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Precalculus Exponential and Logarithmic Functions Flashcards
50 question-and-answer cards covering Exponential and Logarithmic Functions as it is examined in Precalculus. 24 of them are printed below, taken from across the deck — no signup, no paywall on the preview.
24 sample cards from the Exponential and Logarithmic Functions deck
Sampled from the end of the deck, so these are different cards from the ones shown on the syllabus page.
How would you compute $\log_{5}(30)$ on a calculator using change of base?
$\log_{5}(30) = \frac{\ln(30)}{\ln(5)} = \frac{\log(30)}{\log(5)} \approx 2.113$.
Expand $\log_{b}\!\left(\frac{x^{3}y}{z}\right)$ completely.
$3\log_{b}(x) + \log_{b}(y) - \log_{b}(z)$.
Expand $\ln\!\left(\sqrt{\dfrac{x}{y}}\right)$ using logarithm properties.
$\frac{1}{2}\big(\ln(x) - \ln(y)\big) = \frac{1}{2}\ln(x) - \frac{1}{2}\ln(y)$.
Condense $2\log(x) + \log(y) - 3\log(z)$ into a single logarithm.
$\log\!\left(\dfrac{x^{2}y}{z^{3}}\right)$.
When solving an exponential equation, what strategy applies when both sides can be written with the same base?
Rewrite as $b^{m} = b^{n}$, then set the exponents equal: $m = n$, and solve.
How do you solve an exponential equation like $3^{x} = 20$ when the bases cannot be matched?
Take a logarithm of both sides and use the power rule: $x = \frac{\ln(20)}{\ln(3)} \approx 2.727$.
Solve $e^{2x} = 15$ for $x$.
$2x = \ln(15)$, so $x = \frac{\ln(15)}{2} \approx 1.354$.
What is the general strategy for solving a logarithmic equation of the form $\log_{b}(x) = c$?
Rewrite in exponential form: $x = b^{c}$.
To solve $\log_{2}(x) + \log_{2}(x - 2) = 3$, what is the first step and the result?
Condense to $\log_{2}\big(x(x-2)\big) = 3$, rewrite as $x(x-2) = 2^{3} = 8$, giving $x^{2} - 2x - 8 = 0$, so $x = 4$ (reject $x = -2$).
When both sides of an equation are single logs with the same base, $\log_{b}(M) = \log_{b}(N)$, what can you conclude?
Use the one-to-one property: $M = N$, then solve (checking the domain).
What is an extraneous solution in logarithmic equations, and why do they arise?
A value obtained algebraically that fails the equation, arising because logarithm arguments must be positive; substituting it produces the log of zero or a negative number.
After solving a logarithmic equation, why must every candidate solution be checked?
Because a candidate may make an argument $\leq 0$ (undefined log), making it extraneous and requiring rejection.
Solve $\log(x) + \log(x - 3) = 1$ and identify any extraneous root.
$x(x-3) = 10 \Rightarrow x^{2} - 3x - 10 = 0 \Rightarrow x = 5$ or $x = -2$; reject $x = -2$ (extraneous), so $x = 5$.
State the compound interest formula for interest compounded $n$ times per year.
$A = P\left(1 + \frac{r}{n}\right)^{nt}$, where $P$ = principal, $r$ = annual rate, $n$ = compoundings/year, $t$ = years.
State the formula for continuously compounded interest.
$A = P e^{rt}$, where $P$ = principal, $r$ = annual rate, $t$ = time in years.
Using continuous compounding, how long does it take money to double at rate $r$?
Set $2P = Pe^{rt}$, so $t = \frac{\ln 2}{r}$.
State the general exponential growth and decay model, and how to tell growth from decay.
$A(t) = A_{0}e^{kt}$; growth if $k > 0$, decay if $k < 0$ ($A_{0}$ is the initial amount).
For radioactive decay $A(t)=A_0 e^{kt}$ with half-life $T$, how are $k$ and $T$ related?
$\frac{1}{2} = e^{kT}$, so $k = \frac{\ln(1/2)}{T} = -\frac{\ln 2}{T}$ (and $T = -\frac{\ln 2}{k}$).
A population grows so that $A(t) = A_{0}e^{kt}$. If it triples in time $t$, what expression gives $k$?
$3 = e^{kt} \Rightarrow k = \frac{\ln 3}{t}$.
State the logistic growth model and identify the carrying capacity.
$P(t) = \dfrac{c}{1 + a e^{-bt}}$; the carrying capacity is $c = \lim_{t \to \infty} P(t)$, the maximum sustainable value.
How does a logistic growth curve behave differently from pure exponential growth?
It grows nearly exponentially at first, then slows as it approaches the carrying capacity $c$, producing an S-shaped (sigmoidal) curve leveling at the horizontal asymptote $y = c$.
In the logistic model $P(t) = \dfrac{c}{1 + ae^{-bt}}$, what is the initial value $P(0)$?
$P(0) = \dfrac{c}{1 + a}$ (since $e^{0} = 1$).
State Newton's Law of Cooling as a formula.
$T(t) = T_{s} + (T_{0} - T_{s})e^{-kt}$, where $T_{s}$ = surrounding temperature, $T_{0}$ = initial temperature, $k > 0$.
According to Newton's Law of Cooling, what temperature does an object approach as $t \to \infty$, and why?
It approaches the surrounding (ambient) temperature $T_{s}$, because $e^{-kt} \to 0$, leaving $T(t) \to T_{s}$.
What this deck covers
The Exponential and Logarithmic Functions deck follows the Precalculus Exponential and Logarithmic Functions syllabus — 5 chapters and 16 topics — so questions land on material that is genuinely examinable rather than trivia around it. That works out to roughly 10.0 cards per chapter.
Answers are written to be recallable, not just readable — averaging about 91 characters, which is long enough to carry the reasoning and short enough to say out loud.
A deck like this earns its keep on the second and third pass. Read the syllabus first so you know the shape of the subject, then use the cards to find the specific facts that have not stuck.
Exponential and Logarithmic Functions flashcards FAQ
How many Exponential and Logarithmic Functions flashcards are in this Precalculus deck?
50 cards. This page previews 24 of them, sampled evenly across the deck so you can judge the difficulty before installing anything.
Are these Precalculus flashcards free?
Yes. The preview here is free to read with no signup, and the full 50-card deck is free inside the Examius app.
What do the Exponential and Logarithmic Functions cards cover?
They follow the Precalculus Exponential and Logarithmic Functions syllabus — 5 chapters and 16 topics — so the questions track what is actually examinable.
How should I use these flashcards?
Read the syllabus first so you know the shape of the subject, then drill the deck. Examius schedules each card with spaced repetition, so cards you keep missing come back sooner and ones you know drift further apart.