17.3 Exponential & Logarithmic Functions, Base e & Logarithmic Properties

17.3 Exponential & Logarithmic Functions, Base e & Logarithmic Properties

The Natural Base e and Exponential Function Foundations

Exponential functions model dynamic phenomena where the rate of change of a quantity is directly proportional to the quantity itself. The general exponential function is defined as: $f(x) = a \cdot b^x$ where $a \neq 0$ is the initial value (or vertical stretch), $b > 0$ and $b \neq 1$ is the base, and the independent variable $x$ serves as the exponent.

• If $b > 1$, the function models exponential growth: it is strictly increasing on $(-\infty, \infty)$ with $\lim_{x \to \infty} b^x = \infty$ and $\lim_{x \to -\infty} b^x = 0$.
• If $0 < b < 1$, the function models exponential decay: it is strictly decreasing on $(-\infty, \infty)$ with $\lim_{x \to \infty} b^x = 0$ and $\lim_{x \to -\infty} b^x = \infty$.

In both cases with $a > 0$, the domain is $(-\infty, \infty)$, the range is $(0, \infty)$, the horizontal asymptote is the line $y = 0$, and the $y$-intercept is $(0, a)$.

The most important base in advanced mathematics is the natural base $e$ (Euler's number), discovered by Jacob Bernoulli when examining continuous compound interest. It is formally defined by the limit: $e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = \lim_{h \to 0} (1 + h)^{1/h} \approx 2.718281828459\dots$ The number $e$ is irrational and transcendental. Its mathematical significance stems from calculus: $f(x) = e^x$ is the unique non-trivial real-valued function that is its own derivative ($\frac{d}{dx}e^x = e^x$), meaning its instantaneous rate of change at any point equals its current value. Continuous growth or decay is modeled by $f(t) = a e^{kt}$, where $k > 0$ represents continuous growth rate and $k < 0$ represents continuous decay rate.

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Logarithmic Functions as Inverses and Their Geometric Properties

Because exponential functions $f(x) = b^x$ ($b > 0, b \neq 1$) are strictly monotonic, they pass the horizontal line test and are one-to-one (bijective). Consequently, they possess inverse functions called logarithmic functions: $y = \log_b(x) \iff b^y = x$ for $x > 0$, $b > 0$, and $b \neq 1$. A logarithm is fundamentally an exponent: $\log_b(x)$ is the exponent to which the base $b$ must be raised to produce the argument $x$.

• When the base is $e$, the logarithm is the natural logarithm, denoted $\ln(x) = \log_e(x)$.
• When the base is $10$, the logarithm is the common logarithm, denoted $\log(x) = \log_{10}(x)$.

Because inverse functions reflect across the line $y = x$, the domain and range of $f(x) = b^x$ swap for $g(x) = \log_b(x)$:

• Domain: $(0, \infty)$. The argument of a real logarithmic function must be strictly positive.
• Range: $(-\infty, \infty)$. A logarithm can take any real value.
• Vertical Asymptote: The line $x = 0$ ($y$-axis). For $b > 1$, as $x \to 0^+$, $\log_b(x) \to -\infty$.
• Intercept: $(1, 0)$, because $b^0 = 1 \implies \log_b(1) = 0$ for all valid bases $b$.

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Core Algebraic Properties of Logarithms and the Change of Base Formula

From the laws of exponents, logarithms inherit four operational algebraic properties that govern simplification and equation solving:

1. Product Property: The logarithm of a product is the sum of the logarithms: $\log_b(uv) = \log_b u + \log_b v, \quad (u > 0, v > 0)$
2. Quotient Property: The logarithm of a quotient is the difference of the logarithms: $\log_b\left(\frac{u}{v}\right) = \log_b u - \log_b v, \quad (u > 0, v > 0)$
3. Power Property: The logarithm of a power is the exponent times the logarithm of the base argument: $\log_b(u^p) = p \log_b u, \quad (u > 0, p \in \mathbb{R})$
4. Identity and Inversion Properties: $\log_b(1) = 0, \quad \log_b(b) = 1, \quad \log_b(b^x) = x, \quad b^{\log_b(x)} = x \quad (x > 0)$

To compute or manipulate logarithms with an arbitrary base $b$ in terms of natural or common logarithms, we utilize the Change of Base Formula: $\log_b(x) = \frac{\log_a x}{\log_a b} = \frac{\ln x}{\ln b} = \frac{\log_{10} x}{\log_{10} b}$

Critical Algebraic Fallacies to Avoid

On the FTCE exam, test makers construct distractors around classic algebraic misconceptions:

• $\log_b(u + v) \neq \log_b u + \log_b v$ (logarithms do not distribute over addition).
• $\frac{\log_b u}{\log_b v} \neq \log_b(u - v)$ (the quotient of two logarithms is not the logarithm of a difference; by change of base, it equals $\log_v u$).
• $(\log_b u)^p \neq p \log_b u$ (raising the entire logarithm to a power is not equivalent to raising the argument to that power).

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Solving Exponential and Logarithmic Equations and Extraneous Solutions

Solving transcendental equations requires navigating the inverse relationship between exponentials and logarithms:

• Exponential Equations ($b^u = c$):

  1. Isolate the exponential expression on one side.
  2. Take the natural logarithm of both sides: $\ln(b^u) = \ln(c)$.
  3. Apply the power property: $u \ln b = \ln c \implies u = \frac{\ln c}{\ln b}$.
  4. When bases can be matched ($b^u = b^v$), use the one-to-one property: $u = v$.

• Logarithmic Equations ($\log_b u + \log_b v = c$):

  1. Condense the logarithmic expressions into a single logarithm using product, quotient, or power properties: $\log_b(uv) = c$.
  2. Exponentiate both sides to eliminate the logarithm: $uv = b^c$.
  3. Solve the resulting algebraic equation for potential solutions.
  4. Check for Extraneous Solutions: Because algebraic steps like multiplying or squaring can expand the solution space, every potential solution must be substituted back into the original, uncondensed equation. Any value that produces a non-positive argument ($u \le 0$) is an extraneous solution and must be discarded.

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Continuous Mathematical Modeling: Finance, Decay, and Cooling

Transcendental functions govern dynamic physical and financial processes:

1. Continuous Compound Interest: When compound interest is calculated continuously ($m \to \infty$ compounding periods per year), the compound amount formula $A = P(1 + r/m)^{mt}$ converges to: $A(t) = P e^{rt}$ where $P$ is principal, $r$ is annual nominal interest rate, and $t$ is time in years. The doubling time is derived by setting $A = 2P \implies 2P = P e^{rt} \implies 2 = e^{rt} \implies t_{\text{double}} = \frac{\ln 2}{r}$.

2. Radioactive Decay and Half-Life: Radioactive decay follows the continuous exponential model: $N(t) = N_0 e^{-kt}$ where $N_0$ is initial quantity and $k > 0$ is the decay constant. The half-life $t_{1/2}$ is the time required for half of the radioactive nuclei to decay ($N(t_{1/2}) = \frac{1}{2}N_0$): $\frac{1}{2}N_0 = N_0 e^{-k t_{1/2}} \implies \frac{1}{2} = e^{-k t_{1/2}} \implies \ln\left(\frac{1}{2}\right) = -k t_{1/2} \implies -\ln 2 = -k t_{1/2}$ $t_{1/2} = \frac{\ln 2}{k} \iff k = \frac{\ln 2}{t_{1/2}}$

3. Newton's Law of Cooling: The temperature $T(t)$ of an object placed in an ambient medium with constant temperature $T_s$ changes at a rate proportional to the temperature difference: $T(t) = T_s + (T_0 - T_s)e^{-kt}$ where $T_0$ is initial temperature of the object and $k > 0$ is the cooling constant.

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Logarithmic Properties and Formulas Reference

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Worked Exemplar: Radioactive Decay and Half-Life Determination

Problem: A medical isotope utilized in radiation oncology has an initial mass of $80\text{ mg}$. After $24\text{ days}$, the sample has decayed to $10\text{ mg}$.

1. Determine the continuous decay constant $k$ and the half-life $t_{1/2}$.

2. Find the mass remaining after $20\text{ days}$.

3. Determine the time required for the sample to decay to $5%$ of its initial mass ($4\text{ mg}$).

4. Find Decay Constant and Half-Life: Using $N(t) = N_0 e^{-kt}$ with $N_0 = 80$ and $N(24) = 10$: $10 = 80 e^{-24k} \implies \frac{1}{8} = e^{-24k}$ Take the natural logarithm of both sides: $\ln\left(\frac{1}{8}\right) = \ln\left(e^{-24k}\right) \implies -\ln 8 = -24k \implies k = \frac{\ln 8}{24}$ Because $8 = 2^3$, $\ln 8 = 3\ln 2$. Thus: $k = \frac{3\ln 2}{24} = \frac{\ln 2}{8} \approx 0.08664\text{ day}^{-1}$ The half-life is: $t_{1/2} = \frac{\ln 2}{k} = \frac{\ln 2}{\frac{\ln 2}{8}} = 8\text{ days}$ (Verification: $80 \to 40 \to 20 \to 10$ is 3 half-lives; $3 \times 8\text{ days} = 24\text{ days}$.)

5. Mass Remaining After 20 Days: Substitute $t = 20$ and $k = \frac{\ln 2}{8}$: $N(20) = 80 e^{-\frac{\ln 2}{8} \cdot 20} = 80 e^{-2.5 \ln 2} = 80 e^{\ln(2^{-2.5})} = 80 \cdot 2^{-5/2}$ $N(20) = \frac{80}{\sqrt{2^5}} = \frac{80}{4\sqrt{2}} = \frac{20}{\sqrt{2}} = 10\sqrt{2} \approx 14.14\text{ mg}$

6. Time to Decay to 5% of Initial Mass (4 mg): Set $N(t) = 4$: $4 = 80 e^{-kt} \implies \frac{4}{80} = \frac{1}{20} = e^{-kt}$ $-\ln 20 = -kt \implies t = \frac{\ln 20}{k} = \frac{\ln 20}{\frac{\ln 2}{8}} = \frac{8 \ln 20}{\ln 2}$ Evaluating with $\ln 20 \approx 2.99573$ and $\ln 2 \approx 0.69315$: $t = 8 \cdot \frac{2.99573}{0.69315} = 8 \cdot 4.3219 \approx 34.58\text{ days}$ After approximately $34.58\text{ days}$, exactly $4\text{ mg}$ ($5%$) of the isotope remains.