Exponential and Logarithmic Functions

Exponential Functions

• Exponential functions model growth and decay phenomena across mathematics, physics, biology, and finance.

• An exponential function is defined as any function of the form:   $f(x) = b^x$   where $b > 0$, $b \neq 1$, $b$ is a constant base, and $x$ is a variable exponent.

• Distinction Between Exponential Functi5ons and Power Functions:

  • A power function has the form $f(x) = x^b$, where the base is variable and the exponent is constant.

  • An exponential function has the form $f(x) = b^x$, where the base is constant and the exponent is variable.

• Comparing $y = x^2$ and $y = 2^x$:

  • Both functions approach infinity as $x \rightarrow \infty$, but $y = 2^x$ eventually grows much more rapidly than $y = x^2$.

  • As $x \rightarrow -\infty$, $x^2 \rightarrow \infty$, whereas $2^x \rightarrow 0$. The horizontal line $y = 0$ is a horizontal asymptote for $y = 2^x$.

• Table of Numerical Comparison:

  • $x = -3$: $x^2 = 9$, $2^x = 1/8$

  • $x = -2$: $x^2 = 4$, $2^x = 1/4$

  • $x = -1$: $x^2 = 1$, $2^x = 1/2$

  • $x = 0$: $x^2 = 0$, $2^x = 1$

  • $x = 1$: $x^2 = 1$, $2^x = 2$

  • $x = 2$: $x^2 = 4$, $2^x = 4$

  • $x = 3$: $x^2 = 9$, $2^x = 8$

  • $x = 4$: $x^2 = 16$, $2^x = 16$

  • $x = 5$: $x^2 = 25$, $2^x = 32$

  • $x = 6$: $x^2 = 36$, $2^x = 64$

Definitions of Exponents Across Number Systems

• Positive Integer Exponents:

  • For a positive integer $x$, $b^x = b \cdot b \cdots b$ ($x$ factors of $b$).

• Negative Integer Exponents:

  • For a negative integer $-y$ (where $y$ is a positive integer), $b^{-y} = \frac{1}{b^y}$.

• Zero Exponent:

  • For any base $b > 0$, $b^0 = 1$

• Rational Exponents:

  • For rational numbers $x = \frac{p}{q}$, where $p$ and $q$ are integers and $q > 0$:     $b^{p/q} = \sqrt[q]{b^p}$

  • Example: $9^{3/2} = \sqrt{9^3} = \sqrt{729} = 27$

• Irrational Exponents:

  • An expression with an irrational exponent, such as 2^{\text{\sqrt{2}}}, is defined as the limit of values obtained using rational approximations of the exponent.

  • Rational approximations of $\sqrt{2} \approx 1.41421356...$ yield:

  • $x = 1.4 \implies 2^{1.4} \approx 2.639$

  • $x = 1.41 \implies 2^{1.41} \approx 2.65737$

  • $x = 1.414 \implies 2^{1.414} \approx 2.66475$

  • $x = 1.4142 \implies 2^{1.4142} \approx 2.665119$

  • $x = 1.41421 \implies 2^{1.41421} \approx 2.665138$

  • $x = 1.414213 \implies 2^{1.414213} \approx 2.665143$

  • As rational values of $x$ approach $\sqrt{2}$, $2^x$ converges to $2^{\sqrt{2}} \approx 2.665144$

Applications and Properties of Exponential Functions

• Bacterial Population Growth Example:

  • A population doubling every $4\text{\,hours}$ starting with $1000$ bacteria is described by:     $n(t) = 1000 \cdot 2^{t/4}$

  • Bacteria after $6\text{\,hours}$:     $n(6) = 1000 \cdot 2^{6/4} = 1000 \cdot 2^{3/2} \approx 2828\text{\,bacteria}$

  • Bacteria after $10\text{\,hours}$:     $n(10) = 1000 \cdot 2^{10/4} = 1000 \cdot 2^{5/2} \approx 5657\text{\,bacteria}$

  • Bacteria after $24\text{\,hours}$:     $n(24) = 1000 \cdot 2^{24/4} = 1000 \cdot 2^6 = 64,000\text{\,bacteria}$

• Evaluating Exponential Values:

  • For $f(x) = 3^x$:

  • $f(4) = 3^4 = 81$

  • $f(-2) = 3^{-2} = \frac{1}{9}$

• Domain, Range, and Behavior of $f(x) = b^x$:

  • Domain: $(-\infty, \infty)$

  • Range: $(0, \infty)$

  • When $b > 1$: Function is increasing on $(-\infty, \infty)$; $b^x \rightarrow \infty$ as $x \rightarrow \infty$, and $b^x \rightarrow 0$ as $x \rightarrow -\infty$

  • When $0 < b < 1$: Function is decreasing on $(-\infty, \infty)$; $b^x \rightarrow 0$ as $x \rightarrow \infty$, and $b^x \rightarrow \infty$ as $x \rightarrow -\infty$

Laws of Exponents

• For constants $a > 0$, $b > 0$, and all real numbers $x$ and $y$:

  1. $b^x \cdot b^y = b^{x+y}$

  2. $\frac{b^x}{b^y} = b^{x-y}$

  3. $(b^x)^y = b^{xy}$

  4. $(ab)^x = a^x b^x$

  5. $\frac{a^x}{b^x} = \left(\frac{a}{b}\right)^x$

• Simplification Examples:

  • Example A:     $\frac{(2x^{2/3})^3}{(4x^{-1/3})^2} = \frac{2^3 (x^{2/3})^3}{4^2 (x^{-1/3})^2} = \frac{8x^2}{16x^{-2/3}} = \frac{x^2 x^{2/3}}{2} = \frac{x^{8/3}}{2}$

  • Example B:     $\frac{(x^3 y^{-1})^2}{(xy^2)^{-2}} = \frac{(x^3)^2 (y^{-1})^2}{x^{-2} (y^2)^{-2}} = \frac{x^6 y^{-2}}{x^{-2} y^{-4}} = x^6 x^2 y^{-2} y^4 = x^8 y^2$

  • Example C:     $\frac{6x^{-3}y^2}{12x^{-4}y^5} = \frac{6}{12} x^{-3 - (-4)} y^{2 - 5} = \frac{1}{2} x^1 y^{-3} = \frac{x}{2y^3}$

The Number e and the Natural Exponential Function

• Compound Interest and Derivation of $e$:

  • Initial principal $P$ invested at annual interest rate $r$ compounded $n$ times per year for $t$ years:     $A(t) = P\left(1 + \frac{r}{n}\right)^{nt}$

  • Substituting $m = \frac{n}{r}$ gives:     $A(t) = P\left(1 + \frac{1}{m}\right)^{mrt}$

  • As compounding frequency approaches infinity ($n \rightarrow \infty \implies m \rightarrow \infty$):

  • $m = 10 \implies \left(1 + \frac{1}{10}\right)^{10} \approx 2.5937$

  • $m = 100 \implies \left(1 + \frac{1}{100}\right)^{100} \approx 2.7048$

  • $m = 1000 \implies \left(1 + \frac{1}{1000}\right)^{1000} \approx 2.71692$

  • $m = 10,000 \implies \left(1 + \frac{1}{10,000}\right)^{10,000} \approx 2.71815$

  • $m = 100,000 \implies \left(1 + \frac{1}{100,000}\right)^{100,000} \approx 2.718268$

  • $m = 1,000,000 \implies \left(1 + \frac{1}{1,000,000}\right)^{1,000,000} \approx 2.718280$

• Euler's Constant $e$:

  • The limit as $m \rightarrow \infty$ defines the mathematical constant $e$:     $e \approx 2.718281$

  • Notation introduced by Swiss mathematician Leonhard Euler in the 1720s.

• Continuous Compounding Formula:   $A(t) = P e^{rt}$

• The Natural Exponential Function:

  • Function defined by $f(x) = e^x$

  • Uniquely characterized as the only exponential function whose tangent line at $x = 0$ has a slope equal to $1$.

• Worked Examples of Continuous Compounding:

  • Example 1: Principal $P = \$500$, annual interest rate $r = 0.055$ ($5.5\%$):

  • Formula: $A(t) = 500 e^{0.055 t}$

  • After $10\text{\,years}$: $A(10) = 500 e^{0.055 \cdot 10} = 500 e^{0.55} \approx \$866.63$

  • After $20\text{\,years}$: $A(20) = 500 e^{0.055 \cdot 20} = 500 e^{1.1} \approx \$1502.08$

  • Example 2: Principal $P = \$800$, annual interest rate $r = 0.04$ ($4\%$):

  • Formula: $A(t) = 800 e^{0.04 t}$

  • After $5\text{\,years}$: $A(5) = 800 e^{0.04 \cdot 5} = 800 e^{0.2} \approx \$977.12$

Logarithmic Functions

• Since $f(x) = b^x$ (for $b > 0, b \neq 1$) is a continuous, strictly monotonic, one-to-one function with domain $(-\infty, \infty)$ and range $(0, \infty)$, it possesses an inverse function.

• Definition of Logarithmic Function:

  • The logarithmic function with base $b$, denoted $y = \log_b(x)$, is defined as:     $y = \log_b(x) \iff b^y = x$

  • Domain: $(0, \infty)$

  • Range: $(-\infty, \infty)$

• Key Evaluation Examples:

  • $\log_2(8) = 3$ because $2^3 = 8$

  • $\log_{10}\left(\frac{1}{100}\right) = -2$ because $10^{-2} = \frac{1}{100}$

  • $\log_b(1) = 0$ because $b^0 = 1$

• Inverse Cancellation Equations:

  • $\log_b(b^x) = x$ for all $x \in (-\infty, \infty)$

  • $b^{\log_b(x)} = x$ for all $x > 0$

• The Natural Logarithm:

  • Logarithm with base $e$, written as $\ln(x)$ or $\ln x$:     $\ln(x) = \log_e(x)$

  • Key identities: $\ln(e) = 1$, $\ln(e^3) = 3$, $\ln(1) = 0$

  • Inverse cancellation: $\ln(e^x) = x$ and $e^{\ln(x)} = x$

  • The graphs of $y = e^x$ and $y = \ln(x)$ are reflection symmetric across the line $y = x$.

Properties and Equations of Logarithms

• Properties of Logarithms:   For $a, b, c > 0$, $b \neq 1$, and any real number $r$:

  1. Product Property: $\log_b(ac) = \log_b(a) + \log_b(c)$

  2. Quotient Property: $\log_b\left(\frac{a}{c}\right) = \log_b(a) - \log_b(c)$

  3. Power Property: $\log_b(a^r) = r \log_b(a)$

• Solving Exponential Equations:

  • Equation 1: Solve $5^x = 2$     $\ln(5^x) = \ln(2) \implies x \ln(5) = \ln(2) \implies x = \frac{\ln(2)}{\ln(5)}$

  • Equation 2: Solve $e^x + 6e^{-x} = 5$     Multiply by $e^x$:     $e^{2x} + 6 = 5e^x \implies (e^x)^2 - 5(e^x) + 6 = 0$     Factor in $e^x$:     $(e^x - 3)(e^x - 2) = 0 \implies e^x = 3 \text{ or } e^x = 2 \implies x = \ln(3) \text{ or } x = \ln(2)$

  • Equation 3: Solve $3^{x/2} = 11$     $\ln\left(3^{x/2}\right) = \ln(11) \implies \frac{x}{2} \ln(3) = \ln(11) \implies x = \frac{2\ln(11)}{\ln(3)}$

• Solving Logarithmic Equations:

  • Equation 1: Solve $\ln\left(\frac{1}{x}\right) = 4$     $e^4 = \frac{1}{x} \implies x = \frac{1}{e^4} = e^{-4}$

  • Equation 2: Solve $\log_{10}(\sqrt{x}) + \log_{10}(x) = 2$     $\log_{10}\left(x^{1/2} \cdot x^1\right) = \log_{10}\left(x^{3/2}\right) = \frac{3}{2}\log_{10}(x) = 2$     $\log_{10}(x) = \frac{4}{3} \implies x = 10^{4/3} = 10\sqrt[3]{10}$

  • Equation 3: Solve $\ln(2x) - 3\ln(x^2) = 0$     $\ln(2x) - \ln\left(x^6\right) = 0 \implies \ln\left(\frac{2x}{x^6}\right) = \ln\left(\frac{2}{x^5}\right) = 0$     $\frac{2}{x^5} = e^0 = 1 \implies x^5 = 2 \implies x = \sqrt[5]{2}$

  • Equation 4: Solve $\ln(x^3) - 4\ln(x) = 1$     $3\ln(x) - 4\ln(x) = 1 \implies -\ln(x) = 1 \implies \ln(x) = -1 \implies x = e^{-1} = \frac{1}{e}$

Change-of-Base Formulas

• Formulas:   For $a > 0, b > 0$, $a \neq 1, b \neq 1$:

  1. Exponential Change of Base:      $a^x = b^{x \log_b(a)}$      If $b = e$, $a^x = e^{x \log_e(a)} = e^{x \ln(a)}$

  2. Logarithmic Change of Base:      $\log_a(x) = \frac{\log_b(x)}{\log_b(a)}$      If $b = e$, $\log_a(x) = \frac{\ln(x)}{\ln(a)}$

• Algebraic Proofs:

  • Proof 1: By inverse properties, $b^{\log_b(a^x)} = a^x$. Applying the power property to the exponent yields $b^{x \log_b(a)} = a^x$.

  • Proof 2: Let $u = \log_b(a)$, $v = \log_a(x)$, and $w = \log_b(x)$. By definition, $b^u = a$, $a^v = x$, and $b^w = x$. Then $b^w = x = a^v = (b^u)^v = b^{uv}$. Since exponential functions are one-to-one, $w = uv \implies v = \frac{w}{u}$, proving $\log_a(x) = \frac{\log_b(x)}{\log_b(a)}$.

• Change-of-Base Evaluation Examples:

  • $\log_3(7) = \frac{\ln(7)}{\ln(3)} \approx 1.77124$

  • $\log_4(22) = \frac{\ln(22)}{\ln(4)} \approx 2.2295$

• Application: The Richter Scale for Earthquakes:

  • Developed in 1935 by Charles Richter using a base-10 logarithmic scale.

  • Relative amplitude equation between two earthquakes of magnitudes $R_1$ and $R_2$ with wave amplitudes $A_1$ and $A_2$:     $R_1 - R_2 = \log_{10}\left(\frac{A_1}{A_2}\right)$

  • Magnitude 8 vs. Magnitude 7: $8 - 7 = 1 \implies \frac{A_1}{A_2} = 10^1 = 10$ (10 times more intense).

  • Magnitude 8 vs. Magnitude 6: $8 - 6 = 2 \implies \frac{A_1}{A_2} = 10^2 = 100$ (100 times more intense).

  • Magnitude 9.0 (Japan 2011) vs. Magnitude 7.3 (Haiti 2010):     $9.0 - 7.3 = 1.7 = \log_{10}\left(\frac{A_1}{A_2}\right) \implies \frac{A_1}{A_2} = 10^{1.7} \approx 50.1187$     The earthquake in Japan was approximately $50$ times more intense than the earthquake in Haiti.

  • Magnitude 8.5 vs. Magnitude 6.2:     $8.5 - 6.2 = 2.3 \implies \frac{A_1}{A_2} = 10^{2.3} \approx 199.53$ (approximately $200$ times more intense).

Hyperbolic Functions

• Hyperbolic functions represent combinations of $e^x$ and $e^{-x}$ that describe physical systems such as water waves, elastic membrane vibrations, and hanging cables (catenaries).

• Fundamental Definitions:

  • Hyperbolic Cosine:     $\cosh(x) = \frac{e^x + e^{-x}}{2}$

  • Hyperbolic Sine:     $\sinh(x) = \frac{e^x - e^{-x}}{2}$

  • Hyperbolic Tangent:     $\tanh(x) = \frac{\sinh(x)}{\cosh(x)} = \frac{e^x - e^{-x}}{e^x + e^{-x}}$

  • Hyperbolic Cosecant:     \csch(x) = \frac{1}{\sinh(x)} = \frac{2}{e^x - e^{-x}}

  • Hyperbolic Secant:     \sech(x) = \frac{1}{\cosh(x)} = \frac{2}{e^x + e^{-x}}

  • Hyperbolic Cotangent:     $\coth(x) = \frac{\cosh(x)}{\sinh(x)} = \frac{e^x + e^{-x}}{e^x - e^{-x}}$

• Pronunciation Guide:

  • $\cosh(x)$: "kosh" (rhymes with gosh)

  • $\sinh(x)$: "cinch"

  • $\tanh(x)$: "tanch"

  • \sech(x): "seech"

  • \csch(x): "coseech"

  • $\coth(x)$: "cotanch"

• Catenary Equation for Hanging Chains:   $y = a \cosh\left(\frac{x}{a}\right) + b$

• Fundamental Hyperbolic Identity Derivation:   $\cosh^2(t) - \sinh^2(t) = \left(\frac{e^t + e^{-t}}{2}\right)^2 - \left(\frac{e^t - e^{-t}}{2}\right)^2 = \frac{e^{2t} + 2 + e^{-2t}}{4} - \frac{e^{2t} - 2 + e^{-2t}}{4} = \frac{4}{4} = 1$   This identity is the hyperbolic analog of the trigonometric identity $\cos^2(t) + \sin^2(t) = 1$. The point $(\cosh(t), \sinh(t))$ lies on the unit hyperbola $x^2 - y^2 = 1$.

Hyperbolic Identities and Operations

• Complete List of Hyperbolic Identities:

  1. $\cosh^2(x) - \sinh^2(x) = 1$

  2. 1 - \tanh^2(x) = \sech^2(x)

  3. \coth^2(x) - 1 = \csch^2(x)

  4. $\sinh(x \pm y) = \sinh(x)\cosh(y) \pm \cosh(x)\sinh(y)$

  5. $\cosh(x \pm y) = \cosh(x)\cosh(y) \pm \sinh(x)\sinh(y)$

  6. $\sinh(2x) = 2\sinh(x)\cosh(x)$

  7. $\cosh(2x) = \cosh^2(x) + \sinh^2(x) = 2\cosh^2(x) - 1 = 1 + 2\sinh^2(x)$

  8. $\cosh^2(x) = \frac{\cosh(2x) + 1}{2}$

  9. $\sinh^2(x) = \frac{\cosh(2x) - 1}{2}$

• Identity Application Examples:

  • Example A: Simplify $\sinh(5x)\cosh(3x) - \cosh(5x)\sinh(3x)$     Using Identity 4: $\sinh(5x - 3x) = \sinh(2x)$

  • Example B: Given $\tanh(x) = \frac{3}{5}$, evaluate the remaining five hyperbolic functions:

  • 1 - \tanh^2(x) = \sech^2(x) \implies 1 - \frac{9}{25} = \frac{16}{25} = \sech^2(x)

  • Since \sech(x) > 0 for all real $x$, \sech(x) = \frac{4}{5}

  • \cosh(x) = \frac{1}{\sech(x)} = \frac{5}{4}

  • $\sinh(x) = \tanh(x)\cosh(x) = \left(\frac{3}{5}\right)\left(\frac{5}{4}\right) = \frac{3}{4}$

  • \csch(x) = \frac{1}{\sinh(x)} = \frac{4}{3}

  • $\coth(x) = \frac{1}{\tanh(x)} = \frac{5}{3}$

  • Example C: Simplify $\cosh(2x) + \sinh(2x)$     $\frac{e^{2x} + e^{-2x}}{2} + \frac{e^{2x} - e^{-2x}}{2} = \frac{2e^{2x}}{2} = e^{2x}$

Inverse Hyperbolic Functions

• Domain Restrictions for Invertibility:

  • $\sinh(x)$, $\tanh(x)$, \csch(x), and $\coth(x)$ are one-to-one across their entire domains.

  • $\cosh(x)$ and \sech(x) must be restricted to the domain interval $[0, \infty)$ to make them one-to-one and invertibility valid.

• Explicit Logarithmic Formulas and Domains:

  1. $\sinh^{-1}(x) = \ln\left(x + \sqrt{x^2 + 1}\right)$ for domain $(-\infty, \infty)$

  2. $\cosh^{-1}(x) = \ln\left(x + \sqrt{x^2 - 1}\right)$ for domain $[1, \infty)$

  3. $\tanh^{-1}(x) = \frac{1}{2}\ln\left(\frac{1 + x}{1 - x}\right)$ for domain $(-1, 1)$

  4. \csch^{-1}(x) = \ln\left(\frac{1}{x} + \frac{\sqrt{1 + x^2}}{|x|}\right) = \sinh^{-1}\left(\frac{1}{x}\right) for domain $x \neq 0$

  5. \sech^{-1}(x) = \ln\left(\frac{1 + \sqrt{1 - x^2}}{x}\right) = \cosh^{-1}\left(\frac{1}{x}\right) for domain $(0, 1]$

  6. $\coth^{-1}(x) = \frac{1}{2}\ln\left(\frac{x + 1}{x - 1}\right) = \tanh^{-1}\left(\frac{1}{x}\right)$ for domain $(-\infty, -1) \cup (1, \infty)$

• Derivation of $\sinh^{-1}(x)$:

  • Let $y = \sinh^{-1}(x) \implies x = \sinh(y) = \frac{e^y - e^{-y}}{2}$

  • Multiply by $2$: $2x = e^y - e^{-y} \implies e^y - 2x - e^{-y} = 0$

  • Multiply by $e^y$: $(e^y)^2 - 2x(e^y) - 1 = 0$

  • Apply quadratic formula in $e^y$:     $e^y = \frac{2x \pm \sqrt{4x^2 + 4}}{2} = x \pm \sqrt{x^2 + 1}$

  • Since $e^y > 0$ for all $y$ and $\sqrt{x^2 + 1} > x$, select the positive root: $e^y = x + \sqrt{x^2 + 1}$

  • Take natural logarithm: $y = \ln\left(x + \sqrt{x^2 + 1}\right)$

• Inverse Hyperbolic Evaluation Examples:

  • $\cosh^{-1}(2) = \ln\left(2 + \sqrt{2^2 - 1}\right) = \ln(2 + \sqrt{3})$

  • $\tanh^{-1}\left(\frac{1}{2}\right) = \frac{1}{2}\ln\left(\frac{1 + 1/2}{1 - 1/2}\right) = \frac{1}{2}\ln\left(\frac{3/2}{1/2}\right) = \frac{1}{2}\ln(3)$

  • $\sinh^{-1}(1) = \ln\left(1 + \sqrt{1^2 + 1}\right) = \ln(1 + \sqrt{2})$