Lesson Video: Exponential Functions Mathematics

In this video, we will learn how to identify, write, evaluate, and analyze exponential functions.

20:11

Video Transcript

In this video, we will learn how to identify write, evaluate, and analyze exponential functions.

An exponential function is a function with a rule of the form 𝑓 of π‘₯ equals 𝑏 to the π‘₯, where the constant real number 𝑏 is called the base, with 𝑏 greater than zero and 𝑏 not equal to one, and π‘₯ is called the exponent, which can be any real number. Let’s take a look at 𝑓 of π‘₯ for some positive integer values of π‘₯. For π‘₯ equals one, we have 𝑏 to the first power, which is just 𝑏. For π‘₯ equals two, we have 𝑏 to the second power or 𝑏 squared, which is equal to 𝑏 times 𝑏. For π‘₯ equals three, we have 𝑏 to the third power or 𝑏 cubed equal to 𝑏 times 𝑏 times 𝑏. And for π‘₯ equals four, we have 𝑏 to the fourth power equal to 𝑏 times 𝑏 times 𝑏 times 𝑏.

We can see that the previous value of 𝑓 of π‘₯ is multiplied by 𝑏 every time π‘₯ increases by one. So 𝑓 of two is equal to 𝑓 of one times 𝑏. 𝑓 of three is equal to 𝑓 of two times 𝑏, and 𝑓 of four is equal to 𝑓 of three times 𝑏. This alludes to a more general property of the exponential function. For the exponential function 𝑓 of π‘₯ equals 𝑏 to the π‘₯, the value of 𝑓 of π‘₯ is always the product of 𝑓 of π‘₯ minus one and 𝑏, which means that 𝑏 is always the quotient of 𝑓 of π‘₯ and 𝑓 of π‘₯ minus one. That is, 𝑓 of π‘₯ equals 𝑓 of π‘₯ minus one times 𝑏, which in turn means that 𝑏 equals 𝑓 of π‘₯ divided by 𝑓 of π‘₯ minus one. This holds true not only for integer values of π‘₯ but also for any real number value of π‘₯. We can often use the relationship between 𝑏, 𝑓 of π‘₯, and 𝑓 of π‘₯ minus one to find the value of 𝑏 from a graph or table.

Consider the graph which shows 𝑦 equals 𝑓 of π‘₯ equals 𝑏 to the π‘₯ for a specific value of 𝑏. Notice that the graph intersects the 𝑦-axis at one. This will be the case for any real number value of 𝑏 such that 𝑏 greater than zero and 𝑏 not equal to one. We can find the value of 𝑏 in the function that the graph represents by selecting two points that the graph passes through with π‘₯-coordinates that differ by one. It does not matter which points we choose, but both should have π‘₯- and 𝑦-coordinates that are easy to identify. In this case, it is easiest for us to select points with π‘₯-coordinates that are consecutive integers. So let’s pick π‘₯ equals negative three and π‘₯ equals negative two. From the graph, we can see that these have 𝑦-coordinates of eight and four, respectively. So we have 𝑓 of negative three equals eight and 𝑓 of negative two equals four.

Recall from the previous definition that the base 𝑏 is equal to 𝑓 of π‘₯ divided by 𝑓 of π‘₯ minus one regardless of which value of π‘₯ we choose. So we can choose our value of π‘₯ to be negative two. So π‘₯ minus one is negative three. So we have 𝑏 equals 𝑓 of negative two over 𝑓 of negative three. And substituting in these values obtained from the graph, we get 𝑏 equals four over eight. So we have 𝑏 equals one-half. And therefore 𝑓 of π‘₯ equals one-half to the π‘₯.

Next let’s look at a table representing the function 𝑔 of π‘₯ equals 𝑏 to the π‘₯ for another specific value of 𝑏. For π‘₯ equals one, we have 𝑔 of π‘₯ equals two. For π‘₯ equals two, we have 𝑔 of π‘₯ equals four. For π‘₯ equals three, we have 𝑔 of π‘₯ equals eight. And for π‘₯ equals four, we have 𝑔 of π‘₯ equals 16. Since the values of π‘₯ in the table differ by one, we could use the same procedure as before to determine the value of 𝑏, finding either the quotient of four and two or eight and four or 16 and eight. However, this time we will instead substitute the first pair of π‘₯- and 𝑦-values in the table into the rule 𝑔 of π‘₯ equals 𝑏 to the π‘₯.

So substituting one for π‘₯ and two for 𝑔 of π‘₯, this implies that two equals 𝑏 to the first power and therefore 𝑏 equals two. So far, we have only considered exponential functions of this particular form. But exponential functions can take a more general form than this. We can multiply the base by some constant, π‘Ž, and we can multiply the exponent by a constant, 𝛼, and add a constant to it, 𝛽. And then finally, we can add a constant to the end of the function, 𝑐. These are just the standard linear transformations of a function where π‘Ž stretches the graph of the function along the 𝑦-axis by a factor of π‘Ž, 𝛼 squashes the graph of the function along the π‘₯-axis by a factor of 𝛼, 𝛽 translates the function along the π‘₯-axis by negative 𝛽, and 𝑐 translates the function along the 𝑦-axis by 𝑐.

So some other examples of exponential functions might include 𝑔 of π‘₯ equals four times two to the π‘₯ or β„Ž of π‘₯ equals 𝑒 to the π‘₯ plus one or π‘˜ equals nine to the four π‘₯ plus three. For 𝑔 of π‘₯, the base is still two, and the exponent is still π‘₯. For β„Ž of π‘₯, the base is 𝑒 and the exponent is π‘₯. And for π‘˜ of π‘₯, the base is nine and the exponent is four π‘₯ plus three. Let’s look at an example of how to find an exponential function’s base and exponent.

What are the base and exponent of the function 𝑓 of π‘₯ equals five to the power of π‘₯ minus five?

Recall that an exponential function in its simplest form is of the form 𝑓 of π‘₯ equals 𝑏 to the π‘₯, where 𝑏 is the base and π‘₯ is the exponent. We know that the base 𝑏 is a real number, such that 𝑏 is greater than zero and 𝑏 is not equal to one and the exponent π‘₯ can be any real number. In the given function, we have a base of five and this is raised to the power of π‘₯ minus five. This still fulfills the criteria of an exponential function. So the base is five and the exponent is π‘₯ minus five.

In the next problem, we are given a table of values for an exponential function and are asked to find the function’s equation.

Write an exponential equation in the form 𝑦 equals 𝑏 to the π‘₯ for the numbers in the table. π‘₯ equals two, four, and five with corresponding 𝑦-values nine over 16, 81 over 256, and 243 over 1024.

Recall that an exponential function in its simplest form takes the form of 𝑦 equals 𝑏 to the power of π‘₯, where 𝑏 is the base, a real number greater than zero and not equal to one, and π‘₯ is the exponent, which can be any real number. We therefore need to find the value of 𝑏 from the values in the table to solve the problem. Let’s begin by substituting one of the pairs of π‘₯- and 𝑦-values from the table into the equation 𝑦 equals 𝑏 to the π‘₯.

Substituting in two for π‘₯ and its corresponding 𝑦-value nine over 16, we get nine over 16 equals 𝑏 squared. Solving for 𝑏, we get 𝑏 equals plus or minus the square root of nine over 16. And using surd rules, we get plus or minus the square root of nine over the square root of 16. This simplifies to plus or minus three-quarters. And since 𝑏 must be greater than zero, it must be the positive square root. Therefore, 𝑏 equals three-quarters. Substituting this value of 𝑏 into the equation for the function, we get our final answer 𝑦 equals three-quarters to the power of π‘₯.

One way to verify our answer is to check the other values in the table. Substituting in the other values of π‘₯ and 𝑦, we get 81 over 256 equals 𝑏 to the fourth and 243 over 1024 equals 𝑏 to the fifth. If we divide the second equation by the first, we get 243 over 1024 divided by 81 over 256 equal to 𝑏 to the fifth over 𝑏 to the fourth. The right-hand side here simplifies to 𝑏 to the five minus four, and 𝑏 to the five minus four is just 𝑏 to the one, which is just 𝑏. And taking the quotient on the left-hand side, we also get three-quarters.

In the process of checking our answer, we used the quotient rule, which is one of the properties of exponents. And it states that when dividing exponential expressions with the same base, we keep the base and find the difference of the exponents. That is, 𝑏 to the π‘š over 𝑏 to the 𝑛 equals 𝑏 to the π‘š minus 𝑛, where 𝑏 is the base and π‘š and 𝑛 are the exponents. For example, we would use the quotient rule to show that five to the eight over five to the four equals five to the eight minus four, which is equal to five to the four.

In the problem that follows, we are again given a table of values for an exponential function and are asked to find the function’s equation. This time, however, the equation is in a form other than 𝑦 equals 𝑏 to the π‘₯.

Write an exponential equation in the form 𝑦 equals π‘Ž times 𝑏 to the π‘₯ for the numbers in the table. For π‘₯ equal to zero, one, two, and three, the corresponding 𝑦-values are 18, six, two, and two-thirds.

In this problem, we are asked to write an exponential equation in the form 𝑦 equals π‘Ž times 𝑏 to the π‘₯ for the numbers in the table. So we must find the values of both π‘Ž and 𝑏 to solve the problem. Let’s begin by substituting one of the pairs of π‘₯- and 𝑦-values from the table into the equation 𝑦 equals π‘Ž times 𝑏 to the π‘₯. Substituting in the first pair of values, so π‘₯ equals zero and 𝑦 equals 18, we get 18 equals π‘Ž times 𝑏 to the zero. Since any base raised to the power of zero is equal to one, the equation simplifies to π‘Ž equals 18.

We still need to find the value of 𝑏, however, and we can do this by substituting in another pair of values of π‘₯ and 𝑦. So substituting in the next pair of values, π‘₯ equals one and 𝑦 equals six, we get six equals π‘Ž times 𝑏 to the one. Any base 𝑏 raised to the power of one is just itself, so this simplifies to six equals π‘Žπ‘. We already found the value of π‘Ž to be equal to 18. So we can substitute this into the equation to give six equals 18𝑏. Solving for 𝑏 gives 𝑏 equals one-third. Substituting these values of π‘Ž and 𝑏 into the exponential equation gives us our final answer 𝑦 equals 18 times one-third to the π‘₯.

We can verify our answer by substituting the last two π‘₯-values from the table into the equation to make sure that they give the correct 𝑦-values. Substituting in π‘₯ equals two, we get 𝑦 equals 18 times one-third squared, which is equal to 18 times one-ninth which is equal to two, which is correct. And substituting in π‘₯ equals three gives 𝑦 equals 18 times one-third cubed, which is equal to 18 times one over 27, which is equal to two-thirds, which is again correct.

In the process of solving the last problem, we used one of the properties of exponents called the zero-exponent rule, which states that any base raised to the power of zero is equal to one. That is, 𝑏 to the zero equals one, where 𝑏 is the base. For example, three to the zero is equal to one.

In the next problem, we will see how to find the equation of an exponential function from its graph.

Observe the given graph and then answer the following questions. Find the 𝑦-intercept in the shown graph. As this graph represents an exponential function, every 𝑦-value is multiplied by 𝑏 when π‘₯ increases by Ξ”π‘₯. Find 𝑏 for Ξ”π‘₯ equals one. Find the equation that describes the graph in the form 𝑦 equals π‘Žπ‘ to the power of π‘₯ over Ξ”π‘₯.

Let’s begin by finding the graph’s 𝑦-intercept. We can see that the line of the graph passes through the point zero, 10. Therefore, the 𝑦-intercept is 10. Now, let’s find the value of 𝑏 in the equation of the graph. In the problem, we are told that the graph represents an exponential function for which every 𝑦-value is multiplied by 𝑏 when π‘₯ increases by Ξ”π‘₯. And we’re asked to find 𝑏 when Ξ”π‘₯ equals one. We’ve already seen that the graph passes through the point zero, 10. And we can also see that it passes through the point one, 20. The change in π‘₯ between these two points is one, so it satisfies Ξ”π‘₯ equals one.

We are told that the 𝑦-value is multiplied by 𝑏 when π‘₯ increases by Ξ”π‘₯. If the 𝑦-value of our first point is 𝑦 one and the 𝑦-value of our second point is 𝑦 two, then 𝑏 equals 𝑦 two over 𝑦 one which is equal to 20 divided by 10, which is equal to two. And finally, we need to find the equation that describes the graph in the form 𝑦 equals π‘Žπ‘ to the power of π‘₯ over Ξ”π‘₯. We are already given Ξ”π‘₯ equals one, and we have found the value of 𝑏 equal to two. Therefore, 𝑦 equals π‘Ž times two to the π‘₯ over one, which is just equal to π‘Ž times two to the π‘₯. To find the value of π‘Ž, we can substitute the π‘₯- and 𝑦-values of points we know to be on the graph.

We’ve already found the 𝑦-intercept to be 10, so we know that the point zero, 10 lies on the graph. This means we can substitute the values of π‘₯ equals zero and 𝑦 equals 10 into the equation which gives us 10 equals π‘Ž times two to the zero. By the zero-exponent rule, two to the zero is just equal to one. Therefore, this gives π‘Ž equals 10. This gives us our equation and the final answer 𝑦 equals 10 times two to the π‘₯.

In the next problem, we determine the value of an expression by evaluating an exponential function.

Given that 𝑓 of π‘₯ equals four to the π‘₯, determine the value of 𝑓 of π‘₯ over 𝑓 of π‘₯ minus one minus 𝑓 of π‘₯ minus one over 𝑓 of π‘₯.

In this problem, we are asked to find the value of an expression for π‘₯ without being given a value for π‘₯, hinting that this expression has the same value regardless of the value of π‘₯. We are given that 𝑓 of π‘₯ equals four to the π‘₯. Therefore, 𝑓 of π‘₯ minus one equals four to the π‘₯ minus one. Therefore, 𝑓 of π‘₯ over 𝑓 of π‘₯ minus one equals four to the π‘₯ over four to the π‘₯ minus one. Using the quotient rule, this is equal to four to the π‘₯ minus π‘₯ minus one, which is equal to four to the one which is equal to four. This is also the value of the base of the exponential function 𝑓 of π‘₯, which is defined by 𝑓 of π‘₯ over 𝑓 of π‘₯ minus one.

The second part of the expression is 𝑓 of π‘₯ minus one over 𝑓 of π‘₯, which is just the reciprocal of what we just found, one over 𝑓 of π‘₯ over 𝑓 of π‘₯ minus one. Therefore, this is equal to one over four or one-quarter. Therefore, 𝑓 of π‘₯ over 𝑓 of π‘₯ minus one minus 𝑓 of π‘₯ minus one over 𝑓 of π‘₯ equals four minus one-quarter, which equals 15 over four.

The previous problem made use of the negative-exponent rule, which states that any base raised to a negative power is equal to one over the base raised to the exponent’s additive inverse. That is, 𝑏 to the negative 𝑛 equals one over 𝑏 to the 𝑛, where 𝑏 is the base and 𝑛 and negative 𝑛 are the exponents. For example, six to the negative two is equal to one over six to the positive two, or six squared, which is equal to one over 36.

In the final problem, we will determine the values of an exponential function’s base that result in the function decreasing without being told the form of the function.

Consider an exponential function with base π‘Ž. For which values of π‘Ž is the function decreasing?

In this problem, we are given an exponential function 𝑓 of π‘₯ equals π‘Ž to the π‘₯. Recall that the base π‘Ž may be given by 𝑓 of π‘₯ over 𝑓 of π‘₯ minus one. Recall that π‘Ž is a constant real number with π‘Ž greater than zero and not equal to one. This means that regardless of the value of π‘₯, the value of 𝑓 of π‘₯ over 𝑓 of π‘₯ minus one is always a constant. Therefore, for any value of π‘₯, the increase in 𝑓 of π‘₯ between π‘₯ minus one and π‘₯ is always π‘Ž.

Since an exponential function is always positive, 𝑓 of π‘₯ over 𝑓 of π‘₯ minus one being greater than one implies that 𝑓 of π‘₯ is greater than 𝑓 of π‘₯ minus one and, therefore, that the function is increasing. And likewise, 𝑓 of π‘₯ over 𝑓 of π‘₯ minus one being less than one implies that 𝑓 of π‘₯ is less than 𝑓 of π‘₯ minus one and, therefore, that the function is always decreasing.

Since the left-hand side of both of these inequalities is equal to π‘Ž, this means that π‘Ž greater than one implies the function increases and π‘Ž less than one implies it decreases. Remember also that π‘Ž must be greater than zero, so this gives us our final answer. π‘Ž greater than one implies the function increases and π‘Ž between zero and one implies the function decreases.

The last problem alluded to another property of exponential functions, their monotonicity. An exponential function is a monotonic function or a function that always increases or always decreases. If the base 𝑏 is greater than one, it is called the growth factor, and the function will always increase. If the base is greater than zero and less than one, it is called the decay factor, and the function will always decrease.

Let’s finish by recapping some key points from this video. An exponential function has the form 𝑓 of π‘₯ equals 𝑏 to the π‘₯, where 𝑏 is the base, which is a positive real number not equal to one, and π‘₯ is the exponent, which may be any real number. The quotient rule states that when dividing exponential expressions with the same base, we keep the base and find the difference of the exponents. That is, 𝑏 to the π‘š over 𝑏 to the 𝑛 equals 𝑏 to the π‘š minus 𝑛. The zero-exponent rule states that any base raised to the power of zero is equal to one. That is, 𝑏 to the zero equals one.

The negative-exponent rule states that any base raised to a negative power is equal to one over the base raised to the exponent’s additive inverse. That is, 𝑏 to the negative 𝑛 equals one over 𝑏 to the 𝑛. All exponential functions are monotonic functions; that is, they always increase or always decrease. Specifically, if 𝑏 is greater than one, it is called the growth factor, and the function always increases. And if 𝑏 is between zero and one, it is called the decay factor, and the function always decreases.

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