Inverses of Exponential Functions
Why this matters
Imagine your friend Priya shares a new music video with you. You love it and share it with two friends. An hour later, they each share it with two new friends. This pattern continues, with the number of shares doubling every hour. This is exponential growth! You can model it with f(x) = 2^x, where x is the number of hours.
It's easy to calculate that after 10 hours, there will be 2^10 or 1,024 new shares.
But what about the reverse question? How many hours would it take to reach 1 million shares? Now you have to solve 1,000,000 = 2^x. You're solving for the exponent. How do you get x out of that top-right position? You can't just divide by 2. You need a new kind of function—an inverse. That's exactly what we're going to build today.
Concept overview
flowchart TD
A[Exponential Function: y = b^x] --> B{Want to find the inverse};
B --> C[Step 1: Swap x and y];
C --> D[New Equation: x = b^y];
D --> E{Step 2: Solve for y};
E --> F[Use Logarithm Definition];
F --> G[Inverse Function: y = log_b(x)];
Core explanation
Let's get right to it. You have an exponential function like f(x) = b^x. Its job is to take an input, x, and use it as a power to raise a base, b. The inverse function needs to do the opposite: take the output and tell you what power was needed to get there.
Introducing the Logarithm
We call this inverse a logarithmic function, or "log" for short.
Here is the most important relationship you'll learn in this unit:
The exponential equation y = b^x is equivalent to the logarithmic equation x = log_b(y).
Read that log equation out loud like this: "x equals log base b of y." It answers the question, "What exponent x do I need to put on b to get y?"
Think of it as a "rearranging" of the exponential form. A helpful trick is the "base is always the base." In y = b^x, b is the base. In x = log_b(y), b is still the base, just written as a subscript.
The Inverse Relationship in Action
Because f(x) = b^x and g(x) = log_b(x) are inverse functions, they have some special properties.
1. Swapping Coordinates
If a point (s, t) is on the graph of an exponential function, then the point (t, s) is on the graph of its inverse logarithmic function. You just swap the x and y coordinates.
For example, let's use f(x) = 2^x.
- The point
(2, 4)is on the graph because2^2 = 4. - The point
(3, 8)is on the graph because2^3 = 8.
Now, for its inverse, g(x) = log₂(x):
- The point
(4, 2)must be on its graph. (Becauselog₂(4)asks "2 to what power equals 4?" The answer is 2.) - The point
(8, 3)must be on its graph. (Becauselog₂(8)asks "2 to what power equals 8?" The answer is 3.)
2. Composition of Inverses
When you compose a function with its inverse, they cancel each other out, leaving you with just x.
For f(x) = b^x and g(x) = log_b(x), this means:
f(g(x)) = b^(log_b(x)) = xg(f(x)) = log_b(b^x) = x
This is a powerful tool for solving equations.
The Visual Connection: Graphs
What happens when you take every point on a curve and swap its (x, y) coordinates? You get a reflection across the diagonal line y = x.
This is exactly what we see with exponential and logarithmic functions. The graph of y = log_b(x) is a perfect mirror image of the graph of y = b^x, reflected over the line y = x.
Take a look at the graph for y = 2^x and y = log₂(x).
(This is where you would see the visual: y=2^x in blue, y=log₂(x) in red, and the dashed line y=x. The point (2, 4) on the blue curve corresponds to (4, 2) on the red curve.)
How Their Growth Patterns are Inverses, Too
This is a subtle but critical point for the AP exam.
- Exponential functionshave additive inputs that lead to multiplicative outputs. With
f(x) = 2^x, if you increase the inputxby 1 (an additive change), the outputygets multiplied by 2 (a multiplicative change).x = 1 -> y = 2x = 2 -> y = 4(multiplied by 2)x = 3 -> y = 8(multiplied by 2)
- Logarithmic functionshave multiplicative inputs that lead to additive outputs. With
g(x) = log₂(x), if you multiply the inputxby 2 (a multiplicative change), the outputyincreases by 1 (an additive change).x = 2 -> y = 1x = 4 -> y = 2(increased by 1)x = 8 -> y = 3(increased by 1)
The Formal Definition and Rules
The general form of a logarithmic function is f(x) = a log_b(x).
ais a vertical stretch/compression factor.bis the base.
For logs to be well-defined, we have two rules for the base b:
b > 0(The base must be positive).b ≠ 1(The base cannot be 1).
Why? If b were 1, f(x) = 1^x would just be the horizontal line y=1. A horizontal line fails the horizontal line test, meaning it doesn't have a true inverse function. If b were negative, things like (-2)^(1/2) wouldn't be real numbers, creating gaps in the graph. So we stick to positive bases other than 1.
Worked examples
Finding the Inverse Function
Problem: Given the exponential function f(x) = 3^x, find its inverse function, f⁻¹(x).
Solution:
-
Start with the original function.
f(x) = 3^xLet's write this asy = 3^xto make the variables easier to track. -
Swap
xandyto define the inverse. This is the key step for finding any inverse function.x = 3^y -
Solve for the new
y. This is where we need our new tool, the logarithm. The equationx = 3^yasks, "y is the power I need to raise 3 to in order to get x." That is the exact definition of a logarithm!y = log₃(x) -
Write the final answer in function notation.
f⁻¹(x) = log₃(x)
Using Points and Graphs
Problem: The point (4, 625) is on the graph of g(x) = 5^x. What corresponding point must be on the graph of its inverse, h(x) = log₅(x)?
Solution:
-
Recall the property of inverse functions. If
(s, t)is a point on the original function, then(t, s)is the corresponding point on its inverse. -
Identify
sandt. In our given point(4, 625), we haves = 4andt = 625. -
Swap the coordinates. The new point will be
(t, s), which is(625, 4).
Let's check our work: The point (625, 4) should satisfy the inverse function h(x) = log₅(x).
- Plug it in:
4 = log₅(625) - Ask the question: "Does 5 to the power of 4 equal 625?"
5¹ = 55² = 255³ = 1255⁴ = 625- Yes, it does. Our answer is correct.
Additive vs. Multiplicative Change
Problem: An exponential function f(x) has f(2) = 100 and f(4) = 10,000. Let g(x) be the inverse of f(x). What is g(10,000)?
Solution:
-
Translate the
f(x)points into(x, y)pairs.f(2) = 100means the point(2, 100)is on the graph off(x).f(4) = 10,000means the point(4, 10,000)is on the graph off(x).
-
Use the inverse property for points. The function
g(x)is the inverse off(x). This means we can find points ong(x)by swapping the coordinates of the points fromf(x).- Since
(2, 100)is onf(x), the point(100, 2)must be ong(x). This meansg(100) = 2. - Since
(4, 10,000)is onf(x), the point(10,000, 4)must be ong(x).
- Since
-
Answer the question. The problem asks for the value of
g(10,000). Based on our work in step 2,g(10,000) = 4.
This example seems simple, but it's testing a deep understanding of the inverse relationship without even needing to know the base of the exponential function (which is 10 in this case, f(x) = 10^x).
Try it yourself
Ready to try it on your own? Grab a pencil and paper.
- 1ProblemLet
h(x) = 7^x.- Find the inverse function,
h⁻¹(x). - The point
(2, 49)is on the graph ofh(x). What point must be on the graph ofh⁻¹(x)?
- Find the inverse function,
- 2ProblemYou are looking at the graphs of
y = 10^xandy = log₁₀(x)on the same coordinate plane.- Which function passes through the point
(1, 0)? - Which function has a vertical asymptote at
x=0?
- Which function passes through the point
Practice — 8 questions
In simple terms, logarithmic functions are the "undo" button for exponential functions, helping you find the missing exponent in a growth problem.
- 2.10.A: Construct representations of the inverse of an exponential function with an initial value of 1.
- 2.10.A.1
- The general form of a logarithmic function is f(x) = a log_b x, with base b, where b > 0, b ≠ 1, and a ≠ 0.
- 2.10.A.2
- The way in which input and output values vary together have an inverse relationship in exponential and logarithmic functions. Output values of general-form exponential functions change proportionately as input values increase in equal-length intervals. However, input values of general-form logarithmic functions change proportionately as output values increase in equal-length intervals. Alternately, exponential growth is characterized by output values changing multiplicatively as input values change additively, whereas logarithmic growth is characterized by output values changing additively as input values change multiplicatively.
- 2.10.A.3
- f(x) = log_b x and g(x) = b^x, where b > 0 and b ≠ 1, are inverse functions. That is, g(f(x)) = f(g(x)) = x.
- 2.10.A.4
- The graph of the logarithmic function f(x) = log_b x, where b > 0 and b ≠ 1, is a reflection of the graph of the exponential function g(x) = b^x, where b > 0 and b ≠ 1, over the graph of the identity function h(x) = x.
- 2.10.A.5
- If (s, t) is an ordered pair of the exponential function g(x) = b^x, where b > 0 and b ≠ 1, then (t, s) is an ordered pair of the logarithmic function f (x) = log_b x, where b > 0 and b ≠ 1.
flowchart TD
A[Exponential Function: y = b^x] --> B{Want to find the inverse};
B --> C[Step 1: Swap x and y];
C --> D[New Equation: x = b^y];
D --> E{Step 2: Solve for y};
E --> F[Use Logarithm Definition];
F --> G[Inverse Function: y = log_b(x)];
Read what Saavi narrates
(Upbeat, warm intro music fades out)
Hi everyone, I'm Saavi, and welcome to Shrutam.
Let's talk about something you see all the time: things going viral. Imagine your friend Carlos in Chicago posts a really funny video. He sends it to two friends. An hour later, they each send it to two more friends. Then they send it to two more. That doubling every hour is classic exponential growth. We can write it as a function, f of x equals 2 to the power of x, where x is the number of hours.
It's easy to figure out how many people get the video after, say, 5 hours. But what if we ask the opposite question? How long would it take to reach 50,000 shares? Now our equation is 50,000 equals 2 to the power of x. The variable we want is stuck up in the exponent! How do we get it down?
This is where we need an "undo" button. For every exponential function, there's an inverse function that unwraps it. That inverse is called a logarithm. That's our whole focus today: understanding this powerful duo of exponential and logarithmic functions.
Let's walk through an example together. Suppose we have the function f of x equals 3 to the power of x. How do we find its inverse?
First, I'll rewrite it as y equals 3 to the power of x. Now, for any inverse, the big move is to swap the x and the y. So, we write x equals 3 to the power of y.
Okay, now we have to solve for y. This equation, x equals 3 to the power of y, is asking a question: "y is the exponent I need to put on a base of 3 to get an answer of x." That is literally the definition of a logarithm. So we can rewrite it in log form. The new y is equal to log base 3 of x. So, our inverse function is f-inverse of x equals log base 3 of x. See how that works? We just used the log to free the exponent.
Now, one of the most common mistakes I see every year is students mixing up the base. When you convert y equals b to the x, some students write x equals log base y of b. That's wrong because the base of the exponent, the 'b', must stay the base of the logarithm. Always remember: the base stays the base. It's the 'b' in the exponent, so it has to be the little subscript 'b' in the log.
So, the big takeaway is that logarithms are the key to unlocking that variable from the exponent. They are the inverse, the mirror image, the undo button. Keep that relationship in mind, and you'll be in great shape.
Thanks for listening. Keep practicing!
(Outro music fades in)
This is mixing up the function with its inverse. They are opposites, not the same thing. `3^x` is an exponential growth function, while `log₃(x)` is its much slower-growing logarithmic inverse.
Remember that `log_b(x)` is the *exponent* you need. It is the *answer* to the question "b to what power gives me x?". The function `b^x` is the process of *applying* that exponent.
The base of the exponent (`b`) must remain the base of the logarithm. In the incorrect version, the roles of the base and the argument are flipped.
Use the mantra: "The base stays the base." The `b` in `b^x` is the base, so it must be the subscript `b` in `log_b`. The exponent is always what the logarithm is equal to. `y = b^x` correctly becomes `x = log_b(y)`.
The domain of `f(x) = log_b(x)` is `x > 0`. Remember, the log function's inputs are the exponential function's outputs. Since `b^x` (for `b>0`) can never be negative or zero, you can't put a negative number or zero into a log function.
Before you even start a problem, check the domain. If you see `log_b(something)`, remember that `something` must be positive.
The inverse relationship is defined by swapping `x` and `y`, which geometrically corresponds to a reflection across the diagonal line `y = x`, not the axes.
Always visualize (or even lightly sketch) the line `y=x` as your "mirror." Every point on the exponential curve should have a partner on the log curve that is equidistant from this line.