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AP Precalculus · Unit 2 of 4

Exponential & Logarithmic Functions

25–40% of the exam13 lessons · 179 min78 terms
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What this unit covers

The topics below follow the published Precalculus course framework for Unit 2. This unit is worth 25–40% of the exam, so budget your time against that rather than against how long the unit takes to teach.

Growth & decayLog propertiesInversesModeling

Lessons in this unit

Formulas in Unit 2

Exponential model
f(x) = a · bˣ
a = value at x = 0 (initial amount). b > 1 grows, 0 < b < 1 decays. The percent change per step is (b − 1)·100%.
Logarithm laws
log(xy) = log x + log y · log(x/y) = log x − log y · log(xⁿ) = n·log x
Products become sums, quotients become differences, and exponents come out front as multipliers. Note that log(x + y) does NOT simplify.
Inverse composition identity
f(f⁻¹(x)) = x and f⁻¹(f(x)) = x
Composing a function with its inverse (in either order, on the appropriate domain) returns the original input. For example, 2^(log₂ x) = x.
Solving bˣ = c
x = log(c) / log(b)
Take the logarithm of both sides and divide. Any base works (common log or natural log) as long as you use it on both sides.
Explicit formulas
arithmetic: aₙ = a₁ + (n − 1)d · geometric: gₙ = g₁ · r^(n−1)
The exponent and the multiplier are (n − 1), not n, when indexing from a₁. Off-by-one here is the most common error in the topic.
Change of base
log_b(x) = ln(x)/ln(b) = log(x)/log(b) — any common base works
This is a ratio, not a difference. log(x)/log(b) is not log(x) − log(b), and confusing the two is the standard error.
Compound interest, periodic and continuous
A = P(1 + r/n)^(nt) · A = P·e^(rt) as n → ∞
r is the nominal annual rate as a decimal, n the compoundings per year, t the years. The second formula is the first one's limit, not a different model.
Converting between base and continuous rate
b^t = e^(kt) where k = ln b · equivalently b = e^k
k > 0 means growth (b > 1), k < 0 means decay (0 < b < 1). k is often called the continuous rate or the relative growth rate.
Half-life and doubling-time models
A(t) = A₀·(1/2)^(t/h) with half-life h · P(t) = P₀·2^(t/T) with doubling time T
The exponent t/h counts *how many half-lives have elapsed*. That reading makes the formula reconstructable from scratch.
Doubling time
T = ln 2 / ln(1 + r) for periodic growth · T = ln 2 / k ≈ 0.693/k for continuous growth
The familiar "rule of 70" is this formula in disguise: 100·ln 2 ≈ 69.3, so T ≈ 70/(percent rate) is a first-order approximation, good to a few percent for small rates.
Linearizing an exponential
y = a·bˣ ⟹ log y = (log b)·x + log a
Slope = log b, so b = 10^slope. Intercept = log a, so a = 10^intercept. Using ln instead, b = e^slope and a = e^intercept.
Which plot straightens which family
exponential y = a·bˣ → semi-log is linear · power law y = a·xⁿ → log-log is linear, with slope n
A quick discriminator: exponentials eventually beat every power law, and only the exponential has a constant ratio over equal x-steps.
Three logarithmic scales
pH = −log[H⁺] · dB = 10·log(I/I₀) · magnitude difference: I₁/I₂ = 10^(M₁ − M₂)
The minus sign in pH exists so that ordinary concentrations give positive pH values. I₀ for decibels is the threshold of hearing, 10⁻¹² W/m².
The three log rules used to condense
log a + log b = log(ab) · log a − log b = log(a/b) · n·log a = log(aⁿ)
Every one of these requires a > 0 and b > 0. The rules are identities only on the domain where both sides are defined — which is precisely why extraneous solutions appear.
Transformed asymptotes, domains and ranges
y = a·b^(x−h) + k → asymptote y = k, domain all reals, range y > k (if a > 0) · y = a·log_b(x − h) + k → asymptote x = h, domain x > h, range all reals
For a < 0 the exponential's range flips to y < k. The reflection is in the sign of a, not in k.
The log product rule as a transformation identity
log_b(cx) = log_b(c) + log_b(x) — a horizontal scaling equals a vertical shift
Similarly log_b(xⁿ) = n·log_b(x): raising the input to a power is a vertical stretch.
Logistic model
P(t) = L / (1 + a·e^(−kt)), with carrying capacity L
As t → ∞ the term a·e^(−kt) → 0 and P → L. As t → −∞ that term dominates and P → 0. So the curve runs from 0 up to L.
Inflection point of a logistic curve
occurs at P = L/2, at time t = ln(a)/k
Setting L/(1 + a·e^(−kt)) = L/2 requires a·e^(−kt) = 1, giving e^(kt) = a and t = ln(a)/k.

Every term in Unit 2

All 78 terms we publish for Exponential & Logarithmic Functions, with definitions. Reading them through is the fastest way to find the ones you cannot define — then drill those in cram mode until you can produce them without the prompt.

Change of base formula
log_b(x) = log(x)/log(b), using any consistent base. Needed because calculators offer only base 10 and base e.
Semi-log plot
Plotting log(y) against x linearises exponential data. A straight line on a semi-log plot is evidence the underlying model is exponential.
Exponential function
f(x) = ab^x with a ≠ 0 and b > 0, b ≠ 1. Grows by a constant FACTOR per unit input, unlike a linear function's constant difference.
Recognizing exponential data
Equal input intervals produce a constant RATIO of outputs. Constant differences indicate linear; constant second differences indicate quadratic.
The number e
The base for which the function equals its own rate of change, about 2.71828. Arises as the limit of (1 + 1/n)ⁿ.
Exponential growth vs decay
b > 1 gives growth, 0 < b < 1 gives decay. Equivalently, a positive exponent coefficient grows and a negative one decays.
Horizontal asymptote of an exponential
y = 0 for the parent function; a vertical shift by k moves it to y = k. Exponentials approach but never reach it.
Logarithm definition
log_b(x) = y means b^y = x. A logarithm is an exponent — the single most useful sentence in this unit.
Domain of a logarithm
x > 0 strictly. The log of zero or a negative number is undefined over the reals, which is why solutions must be checked after solving.
Product and quotient rules for logs
log(MN) = log M + log N and log(M/N) = log M − log N. Valid only when M and N are both positive.
Power rule for logs
log(Mᵖ) = p·log M. The rule that turns an unknown exponent into a coefficient, which is how exponential equations get solved.
Solving exponential equations
Take the log of both sides and apply the power rule, or rewrite both sides with a common base and equate exponents.
Solving logarithmic equations
Condense to a single log, exponentiate, then CHECK every solution — squaring and condensing can introduce values outside the domain.
Extraneous solutions
Values satisfying the transformed equation but not the original. Common with logs (negative arguments) and with squaring both sides.
Inverse relationship of exp and log
b^(log_b x) = x and log_b(b^x) = x. Their graphs are reflections across y = x, so the exponential's horizontal asymptote becomes the log's vertical one.
Log-log plot
Plotting log(y) against log(x) linearises a power function y = axⁿ, and the slope of that line is n.
Compound interest
A = P(1 + r/n)^(nt) for n compoundings per year; A = Pe^(rt) compounded continuously.
Half-life and doubling time
The constant time for a quantity to halve or double, independent of starting amount — the defining property of exponential change.
Logistic model
Growth that is nearly exponential at first and levels off at a carrying capacity, giving an S-shaped curve with an inflection point at half the capacity.
Newton's law of cooling
Temperature difference from the surroundings decays exponentially, so the object cools fastest when the difference is largest.
Exponent rules
b^m·b^n = b^(m+n), b^m/b^n = b^(m−n), (b^m)^n = b^(mn), b^(−n) = 1/b^n, b^0 = 1.
Fractional exponents
b^(m/n) is the nth root of b^m. Converting radicals to exponents makes the exponent rules available.
Why the base cannot be negative
A negative base would make b^(1/2) undefined over the reals, so exponential functions require b > 0.
Transformations of exponentials
y = a·b^(x − h) + k: k moves the horizontal asymptote, h shifts horizontally, and a reflects when negative.
Natural log
ln x is log base e. Its derivative properties make it the base of choice in calculus, and it obeys all the ordinary log rules.
Common modeling error
Fitting an exponential to data that only looks curved. Check whether ratios of successive outputs are constant before choosing the model.
Residuals
Observed minus predicted. A good model leaves residuals scattered randomly about zero; a pattern in the residuals means the model is the wrong shape.
Interpreting model parameters in context
In y = ab^t, a is the initial amount and b − 1 is the fractional change per unit time. Both need units and a sentence.
Exponential vs power function
In an exponential the variable is the exponent; in a power function it is the base. 2^x and x² behave completely differently for large x.
Inverse of an exponential model
Solving y = ab^t for t gives t = log(y/a)/log(b) — the function that answers "when does it reach this value".
Common difference
The fixed amount added at each step of an arithmetic sequence. Found by subtracting any term from the next, and constant throughout if the sequence really is arithmetic.
Common ratio
The fixed factor multiplied at each step of a geometric sequence. Found by dividing any term by the previous one.
Recursive vs explicit formula
Recursive gives the first term and a rule for the next from the current — it describes the process. Explicit gives the nth term directly — it lets you reach term 40 without computing 39 others.
Off-by-one in a sequence formula
Indexing from a₁ puts (n − 1) in the formula, not n: aₙ = a₁ + (n − 1)d and gₙ = g₁·r^(n−1). Reaching term 4 from term 1 takes three steps, not four.
Sequence as a function on the integers
A sequence is a function whose domain is the positive integers. Plotting aₙ = 4n + 3 gives points on the line y = 4x + 3 — the same rule, sampled.
Arithmetic sequence and linear function
Both add a constant per unit step, so every arithmetic sequence lies on a line whose slope is the common difference.
Geometric sequence and exponential function
Both multiply by a constant per unit step, so every geometric sequence lies on an exponential curve whose base is the common ratio.
Percentage depreciation
Losing 15% per year multiplies by 0.85 each year — geometric. Losing a fixed dollar amount per year is arithmetic instead, and the two diverge sharply over time.
Matching bases to solve
If both sides can be written as powers of the same number, equate the exponents. 2^(3x−1) = 32 = 2⁵ gives 3x − 1 = 5 with no logarithms and an exact answer.
Isolating the exponential first
Divide off any coefficient before taking logs. From 5·3^(2x) = 400, get 3^(2x) = 80 first; logging the product invites the error ln(5·3^u) = 5·ln(3^u).
Quadratic in an exponential
4^x − 5·2^x + 4 = 0 becomes u² − 5u + 4 = 0 under u = 2^x, because 4^x = (2^x)². Factor, solve for u, then back-substitute.
Undoing a substitution
After solving for u you must return to x. Reporting u values as answers is the standard trap: u = 4 with u = 2^x means x = 2, not x = 4.
Origin of e
The limit of (1 + 1/n)ⁿ as n grows — the ceiling that ever-more-frequent compounding approaches. e ≈ 2.718281828, and it comes from the mathematics rather than from a convention.
Nominal vs continuous rate
A nominal 6% compounded quarterly is not the same as a continuous 6%. The continuous rate k that matches a base b satisfies k = ln b, and for growth k is slightly smaller than the periodic percentage.
Converting a base to a continuous rate
b^t = e^(kt) with k = ln b. So (1.07)^t equals e^(0.0677t): a 7% annual increase corresponds to a continuous rate of about 6.77%.
Why e is the natural base
y = eˣ is the unique exponential whose rate of change at every point equals its own value. Any process whose rate is proportional to its current amount is simplest in base e.
Frequency of compounding
More frequent always yields more, but the gains shrink fast. At a nominal 5% over 20 years, annual gives about $2,653 per $1,000 and continuous about $2,718 — a spread of only 2.5%.
Half-life independence from amount
Setting A₀b^(t+h) = ½A₀b^t cancels A₀, leaving b^h = ½. So half-life depends only on the base — which is why it is quoted as a property of an isotope, not of a sample.
Counting half-lives
Compute t/h before anything else. If the answer says 3.4 half-lives but 60% of the material remains, something is wrong: 3.4 half-lives would leave under 10%.
Rule of 70
Doubling time ≈ 70/(percent growth rate). It is T = ln 2/ln(1 + r) approximated, since 100·ln 2 ≈ 69.3 — good to a few percent for small rates.
Doubling time from a continuous rate
T = ln 2/k ≈ 0.693/k. Doubling time and growth rate are inversely related, so doubling the rate halves the time.
Why nothing exponential reaches zero
b^t is positive for every real t, so exponential decay approaches zero without ever attaining it. A decay model never predicts that the last atom is gone.
Linearizing an exponential
Taking a log of y = a·bˣ gives log y = (log b)x + log a — a line. Straightness is easy to judge by eye; exponentiality is not.
Slope of a semi-log fit
Equals log b, so b = 10^slope (or e^slope if natural logs were used). A base-10 semi-log slope near 0.301 is the fingerprint of doubling.
Recovering a from a semi-log intercept
The vertical intercept is log a, so a = 10^intercept (or e^intercept for natural logs). An intercept of 1.70 gives a ≈ 50.
Semi-log vs log-log
Semi-log (log y vs x) straightens exponentials. Log-log (log y vs log x) straightens power laws, and there the slope IS the exponent. Which plot straightens the data identifies the family.
Reading a logarithmic axis
Equal vertical distances mean equal RATIOS. The gap from 1 to 10 is the same size as from 1,000 to 10,000. Reading it as linear misjudges values by orders of magnitude.
pH definition
pH = −log[H⁺], with the minus sign there so ordinary concentrations give positive values. Neutral is pH 7, corresponding to [H⁺] = 10⁻⁷ mol/L.
One pH unit
A factor of 10 in [H⁺]. So pH 3 is a thousand times as acidic as pH 6, not three times — reading a difference of 3 as "3 times" understates the gap by over 300-fold.
Decibel definition
dB = 10·log(I/I₀) with I₀ = 10⁻¹² W/m², the threshold of hearing. The factor of 10 means it takes 10 dB to multiply intensity by ten.
Adding decibels
Not additive in intensity. Two identical 70 dB sources give about 73 dB, because doubling intensity adds 10·log 2 ≈ 3 dB. Summing decibel values directly is meaningless.
Earthquake magnitude and amplitude
A 2-unit magnitude difference is 10² = 100 times the wave amplitude. Energy scales differently, roughly 10^1.5 per unit, so the same 2 units is about 1,000 times the energy.
Why perception is logarithmic
Loudness and brightness scale with equal ratios of intensity, not equal differences. Doubling a speaker's power does not sound twice as loud, which is why sound is quoted in decibels.
Condensing before exponentiating
Combine logarithms into one using the product, quotient and power rules, then remove the logarithm. Exponentiating a sum of logs term by term is invalid.
Why extraneous solutions appear
log a + log b requires a > 0 AND b > 0, but log(ab) only requires the product positive. Condensing widens the domain, so the algebra can produce roots the original equation rejects.
Checking in the original equation
Always the original, never the condensed form. The condensed version has a wider domain by construction, so it accepts the extraneous root without complaint.
One-to-one property of logarithms
log_b(A) = log_b(B) implies A = B. Often faster than exponentiating — but it does not enforce positivity, so the domain still has to be checked separately.
Asymptote of a transformed exponential
y = a·b^(x−h) + k has horizontal asymptote y = k. The asymptote travels with the vertical shift; h and a do not move it.
Range of a reflected exponential
For a > 0 the range is y > k; for a < 0 it is y < k. So y = −2·5^x + 7 has range y < 7, entirely below its asymptote.
Domain of a transformed logarithm
Set the argument greater than zero and solve. For log₃(2x − 6) + 4 the domain is x > 3. The vertical shift is irrelevant — only what is inside the logarithm restricts it.
Log of a scaled input
log(2x) = log 2 + log x, so horizontally compressing a logarithm is the same as shifting it up. No other elementary function has this property.
−log x versus log(−x)
The first reflects across the x-axis and keeps domain x > 0. The second reflects across the y-axis and has domain x < 0. Whether the sign is inside or outside decides the domain.
Anchor point of an exponential graph
y = bˣ always passes through (0, 1). Tracking that one point through a transformation catches sign and direction errors immediately.
Anchor point of a logarithmic graph
y = log_b x always passes through (1, 0). Its asymptote is the y-axis, and both features move together under transformation.
Carrying capacity
The ceiling L in P(t) = L/(1 + a·e^(−kt)). As t grows the exponential term vanishes and P approaches L without reaching it.
Logistic inflection point
At P = L/2, exactly half the carrying capacity, at time t = ln(a)/k. It is the moment of fastest growth — plenty reproducing and plenty of room still left.
Early logistic looks exponential
While P is far below L the ceiling is not yet binding, so the two models are nearly indistinguishable. This is why extrapolating early epidemic or product-launch data overpredicts.
Choosing logistic over exponential
Ask whether a ceiling exists and whether the data has reached enough of it to be visible. Systematic overprediction at the high end of an exponential fit is the tell.

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Questions about this unit

How much of the AP Precalculus exam is Unit 2?

Unit 2, Exponential & Logarithmic Functions, is worth 25–40% of the Precalculus multiple-choice section according to the published course framework. Across all 4 units that makes it one of the heaviest units on the exam, and worth front-loading.

What topics are covered in Precalculus Unit 2?

Exponential & Logarithmic Functions covers Growth & decay, Log properties, Inverses and Modeling. We publish 78 terms with definitions for this unit, all of them on this page.

How should I study Precalculus Unit 2?

Read the 13 lessons below first — about 180 minutes — then drill the 78 terms in cram mode until you can produce each definition from memory rather than just recognize it. Recognition is what makes a unit feel finished when it is not. Finish with practice questions and read the explanation for every one you get right by elimination as well as the ones you miss.

All 4 units of AP Precalculus

  1. Unit 1 · Polynomial & Rational Functions
  2. Unit 2 · Exponential & Logarithmic Functions
  3. Unit 3 · Trigonometric & Polar Functions
  4. Unit 4 · Functions Involving Parameters, Vectors & Matrices

Unit names, topics and exam weights follow the published College Board course framework for AP Precalculus. AP® is a trademark registered by the College Board, which does not endorse this site.