Precalculus Exponential and Logarithmic Functions Syllabus

State the Product Rule for logarithms.

$\log_{b}(MN) = \log_{b}(M) + \log_{b}(N)$.

State the Quotient Rule for logarithms.

$\log_{b}\left(\frac{M}{N}\right) = \log_{b}(M) - \log_{b}(N)$.

State the Power Rule for logarithms.

$\log_{b}(M^{p}) = p\,\log_{b}(M)$.

What do the identities $\log_{b}(b^{x})$ and $b^{\log_{b}(x)}$ simplify to?

$\log_{b}(b^{x}) = x$ and $b^{\log_{b}(x)} = x$ (inverse property).

Evaluate $\log_{b}(1)$ and $\log_{b}(b)$ for any valid base $b$.

$\log_{b}(1) = 0$ and $\log_{b}(b) = 1$.

State the Change-of-Base Formula for evaluating $\log_{b}(x)$ with a new base $a$.

$\log_{b}(x) = \frac{\log_{a}(x)}{\log_{a}(b)}$.

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 $T(t) \to T_{s}$.