## Introduction

In the fascinating world we live in, populations vary all the time from city to city, province to province, and country to country. You too can become an expert in understanding population changes. Let’s get started by examining a table of the province of Ontario’s population estimates from 2006 to 2016.

Have you ever wondered if there is a way to model and predict population growth?

Here is a graph of the population estimates from 2006 to 2016.

Now examine the estimated population information in the scatterplot graph to see if there is a pattern and if it looks linear.

Notice that the scale for the population has been converted to scientific notation. As a refresher, scientific notation is all about using decimal forms for extremely small and large number.

Scientific notation also expresses the large number as a number between 1 and 10 and multiplied by a power of 10. For example 12,660,000 can be expressed as 1.266 × 107.

In examining the scatterplot graph, you can also draw a line of best fit and even find the equation. A line of best fit is exactly what we may think it is! A line that is best at representing any type of data based on a scatter plot. Let's achieve some more expertise on this subject by understanding how we identify the best line of best fit!

Let’s examine two more examples. Which one do you think is a better fit?

When analyzing the examples, you also have to consider whether, or not, the population is growing at an exponential rate. Perhaps using an exponential equation would be a better fit than a line. Looking at both possibilities it is hard to tell which one is better.

Line of best fit: $\phantom{\rule{0ex}{0ex}}y=132198x-2.5257×{10}^{8}\phantom{\rule{0ex}{0ex}}$

Exponential curve of best fit:

NOTE: Graphs were drawn in Desmos using the regression formulas ${y}_{1}\sim m{x}_{1}+b$ and ${y}_{1}\sim a{b}^{{x}_{1}}$.

Mathematicians pondered the same question. They developed a procedure called regression that not only figures out the formula but also decides which type of function is the best.

The last unit in this course is all about finding these formulas to represent trends and how to decide which one is the best. We will revisit regression and this problem later at the end of this course to find out which formula is the best.

## Exponent Laws

As you have achieved expertise in the Minds On section about scatterplots, let’s now explore Exponent Laws.

Exponent laws are algebraic ‘short cuts’ that are based on patterns that we see often while working with powers. You may be familiar with exponent laws, but as a refresher, let’s review them and try some examples.

${x}^{a}×{x}^{b}={x}^{a+b}$

${x}^{a}÷{x}^{b}={x}^{a-b}$

${\left({x}^{a}\right)}^{b}={x}^{ab}$

${\left(xy\right)}^{a}={x}^{a}{y}^{a}$

${\left(\frac{x}{y}\right)}^{a}=\frac{{x}^{a}}{{y}^{a}}$

${x}^{0}=1 \left(x\ne 0\right)$

${x}^{-a}=\frac{1}{{x}^{a}} \left(x\ne 0\right)$

${\left(\frac{x}{y}\right)}^{-a}={\left(\frac{y}{x}\right)}^{a}$

### Try it!

Now it’s time to test your math skills on Exponent Laws. Try your best and know that solutions are available to compare your answers! Just remember to review the solutions to build on your expertise in following the correct format. Let’s begin!

Simplify the following.

Now that you have built on your knowledge on the exponent rules, which selection would best express your understanding of exponential rules? Remember, it is perfectly alright to select either weak, okay, good, or excellent, as you will be learning to become an expert in the world of Exponent Rules!

Agree or Disagree statements
Weak Okay Good Excellent

## Rational Exponents

It is time to move to the next level and learn about exponents with rational values! As we have covered information on both exponents that have integer values, let’s transition to the next level of learning, which is …

### Exponents as fractions!

The fractions are used to express radical numbers. For instance, the exponent of $\frac{1}{2}$ implies that you want the square root of the base.

#### Example:

Evaluate ${\left(25\right)}^{\frac{1}{2}}$.

${\left(25\right)}^{\frac{1}{2}}$ is asking you to find $\sqrt{25}$ which is $±5$.

Therefore ${\left(25\right)}^{\frac{1}{2}}=\sqrt{25}=±5$.

Did you know that rational exponents can also be used for cube roots? Not only that, let’s check out four additional rules for Rational Exponents together!

${\left(\frac{x}{y}\right)}^{-a}={\left(\frac{y}{x}\right)}^{a}$

${x}^{\frac{1}{2}}=\sqrt{x}$

${x}^{\frac{1}{a}}=\sqrt[a]{x}$

${x}^{\frac{a}{b}}={\left(\sqrt[b]{x}\right)}^{a}=\sqrt[b]{\left({x}^{a}\right)}$

### Example:

First Example

Evaluate the following.

$8{1}^{\frac{3}{4}}$

$8{1}^{\frac{3}{4}}={\left(\sqrt[4]{81}\right)}^{3}={\left(3\right)}^{3}=27$

Usually finding the root first and then raising it to the exponent is the preferred choice because finding the root often results in a number that is smaller and more familiar.

Second Example

Check out this alternative method.

$8{1}^{\frac{3}{4}}=\sqrt[4]{{\left(81\right)}^{3}}=\sqrt[4]{531441}=27$

You’ll find that the answers are exactly the same.

The first method will be less reliant on a calculator due to the simpler mathematic expressions in keeping the numbers ‘small’

${\left(-81\right)}^{\frac{3}{4}}$

${\left(-81\right)}^{\frac{3}{4}}={\left(\sqrt[4]{\left(-81\right)}\right)}^{3}$

Did you know that you cannot take the square root of a negative number or the 4th root of a negative number? In fact, you can never take an even numbered root of a negative number. Which leads us to answering the question, “How do you take the 4th root of a negative number?”

Surprise! The question cannot be answered, as there is no real answer to this question.

$-{\left(81\right)}^{\frac{3}{4}}$

$-{\left(81\right)}^{\frac{3}{4}}=-{\left(\sqrt[4]{81}\right)}^{3}=-{\left(3\right)}^{3}=-27$

Since the negative was outside the bracket it did not affect your ability to find the 4th root.

$8{1}^{-\frac{3}{4}}$

This time the negative is in the exponent. The seventh exponent rule $\left({x}^{-a}=\frac{1}{{x}^{a}} \left(x\ne 0\right)\right)$will have to be used in addition to the rules for rational exponents.

Since the negative exponent is going to create a fraction, it is best to leave it until the end.

$8{1}^{-\frac{3}{4}}={\left(\sqrt[4]{81}\right)}^{-3}={\left(3\right)}^{-3}$

You have two choices at this point in the solution.

You can take two steps: ${\left(3\right)}^{-3}={\left({3}^{3}\right)}^{-1}=2{7}^{-1}=\frac{1}{27}$.

Or you can take one step: ${\left(3\right)}^{-3}=\frac{1}{{\left(3\right)}^{3}}=\frac{1}{27}$.

### Try it!

Now it’s your turn to test your math skills. Try your best and know that solutions are available to compare your answers. Just remember to review the solutions to build on your expertise in following the correct format. Let’s begin!

## Evaluating Exponential Expressions vs Simplifying Exponential Expressions

Sometimes you are asked to evaluate an expression and sometimes you are asked to simplify.

Brainstorming Time: What do you think the difference may be between evaluating and simplifying exponential expressions?

What’s the difference between evaluating an expression and simplifying an expression? Let’s find out together!

### Similarities:

Both simplify the expression using exponent rules.

### Differences:

The evaluate questions are dealing with specific values and can calculate the exact answer.

#### Example

A carpenter is creating rectangular tables. The design she is working with has the length of the table being twice the width.

The cost to build each table depends partly on the area of the top. Amy creates a sketch of the design and determines a formula for the area of the table top.

A woodworker knows that the area of a rectangle is calculated by multiplying length by width.

$A=w×2w$

The woodworker simplifies the right side of the equation:

$A=2{w}^{2}$

A client wants to have a table that has a width of 50 cm. The woodworker uses her formula to determine the area of the table:

$A=2{\left(50\right)}^{2}$

$A=5000$

The woodworker evaluated $2{\left(50\right)}^{2}$ and determined that the area of the table top is $5000$cm².

Evaluate $\frac{\left({2}^{3}\right)\left({2}^{5}\right)}{\left({2}^{4}\right)\left({2}^{2}\right)}$

$\frac{\left({2}^{3}\right)\left({2}^{5}\right)}{\left({2}^{4}\right)\left({2}^{2}\right)}=\frac{{2}^{3+5}}{{2}^{4+2}}=\frac{{2}^{8}}{{2}^{6}}={2}^{8-6}={2}^{2}=4$

#### Example

Simplify $\frac{{x}^{3}{x}^{5}}{{x}^{4}{x}^{2}}$

$\frac{{x}^{3}{x}^{5}}{{x}^{4}{x}^{2}}=\frac{{x}^{3+5}}{{x}^{4+2}}=\frac{{x}^{8}}{{x}^{6}}={x}^{8-6}={x}^{2}$

Sometimes when evaluating an expression, it is more efficient to simplify and then substitute the given value. Let’s look at this same expression when $x=2$.

#### Example

Simplify $\frac{{x}^{3}{x}^{5}}{{x}^{4}{x}^{2}}$ and then evaluate when $x=2$.

Simplify: $\frac{{x}^{3}{x}^{5}}{{x}^{4}{x}^{2}}=\frac{{x}^{3+5}}{{x}^{4+2}}=\frac{{x}^{8}}{{x}^{6}}={x}^{8-6}={x}^{2}$

Evaluate: ${x}^{2}=\left(2{\right)}^{2}=4$

## Exponential Equations

Questions often require problem solving. How would you approach solving exponential equations?

Exponential equation questions will always have equals signs since they are actually equations. To solve these equations means you are finding the value(s) for x that will satisfy the conditions of the equation.

### Example

Solve the following equations.

## Applications

As we explored population growth at the beginning of this lesson, let’s explore deeper into understanding population in the world of Exponential Rates.

Whether it’s the growth of a rabbit population in a forest, or bacteria in a petri dish, these populations will typically grow at an exponential rate. This type of growth assumes there are no limiting factors, such as habitat or food limitations, and other limits on populations such as predators and mass migrations.

### Examples

A population of cells is being studied to better understand a disease. This particular type of cell doubles every day. If a sample of these cells is estimated to consist of 1000 cells initially, the size of the sample after n days can be modelled with the equation $S=1000\left(2{\right)}^{n}$, where S represents the sample size of cells, and n represents the number of days that have passed. How long would it take to reach 512 000 cells?

Since S represents the sample size, $S=512000$.

Using the equation that models the growth we know

$512000=1000\left(2{\right)}^{n}$

Divide both sides by 1000.

$\frac{512000}{1000}=\frac{1000\left(2{\right)}^{n}}{1000}$

${2}^{n}=512$

Get a common base.

${2}^{n}={2}^{9}$

Since the bases are now equal, the exponents must also be equal.

n = 9

Therefore, the cell population will reach 512 000 in 9 days.

Exponent Rules

${x}^{a}×{x}^{b}={x}^{a+b}$

${x}^{a}÷{x}^{b}={x}^{a-b}$

${\left({x}^{a}\right)}^{b}={x}^{ab}$

${x}^{a}÷{x}^{b}={x}^{a-b}$

${\left({x}^{a}\right)}^{b}={x}^{ab}$

${\left(xy\right)}^{a}={x}^{a}{y}^{a}$

${\left(\frac{x}{y}\right)}^{a}=\frac{{x}^{a}}{{y}^{a}}$

${x}^{0}=1 \left(x\ne 0\right)$

Here are some questions for you to try on paper. Answers and solutions are available for comparison when you are ready. Please review the solutions to ensure you are following the correct format. These questions will be similar to those on the assessment at the end of this unit.

## Conclusion

Congratulations! You have now completed Learning Activity 1. By working through all of the examples… you’ll likely feel very comfortable with:

• evaluating powers with rational exponents,
• simplifying algebraic expressions involving exponents, and
• solving problems involving exponential equations using common bases.

## Next Steps

Let’s get energized! Next, in Learning Activity 2, we will learn to interpret graphs and use graphical models.

Stay tuned to see why graphs can really help us in the real world!