Exponential and Logarithmic Functions: Chapter Review

Vocabulary and notation

Growth factor $b$ and growth rate $r$

The rate is the fraction added per step; the factor is what you multiply by, with $b = 1 + r$. A factor of $0.85$ means a $15\%$ decrease.

Effective annual rate

The once-a-year rate matching a schedule, $\left(1 + \tfrac{r}{n}\right)^{n} - 1$ or $e^{r} - 1$, which is how two offers are compared.

Common logarithm $\log x$

Base $10$, written with no base at all; a calculator's $\log$ key.

$e$

The irrational number $2.718281828\ldots$, so every decimal you write for it is a rounding.

Argument of a logarithm

The quantity inside it. Positivity restricts the ARGUMENT, never the variable.

Extraneous solution

A candidate satisfying an equation you derived but not the one you were given: a step widened the domain.

Formulas and theorems

• Exponential function, and the constant-ratio test

  $$f(x) = a \cdot b^{x}, \qquad \frac{f(x+s)}{f(x)} = b^{s}$$

  Use when $a \neq 0$, $b > 0$, $b \neq 1$. Test a table by DIVIDING neighbours: with spacing $s$ the constant quotient is $b^{s}$, the base only when $s = 1$. Then $b^{0} = 1$, $b^{-n} = \tfrac{1}{b^{n}}$, $b^{p/q} = \sqrt[q]{b^{\,p}}$ and $b^{x} > 0$ are forced.

  e.g. Outputs $3, 12, 48$ at $x = 0, 2, 4$: $b^{2} = 4$, so $b = 2$.

• Graph of $y = a \cdot b^{\,x-h} + k$

  $$y = a \cdot b^{\,x-h} + k, \qquad b^{\,x-h} = b^{-h} \cdot b^{x}$$

  Use when Parent $y = b^{x}$: domain all reals, range $y > 0$, points $(0,1)$ and $(1,b)$, asymptote $y = 0$, increasing exactly when $b > 1$. Shifted: asymptote $y = k$, range $y > k$ for $a > 0$, $y < k$ for $a < 0$. Intercepts: substitute $x = 0$, or solve $y = 0$.

• Equating exponents

  $$b^{u} = b^{v} \iff u = v$$

  Use when $b > 0$, $b \neq 1$, with no restriction on $u$ and $v$. It fails at $b = 1$ ($1^{2} = 1^{5}$) and needs ONE shared base: $3^{x} = 5^{x-1}$ does not give $x = x - 1$.

• Definition of a logarithm

  $$y = \log_b x \iff b^{y} = x$$

  Text description

  The exponential and the logarithm are mirror images across the line y = x, so their points, intercepts, and asymptotes all swap coordinates.

  Use when $b > 0$, $b \neq 1$, $x > 0$. Inverse of $b^{x}$: range all reals, crosses $(1,0)$, passes $(b,1)$, vertical asymptote $x = 0$, increasing exactly when $b > 1$. Free: $\log_b 1 = 0$, $\log_b b = 1$. An unknown BASE gives $x^{k} = N$: keep only the positive root, never $1$.

  e.g. $\log_5 \tfrac{1}{25} = -2$, because $5^{-2} = \tfrac{1}{25}$.

• Cancellation laws

  $$\log_b\!\left(b^{t}\right) = t, \qquad b^{\log_b x} = x$$

  Use when The first for EVERY real $t$; the second only for $x > 0$, since otherwise $\log_b x$ does not exist.

  e.g. $9^{\log_3 5} = \left(3^{\log_3 5}\right)^{2} = 25$.

• Product, quotient, and power rules

  $$\begin{gathered} \log_b(MN) = \log_b M + \log_b N \\ \log_b\!\left(\tfrac{M}{N}\right) = \log_b M - \log_b N \\ \log_b\!\left(M^{p}\right) = p\log_b M \end{gathered}$$

  Use when $M > 0$ and $N > 0$ SEPARATELY, not merely $MN > 0$; $p$ any real. Also $\log_b \tfrac{1}{N} = -\log_b N$, and $\log_b\!\left(x^{2}\right) = 2\log_b |x|$ is good for every $x \neq 0$. The power rule never fires on $\left(\log_b M\right)^{p}$.

  e.g. $\log_2 6 + \log_2 \tfrac{2}{3} = \log_2 4 = 2$.

• Change of base

  $$\log_b M = \frac{\log_a M}{\log_a b} = \frac{\log M}{\log b} = \frac{\ln M}{\ln b}$$

  Use when $M > 0$, both bases positive and unequal to $1$, which keeps $\log_a b \neq 0$. Argument on top, old base underneath. Special case $\log_b M = \dfrac{1}{\log_M b}$, needing $M \neq 1$.

  e.g. $\log_8 32 = \dfrac{\log_2 32}{\log_2 8} = \dfrac{5}{3}$.

• Bracketing a logarithm

  $$\begin{gathered} b^{m} < x < b^{m+1} \\ \implies m < \log_b x < m+1 \end{gathered}$$

  Use when Needs $b > 1$; for $0 < b < 1$ the logarithm decreases and the inequalities reverse. In base $10$, a positive integer $N$ has $d$ digits exactly when $d - 1 \le \log N < d$.

  e.g. $32 < 50 < 64$ gives $5 < \log_2 50 < 6$; $\log\!\left(2^{40}\right) \approx 12.04$ means $13$ digits.

• Constant-percent model

  $$A(t) = P(1 + r)^{t} = P\,b^{\,t}$$

  Use when $P > 0$, $b > 0$. Grows when $b > 1$, holds at $b = 1$, decays toward zero without reaching it when $0 < b < 1$. Chained percent changes multiply their factors rather than adding.

  e.g. Falling $12\%$ a year gives $A = P(0.88)^{t}$, not $P(0.12)^{t}$.

• Compounding $n$ times a year

  $$A = P\left(1 + \frac{r}{n}\right)^{nt}$$

  Use when $r$ as a decimal ($6\%$ is $0.06$), $t$ in the same unit as $r$. Divide the rate AND multiply the time; $n = 1$ recovers $A = P(1+r)^{t}$.

  e.g. $2000$ dollars at $6\%$ monthly for $5$ years: $2000(1.005)^{60} \approx 2697.70$.

• Doubling time and half-life

  $$\begin{gathered} \text{doubling:}\ t = \frac{\log 2}{\log b} \\ \text{half-life:}\ t = \frac{\log \tfrac{1}{2}}{\log b} \\ A(t) = P \cdot 2^{\,t/d} \\ \text{or}\ \ A(t) = P\left(\tfrac{1}{2}\right)^{t/h} \end{gathered}$$

  Use when $b > 0$, $b \neq 1$; doubling needs $b > 1$, half-life $0 < b < 1$. $P$ cancels first, so neither time depends on the starting amount. Both logarithms in the decay quotient are negative, so the time is positive. Elapsed time goes in the NUMERATOR, over $d$ or $h$, and the two model forms are ALTERNATIVES, never equal.

  e.g. At $4\%$ a year, $t = \dfrac{\log 2}{\log 1.04} \approx 17.7$ years.

• Continuous growth model

  $$\left(1 + \frac{1}{n}\right)^{n} \longrightarrow e, \qquad A = Pe^{rt}$$

  Use when $r$ as a decimal, $t$ in the same unit. Those values rise with $n$ yet never reach $3$, so $Pe^{rt}$ is a CEILING no schedule beats. Decay uses a negative exponent coefficient.

  e.g. $2000$ dollars at $5\%$ for $8$ years: $2000e^{0.4} \approx 2983.65$, against $2976.26$ quarterly.

• Natural logarithm

  $$\begin{gathered} \ln x = \log_e x \\ e^{\ln x} = x \\ \ln\!\left(e^{x}\right) = x \end{gathered}$$

  Use when The first two need $x > 0$; the third holds for every real $x$. Base $e$, so every logarithm rule carries over, with $\ln 1 = 0$, $\ln e = 1$.

  e.g. $e^{0.04t} = 2$ gives $0.04t = \ln 2$, so $t \approx 17.33$.

• Exponentials outrun every polynomial

  $$2^{n} \ge n^{2} \quad \text{for every integer } n \ge 4$$

  Use when For $x \ge 0$, $y = 2^{x}$ and $y = x^{2}$ meet exactly twice, at $(2,4)$ and $(4,16)$, and the parabola leads only between them. Higher degree merely delays the crossing.

  e.g. $2^{10} = 1024$ while $10^{2} = 100$.

Problem types, step by step

Exam traps

• Trap Splitting the logarithm of a sum: writing $\log(A + B) = \log A + \log B$.

  Fix The product rule wants a PRODUCT inside; a sum inside does not break up at all. Test it: $\log_2(4+4) = 3$, while $\log_2 4 + \log_2 4 = 4$.

• Trap Reading $\dfrac{\log M}{\log N}$ as $\log M - \log N$.

  Fix A quotient of logarithms is a change of base, equal to $\log_N M$; the difference is $\log \tfrac{M}{N}$. Compare $\dfrac{\log 40}{\log 5} \approx 2.292$ with $\log 8 \approx 0.903$.

• Trap Skipping the check after combining logarithms, or discarding a candidate for being negative.

  Fix Combining widens the domain, so test each candidate in the original: every ARGUMENT must be positive. In $\log_2\!\left(x^{2} - 9\right) = 4$ the root $x = -5$ is genuine.

• Trap Applying $\log_b\!\left(x^{2}\right) = 2\log_b x$ without knowing the sign of $x$.

  Fix It is false for every $x < 0$, so it silently drops the negative half of a solution set. Use $2\log_b |x|$ when $x$ could be negative.

• Trap Putting the rate where the factor belongs.

  Fix "Grows $6\%$ a year" gives $b = 1.06$. The model $A = P(0.06)^{t}$ describes something losing $94\%$ of itself yearly.

• Trap Adding percent changes instead of multiplying the factors.

  Fix A $30\%$ rise then a $30\%$ fall is $1.30 \times 0.70 = 0.91$, a $9\%$ net loss rather than a wash, and $(1+r)(1-r) = 1 - r^{2} < 1$ for every $r \neq 0$.

• Trap Dividing an inequality by $\log b$ without flipping it, for a decaying model.

  Fix For $0 < b < 1$, $\log b$ is negative, so the inequality REVERSES: $(0.75)^{t} < 0.05$ gives $t > \dfrac{\log 0.05}{\log 0.75} \approx 10.41$, so the first whole year is $11$.

• Trap Reading $-2 \cdot 3^{x}$ as $(-6)^{x}$, or insisting every exponential has range $y > 0$.

  Fix An exponent binds only to the base beneath it, so evaluate $3^{x}$ then multiply. A negative $a$ makes every value negative: $y = -2^{x}$ stays below the axis.

• Trap Mixing up $\ln$ and $\log$, or feeding a percent straight into $e^{rt}$.

  Fix $\log 100 = 2$ but $\ln 100 \approx 4.605$. And $6\%$ enters as $r = 0.06$: $e^{6t}$ inflates one year's growth from about $1.06$ to over $400$.