Exponent Basics at a Glance

Exponential Basic Training

Okay Shmoopers, let's begin by getting the basics out of the way. This is the stuff you might already know. To be sure everyone is on the same page, let's do this thing.

Exponential expressions have two basics parts: a base and an exponent. So if we consider the expression 3⁵, 3 would be the base while 5 is the exponent. In this example, we can think of 3 being multiplied by itself 5 times.

3⁵ = 3 · 3 · 3 · 3 · 3 = 243

For those of you playing at home, this does not mean that you need to multiply 3 times 3 times 3 times 3 times 3 on your calculator every single time you evaluate an exponent. Instead, most calculators have a button that either looks like y ^x or ^ that you can use to evaluate exponents. This makes life much, much easier. The ^ thing is called a "caret," by the way.

Now that you know what these things are, let's simplify 'em. Here are a few rules to get you started:

If you multiply like bases, add the exponents. This is the product rule.

x³ · x⁴ = x^{(3 + 4)} = x⁷

Think about this for a second: x³ is really like x · x · x, and x⁴ is really like x · x · x · x. Multiplied together, we have (x · x · x) · (x · x · x · x). That's 7 x's. So our simplified answer has got to be x⁷.

If one exponent is raised to another, you multiply the exponents. This is the power rule.

(x³)² = x^{(3 · 2)} = x⁶

If you understand the product rule, the power rule really isn't too terrible. In this case, x³ is squared. This means we have two x³'s multiplied together: x³ · x³. Using the rule we've already established, we can get our answer.

Two other pieces of info that might interest you are what happens when the exponent is either 1 or 0. For instance, imagine you were asked to simplify x¹. Luckily, this is just x. This is the case for any number or variable. The way we like to remember it is that a blob to the first power is just a blob. Or a labradoodle to the first power is just one labradoodle.

On the other hand, a blob raised to the power of 0 is, believe it or not, 1. There are many ways that we could use various patterns and tricks to convince you that this is true. But for now, we're going to ask you to trust us. After all, who has time for those shenanigans? Anyway, anything raised to the power of 0 is 1.

x⁰ = 1
5⁰ = 1
79,874⁰ = 1

And just in case you need these rules in rap form in order for them to make sense, we've got your back.

Sample Problem

Simplify the following: (a²a³)⁴.

This is a pretty basic sample problem to get things started. First we'll use the product rule by adding exponents in the parentheses.

(a²a³)⁴ = (a⁵)⁴

Next, we apply our power rule by multiplying exponents.

(a⁵)⁴ = a²⁰

No big deal. Let's see what else we have in our bag of Shmooper dooper tricks.

Sample Problem

Simplify the following: (x⁵y³x²y⁷)⁴.

Multiple variables, eh? Not so bad. We just need to remember to keep the x's with the x's and the y's with the y's. We can start inside the parentheses by adding the exponents of the terms with like bases.

(x⁵y³x²y⁷)⁴ = (x⁷y¹⁰)⁴

Next, we apply our power rule by multiplying exponents.

(x⁷y¹⁰)⁴ = x²⁸y⁴⁰

Are the exponents big? Yes. Are they correct? Yes.

Sample Problem

Simplify the following: (2x³)⁴(3xy⁴).

A few coefficients never hurt anyone. Just like we dealt with the different variables individually, we can do the same for the coefficient. A coefficient is nothing more than a constant in front of or multiplied by a variable.

To solve, we must follow the good ol' PEMDAS order of operation rules. Since everything's already simplified in the parentheses, we head to the exponent. Remember to treat 2⁴ like the rascal of a regular number it really is.

(2x³)⁴(3xy⁴) = 16x¹²(3xy⁴)

Now, we multiply coefficients and add the exponents…

16x¹²(3xy⁴) = 48x¹³y⁴

Done and done.

Sample Problem

Simplify the following: (-5x⁰y³z)²(2xy⁸y³z²)³.

Just one more challenge before we send you packing. This time there's a lot going on, so we'll start by simplifying in the parentheses. Pay special attention to the x⁰ in there.

(-5x⁰y³z)²(2xy⁸y³z²)³ = (-5(1)y³z)²(2xy¹¹z²)³ = (-5y³z)²(2xy¹¹z²)³

Next it's time to take care of the coefficients and multiply those exponents.

(-5y³z)²(2xy¹¹z²)³ = (25y⁶z²)(8x³y³³z⁶)

Finally, we  multiply our two coefficients while adding our exponents wherever our bases are the same.

(25y⁶z²)(8x³y³³z⁶) = 200x³y³⁹z⁸

Cowabunga, dude.