Difficulty: Intermediate | Prerequisites: Inverse Functions, Logarithms, and Log Laws (the companion notes to this set)
This picks up where the log-laws notes left off. Once you know what logarithms are and how their rules work, the natural next question is: how do you use them to solve equations where the unknown is in the exponent? And a more subtle question: when does an inverse function even exist? The horizontal line test answers that, and it matters for every inverse you will meet in calculus, including the inverse trigonometric functions.
When the variable you need is stuck inside an exponent, taking the log of both sides brings it down where you can isolate it. Before building an inverse function, though, you need to check that the original function is one-to-one (passes the horizontal line test). Functions that fail, like sin(x), can still have inverses if you restrict their domain to a portion where the test passes.
One-to-one function
A function where every output comes from exactly one input. Formally, if f(a) = f(b) then a = b.
Think of it as: you can always work backwards from the output to a single, unambiguous input.
Horizontal line test
A visual check for whether a function is one-to-one. Slide a horizontal line (a ruler held flat) up and down the graph. If it never hits the curve in more than one place, the function is one-to-one and its inverse is also a function.
In simple terms: it is the vertical line test, rotated 90 degrees, applied to decide whether an inverse function can exist.
Vertical line test
The standard check for whether a graph represents a function at all. A vertical line dragged across the graph should hit the curve at most once at any position.
Domain restriction
The practice of limiting the inputs of a function so that it becomes one-to-one on that smaller domain, making an inverse possible. For example, restricting sin(x) to [-π/2, π/2].
Think of it as: zooming in on just the piece of the graph that passes the horizontal line test.
Arcsin, arccos, arctan (inverse trigonometric functions)
The inverse functions of sin, cos, and tan on their restricted domains. sin⁻¹(x) = arcsin(x), cos⁻¹(x) = arccos(x), tan⁻¹(x) = arctan(x). The -1 superscript means "inverse," not "reciprocal."
Change-of-base formula
A rule that converts any logarithm to a ratio of natural logs: log_b(x) = ln(x) / ln(b). Useful when your calculator only has an ln button, or when you need to solve for a variable in the exponent.
The basic idea: if the variable is trapped in the exponent, apply a log to both sides to bring it down.
Simple example (kindness acts). Suppose k = 2^x, where k is the number of kindness acts and x is the number of days. To find k when you know x, you just plug in. To find x when you know k, you need logs.
Start with k = 2^x.
Apply log base 2 to both sides: log_2(k) = log_2(2^x) = x.
So x = log_2(k). Plugging in k = 7 × 10⁹ (seven billion): x = log_2(7 × 10⁹) ≈ 32.7 days.
Using the natural log instead. You do not need a log base 2 button. Apply ln to both sides, then use the power rule.
ln(k) = ln(2^x) = x · ln(2).
Divide both sides by ln(2): x = ln(k) / ln(2).
Plugging in k = 7 × 10⁹ gives the same answer, ≈ 32.7 days.
This is the change-of-base formula in action: log_b(x) = ln(x) / ln(b).
A harder problem. Solve for x in y = A · 2^(x + B) where A = 7, B = -10, and y = 1,000.
Take ln of both sides: ln(y) = ln(A · 2^(x + B)).
Product rule: ln(y) = ln(A) + ln(2^(x + B)).
Power rule: ln(y) = ln(A) + (x + B) · ln(2).
Distribute: ln(y) = ln(A) + x · ln(2) + B · ln(2).
Isolate x: x · ln(2) = ln(y) - ln(A) - B · ln(2).
Divide: x = [ln(y) - ln(A) - B · ln(2)] / ln(2).
Plug in: x = [ln(1000) - ln(7) + 10 · ln(2)] / ln(2) ≈ 17.2.
Vertical line test (review). Slide a vertical line across the graph. If it hits the curve at most once everywhere, the graph is a function: every input x has at most one output y.
Horizontal line test. Slide a horizontal line up and down the graph. If it hits the curve at most once everywhere, the function is one-to-one: every output y comes from at most one input x.
Why this matters: an inverse function swaps inputs and outputs. If a function maps two different inputs to the same output (fails the horizontal line test), its inverse would map that output to two inputs, which means the inverse is not a function.
y = x² passes the vertical line test (it is a function), but fails the horizontal line test. For example, both x = -2 and x = 2 give y = 4.
The inverse of x² would need to send 4 back to a single input, but there are two candidates: -2 and 2. So the unrestricted inverse is not a function.
Fix: restrict the domain to x ≥ 0 (or x ≤ 0). On that restricted domain, x² is one-to-one and its inverse, √x, is a function.
The graph of e^x passes both the vertical and horizontal line tests. It is strictly increasing: each output comes from exactly one input.
Its inverse, ln(x), is therefore a well-defined function.
sin(x) passes the vertical line test (it is a function), but fails the horizontal line test because it oscillates. Infinitely many inputs produce the same output.
To define an inverse, restrict the domain of sin(x) to [-π/2, π/2]. On that interval, sin is one-to-one.
The inverse on this restricted domain is sin⁻¹(x) = arcsin(x). The -1 is not a reciprocal; it denotes the inverse function.
The same approach gives arccos(x) from cos(x) restricted to [0, π], and arctan(x) from tan(x) restricted to (-π/2, π/2).
You must restrict the domain before defining the inverse. This is a requirement, not an optional step.
General solution for y = A · 2^(x + B):
x = \frac{\ln(y) - \ln(A) - B\ln(2)}{\ln(2)}Change-of-base formula:
\log_b(x) = \frac{\ln(x)}{\ln(b)}Restricted domains for inverse trig functions:
arcsin(x) is the inverse of sin(x) on [-π/2, π/2]
arccos(x) is the inverse of cos(x) on [0, π]
arctan(x) is the inverse of tan(x) on (-π/2, π/2)
Solving for a variable in the exponent is how scientists determine the age of archaeological samples (radiocarbon dating) and how pharmacologists calculate the half-life of a drug. The horizontal line test, meanwhile, comes up any time you need to invert a sensor reading or undo a signal transformation in engineering: if the transformation is not one-to-one, you cannot uniquely recover the original signal.
Students often think the horizontal line test tells you whether something is a function. It does not. The vertical line test checks that. The horizontal line test checks whether the function is one-to-one, which is a separate and stricter condition.
Students sometimes assume that if a function fails the horizontal line test, it has no inverse at all. It does, once you restrict the domain to a one-to-one portion. The inverse exists on the restricted piece.
The notation sin⁻¹(x) trips people up. Students read it as 1/sin(x), which is actually csc(x). The superscript -1 on a function name means "inverse function," a completely different operation from taking a reciprocal.
When solving multi-step exponential equations, students sometimes forget to distribute after applying the power rule, or they isolate the wrong term. Write each algebraic step on its own line and check it before moving on.
⚠️ "Solve for x" problems where x is in the exponent are a staple of calculus exams. Know the steps: take ln of both sides, apply log laws, isolate x, divide.
⚠️ The horizontal line test is frequently asked as a conceptual question. Be prepared to sketch a graph and explain whether its inverse exists, and if not, what domain restriction would fix it.
⚠️ Know the restricted domains for arcsin, arccos, and arctan by heart. These will reappear when you differentiate inverse trig functions later in the course.
⚠️ The harder worked example (y = A · 2^(x + B)) is a template for multi-step problems. Practise the algebra until each step feels automatic.
True or false: a function that passes the horizontal line test always has an inverse that is also a function.
Fill in the blank: sin⁻¹(x) is the inverse of sin(x) restricted to the interval ______.
True or false: y = x² has an inverse function on the domain of all real numbers.
Fill in the blank: to solve 5^x = 200, take ln of both sides to get x = ______ / ______.
True or false: the -1 in cos⁻¹(x) means "one divided by cos(x)."
Answers: 1. True. 2. [-π/2, π/2]. 3. False (it fails the horizontal line test). 4. ln(200) / ln(5). 5. False (it means "inverse function").
Q: Solve for x: 3^x = 500.
A: Take ln of both sides: x · ln(3) = ln(500). Divide: x = ln(500) / ln(3) ≈ 5.66.
Q: Does the function f(x) = e^x have an inverse that is a function? Justify using the horizontal line test.
A: Yes. e^x is strictly increasing, so every horizontal line hits the curve at most once. It passes the horizontal line test, meaning its inverse ln(x) is a function.
Q: Why can you not define arcsin(x) as the inverse of sin(x) on the full real line?
A: sin(x) on the full real line fails the horizontal line test: each output between -1 and 1 corresponds to infinitely many inputs. The inverse would send one output to multiple inputs, violating the definition of a function. Restricting to [-π/2, π/2] makes sin one-to-one.
Q: Solve for x in y = 5 · 3^(x - 2) when y = 400.
A: Take ln of both sides: ln(400) = ln(5) + (x - 2) · ln(3). Isolate x: (x - 2) = [ln(400) - ln(5)] / ln(3) = ln(80) / ln(3) ≈ 3.99. So x ≈ 5.99.
Q: Give an example of a function that passes the vertical line test but fails the horizontal line test.
A: y = x². It is a valid function (one output per input), but f(-2) = f(2) = 4, so the horizontal line y = 4 hits the curve twice.
The inverse trig functions introduced here (arcsin, arccos, arctan) will come back prominently when you study their derivatives. The derivative of arcsin(x) is 1/√(1 - x²), and these show up in integration techniques (trig substitution) as well. The solving-for-x-in-the-exponent technique is also the foundation for working with exponential growth and decay models in applied calculus.
horizontal line test, vertical line test, one-to-one function, injective function, inverse function, arcsin, arccos, arctan, inverse trig functions, domain restriction, solving exponential equations, log both sides, change of base, natural log, ln, e^x, x², restricted domain, Calculus I Section 1.5, Brief Calculus ASU