Exponential and Logarithmic Functions

An exponential function has the form

The number $b$ is the base. For the parent function $f(x)=b^{x}$:

• Domain: all real numbers, $(-\infty,\infty)$.   Range: $(0,\infty)$.
• Horizontal asymptote $y=0$ (the $x$-axis).
• $y$-intercept $(0,1)$, since $b^{0}=1$; no $x$-intercept.
• If $b>1$ the graph grows; if $0<b<1$ it decays. Note $b^{-x}=(1/b)^{x}$.

The natural base is the irrational constant

The natural exponential function is $f(x)=e^{x}$; it behaves like any base $b>1$ (asymptote $y=0$, intercept $(0,1)$) and is the base used for continuous growth and decay. Any base can be rewritten with $e$: $\;b^{x}=e^{(\ln b)x}.$

For $b>0,\ b\neq 1$, the logarithm is the inverse of the exponential:

• Common log: $\log x=\log_{10}x$.   Natural log: $\ln x=\log_{e}x$.
• Domain of $\log_b x$ is $(0,\infty)$: you may only take a log of a positive number.
• Range: $(-\infty,\infty)$. Vertical asymptote $x=0$; $x$-intercept $(1,0)$.
• Inverse identities: $b^{\log_b x}=x$ and $\log_b b^{x}=x$.

For a shifted log $\log_b(x-h)$, the domain is $x>h$ and the asymptote moves to $x=h$.

Change of base: \( _b M=$\frac{\log M}{\log b}$=$\frac{\ln M}{\ln b}$.\)

• Exponential: isolate the power, then take a log of both sides (use $\ln$, or $\log_b$ directly). $\;b^{x}=c\Rightarrow x=\log_b c=\dfrac{\ln c}{\ln b}.$
• Logarithmic: condense to one log, rewrite in exponential form, then solve.
• Always check the domain: any solution making an argument $\le 0$ inside a log is extraneous and must be rejected.
• Give an exact answer (logs) and, if asked, an approximate decimal.

Change of base can be strung into a chain. Because

a product of logs whose bases and arguments interlock telescopes:

Two consequences worth memorizing:

• Reciprocal rule: $\log_a b=\dfrac{1}{\log_b a}$, since $\log_a b\cdot\log_b a=1$.
• Swap-the-power rule: $a^{\log_b c}=c^{\log_b a}$ (take $\ln$ of both sides to verify).

A sum of logs telescopes through the quotient rule rather than change of base:

Likewise $\displaystyle\sum_{k=2}^{n}\log_b\!\Big(1-\tfrac1{k^2}\Big)=\log_b\!\frac{n+1}{2n}$, because $1-\tfrac1{k^2}=\tfrac{(k-1)(k+1)}{k\cdot k}$ collapses as a product. The trick: turn each term into a ratio, multiply, and watch the middle die.

If you are told, e.g., $\log_2 3=a$ and $\log_2 5=b$, then every log whose argument factors over $\{2,3,5\}$ can be written in $a,b$ using the log laws --- and change of base for a different outer base.

• Factor the argument into powers of the “known” primes.
• Apply product/quotient/power rules to split it.
• $\log_2 2=1$ supplies any factor of $2$ for free.

For a different base, e.g. $\log_5(\cdot)$, first change to base $2$: $\log_5 x=\dfrac{\log_2 x}{\log_2 5}=\dfrac{\log_2 x}{b}$.

When an equation contains a base raised to $x$ and to $2x$ (or $-x$), substitute $u=b^{x}$ (with $u>0$) to expose a quadratic:

Solve for $u$, discard any $u\le 0$ (an exponential is never negative), then recover $x=\log_b u$. Equations mixing $e^{x}$ and $e^{-x}$ clear to a quadratic after multiplying through by $e^{x}$.

Treat the logs as unknowns. A system such as

is linear in $u=\log_2 x$ and $v=\log_2 y$: add/subtract to get $u=3,\ v=2$, hence $x=8,\ y=4$. When the equations mix a log-sum with an ordinary equation (e.g. $x+y=\text{const}$), condense the logs into one, convert to exponential form to get a product $xy$, and solve the resulting system by substitution. Always re-check each solution against every domain restriction.

Solve the corresponding equation to find boundary points, then respect monotonicity and the domain:

• $\log_b$ with $b>1$ is increasing: $\log_b u<\log_b v\iff 0<u<v$. Never forget $u>0$.
• $\log_b$ with $0<b<1$ is decreasing: the inequality flips.
• $b^{x}<c$ with $b>1$: take $\log_b$ to get $x<\log_b c$ (flip if $0<b<1$).

The domain condition is part of the answer --- intersect it with the inequality's solution.

Pure continuous growth $A=A_0e^{kt}$ is unbounded, so it only models the early phase of real populations. The logistic model

adds a ceiling: $P(0)=\dfrac{c}{1+a}$, and $P(t)\to c$ (the carrying capacity) as $t\to\infty$. Growth is nearly exponential while $P\ll c$, fastest at $P=\tfrac{c}{2}$ (the inflection point), then levels off. To fit or solve, isolate the exponential: