

Below are some basic ideas.

##### Idea 1

The ± rules for cubes are different.

Remember that (positive)^{3} = positive and (negative)^{3} = negative.

Therefore, **x ^{3 }= positive** has only one

**positive**solution and

**x**has only one

^{3}= negative**negative**solution.

With square roots, we can only find the square roots of positives and cannot take the square root of a negative.

By contrast, we can take the cube root of any number on the number line, positive or zero or negative.

^{3}√8 = 2

^{3}√0 = 0

^{3}√-8 = -2

##### Idea 2

For quick computations, it’s good to memorize the cubes of up to 10.

1^{3} = 1

2^{3 }= 8

3^{3} = 27

4^{3} = 64

5^{3} = 125

6^{3} = 216

7^{3} = 343

8^{3} = 512

9^{3} = 729

10^{3} = 1000

Then, we automatically have some cube roots memorized as well

^{3}√1 = 1

^{3}√8 = 2

^{3}√27 = 3

^{3}√64 = 4

^{3}√125 = 5

^{3}√216 = 6

^{3}√343 = 7

^{3}√512 = 8

^{3}√729 = 9

^{3}√1000 = 10

##### Idea 3

The square root √a, the fourth root ^{4}^{√a, the sixth root 6}√a etc are called even roots.

The cube root ^{3}√a, the fifth root ^{5}√a, the seventh root ^{7}√a etc are called odd roots.

As with square roots, we can take ant even root of positive numbers which results in a positive output but we cannot take any even root of a negative number.

As with cube roots, any odd root of a positive number is positive and any odd root of a negative number is negative.

##### Idea 4

For all n,

^{n}√0 = 0 and ^{n}√1 = 1

All roots preserve the order of inequality,

0 < a < b < c, then 0 < ^{n}√a < ^{n}√b < ^{n}√c.

Suppose we have to estimate ^{4}√50.

Because 16 < 50 < 81, we ^{4}√16 < ^{4}√50 < ^{4}√81 → 2 < ^{4}√50 < 3.

We can also compare size of different roots.

If x > 1 and if n > m, then 1 < ^{n}√x < ^{m}√x < x.

If 0 < x < 1 and if n > m, then 0 < x < ^{m}√x < ^{n}√x < 1.

**Thus the higher the order of the root, the closer the result is to 1.**

### Properties of Roots

Fundamentally, roots are a special case of exponents, so some of the properties of exponents are the same as the properties of roots.

We will discuss under this subtopic the link between roots and exponents.

We are going to use square roots to discuss properties of roots and assume the numbers A and B are positive.

Even though we are using square roots, all these properties also work for all higher-order roots and for the odd roots, the numbers involved don’t have to be positive.

##### Distributive Property Of Roots

Roots distribute over multiplication and division √AB = √A * √B, √(A/B) = √A / √B.

Suppose we have √N.

If N has, as one of its factors, a perfect square A, then we can replace N with A*B for some B, and take the square roots of the factors separately.

The square root of the perfect square A will come out evenly.

e.g √63 = √(9*7) = 3√7

For large numbers, we can find the prime factorization and it will then be easy to find the simplified square root.

Every pair of prime factors and every even power of a prime factor is a perfect square and can be simplified.

E.g suppose we have (2^{6})(3^{5})(5^{2})(7) as the prime factorization of a number, we can express √N in simplified form.

√N = √2^{6} * √3^{5} * 5^{2 }* 7

= √(2^{3} * 2^{3}) * √(3^{2} * 3^{2} * 3^{1}) * √5^{2 }* √7

= 2^{3 }* 3^{2}√3 * 5 * √7

= 360(√3)(√7)

= 360√21

##### Addition and Subtraction of Roots

We can combine two terms only if they have the same radical factor.

E.g √72 – √32 = √(36*2) – √(16*2) = 6√2 – 4√2 = 2√2

But the expression 6√2 – 4√3 cannot be simplified further.

##### Multiplication and Division of Roots

We can use the distribute property to simplify products or divisions of roots

E.g (2√42) * (4√63) = 8√(42*63) = 8√(2*3*7*3*3*7) = 8√(3*3) * √(7*7) * √(2*3) = 8*3*7*√6 = 168√6.

##### Powers of Roots

Radical expressions raised to powers can be simplified. Keep in mind that any even power of a square root can be written as the power of a whole number.

E.g (√2)^{48} = ((√2)^{2})^{24 }= 2^{24.}

We can’t compute that numerical value off-hand but we can still compare the size to that of something else. For example, we can see (√2)^{48 }> 2^{20}.

When we raise a radical expression to a power, we distribute the exponent to each factor. Any even power of a radical is the power of a whole number.

### Equations with Square Roots

We might have an equation involving square root like so √(x+2) = 3.

Since we can undo a square root by squaring, all we have to do is square both sides and solve.

√(x+2) = 3 → x+3 = 9 → x = 7.

Since we know that √x^{2} = x is not always true. It is true for positive numbers and zero, but not for negative numbers.

This suggests that we may run into some kind of problem when negative values arise.

It turns out that, in radical equations, we have to beware of **extraneous roots.**

When we do all of our algebra correctly including squaring both sides of the equation, the algebra can lead to answers that do not actually work in the original equation. These are extraneous roots.

Let’s look at an example:

√(x+3) = x-3 → x+3 = (x-3)^{2 }→ x+3 = x^{2}-6x+9

→ 0 = x^{2}-7x+6 = (x-6)(x-1)

x = 1 or 6.

Now we have to check whether these values work in the original equation.

Checking x = 1;

√(x+3) = √(1+3) = √4 = 2

x-3 = 1-3 = -2

x = 1 does not work since 2 != -2.

Checking x = 6;

√(x+3) = √9 = 3

x-3 = 6-3 = 3

x = 6 works and so this is the only correct value of x.

If we get a quadratic after squaring both sides of an equation with a radical part, the algebra will lead to 2 roots.

Sometimes both roots work.

Sometimes one root works and one is extraneous.

Sometimes both are extraneous and the equation has no solutions.

For example √(2x-4) = √(x-4)

2x – 2 = x -4 → x-2 = -4 → x = -2

When we plug this in the original equation, this results in the square root of a negative on both sides. Therefore the equation has NO solution.

Finally, keep in mind that we should square both sides only when the radical is by itself on one side of the equation.

If the radical appears with other terms on the side, we will have to isolate the radical on one side before it would make sense to square both sides.

E.g 2+√(4-3x) = x

√(4-3x) = x-2 → 4-3x = (x-2)^{2 }→ 0 = x^{2 }– x = x(x-1)

The algebra leads to two roots x = 0 and x = 1.

When we check both roots in the original equation, none works therefore the equation has NO solution.

### Rationalization

Rationalization is the process of changing a form of an equation or expression containing radical in the denominator so that the denominator no longer contain the radical. The process is called rationalizing the denominator.

In other words, to eliminate roots from the denominator of a fraction is called rationalizing.

If the fraction has a single root in the denominator, we rationalize by multiplying by that root over itself.

For example:

Simplify (4-√6) / 2√3

( (4-√6) / 2√3 ) * √3/√3

( 4√3 – √6 * √3 ) / (2*3)

(4√3 – 3√2) / 6

If the denominator of the fraction contains addition or subtraction involving a radical expression, to rationalize, we need to multiply by the conjugate of the denominator over itself.

Using the difference of two squares i.e (x-y)(x+y) = x^{2} – y^{2}, we can eliminate roots from such denominators.

E.g Simplify 2/(√5 – 1)

We simply multiply and divide by the conjugate of √5 – 1

(2/(√5 – 1)) * ( (√5 + 1)/(√5 + 1) )

2(√5 + 1) / ((√5)^{2} – 1^{2})

(√5 + 1)/2

## Fractional Exponents

So far, we have only considered exponents that are integers, positive and negative and zero.

Fractional exponents make explicit the link between roots and exponents.

E.g 2^{1/2} = A, if we square A, we will get (2^{1/2})^{2} = A^{2 }→A^{2 }= 2 → A = √2.

Therefore 2^{1/2} = √2.

In general X^{1/2} = √X.

The same thing is applicable to X^{1/3} = ^{3}√X.

So generally X^{1/m} = ^{m}√X.

Also X^{n/m} = (^{m}√X)^{n.}

E.g 8^{4/3 }= (^{3}√8)^{4}

## Exponential Equations

An exponential equation is one in which variables are in the exponents.

If two powers with the same base are equal, then the exponents must be equal.

A^{x} = A^{y} → x = y

This rule works for all bases other than 0 or ±1.

To solve exponential equations, we have to get equal bases on both sides. This may involve expressing the given bases as powers of smaller bases.

Once the bases on both sides are equal, we can equate the exponents and solve.

Let’s look at some examples.

- If (
^{5}√3)^{3x+7}= 3^{2x}, find x?

(3^{1/5})^{3x+7} = 3^{2x} → (3x+7)/5 = 2x → x = 1.

Sometimes, neither of the bases will be written as a power of the other. Instead, to solve, both cases have to be written as a power of some other, smaller number.

- If 27
^{2x-2}= 81^{x+1}, find the value of x.

(3^{3})^{2x-2} = (3^{4})^{x+1}

3(2x-2) = 4(x+1) which leads to x = 5.

## Conclusion

In this article, we have discussed some powers, exponents and roots and their properties. We also discussed the link between exponents and roots.

In the next article, we will discuss algebra, equations and inequalities.