Logarithm functions

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Recall

Definition. A function f is a rule, relationship, or correspondence that assigns to each element x in a set D, x ∈ D (called the domain) exactly one element y in a set E, y ∈ E (called the codomain or range).

• The pair (x, y) is denoted as y = f(x) where: x is the independent variable (input) and y is the dependent variable (output). Often, both the domain D and codomain E are the set of real numbers ℝ or subsets of ℝ. A mathematical function can be thought of as a black box (or machine) that takes an input from its domain and produces exactly one output in its codomain. Inside the machine lives a specific rule (formula, procedure, or mapping) that dictates or tells you which output corresponds to each input, and a key property is uniqueness —each input maps to a single, deterministic output. No input can ever produce two different results (Figure E). The function f(x) = x² accepts any real number x (domain: ℝ) and returns exactly one non-negative value x² (output in codomain: [0,∞)). The input 3 always yields 9, never any other value.

• D is the domain, the set of all possible inputs. E is the codomain or range, the set of all possible outputs.

• Key property💡: Each input has exactly one output. (No input is assigned two different outputs — this is the vertical line test!)

• Examples: constant, f(x) = c, horizontal line, slope = 0; linear, f(x) = mx + b, straight line, constant slope m and y-intercept b; quadratic $f(x) = ax^2 + bx + c$, u-shaped or inverted U, opens up (a > 0) or down (a < 0), vertex at x = $\frac{-b}{2a}$, symmetry about vertical axis through vertex; polynomial, $f(x) = a_n x^n + \dots + a_0$, a smooth and continuous curve, n roots (counting multiplicity), end behaviour determined by its leading term $a_n x^n$; exponential function, $f(x) = a \cdot b^x, a \ne 0, b \gt 0$, rapid growth (b > 1) or decay (0 < b < 1); trigonometric functions, $\sin(x), \cos(x), \tan (z)$ oscillatory, periodic behavior (period 2π for sin/cos, π for tan), sin and cos are bounded between -1 and 1, but tan is unbounded; step function $f(x) = \lfloor x \rfloor$, greatest integer ≤ x, constant on intervals [n, n+1), jumps at integers, its graph is a staircase shape; absolute value f(x) = |x|, V-shaped graph, slope changes at 0.

• Functions can be expressed in multiple forms, each useful in different contexts: verbal description, table of values (list of pairs), algebraic formula, graph, piecewise definition, recursive definition, parametric or integral form, and series representation.

• Evaluating a function means finding or computing the output value f(x) for a given input value x. f(x) = $x^2-2x +4, f(2) = 2^2 -2\cdot 2 + 4 = 4 - 4 + 4 = 4, f(0) = 0^2 -2\cdot 0 + 4 = 4, f(1) = 1^2 -2\cdot 1 + 4 = 1 -2 +4 = 3$

• The x-intercept is any point on the graph that intersects or crosses the x-axis. In other words, it is the value of x when the function (y-coordinate or y-value) is zero. The y-intercept is the point where the graph intersects or crosses the y-axis. y-coordinate of the point whose x-coordinate is 0, e.g., 2x - 3y = 6. x-intercept: set y = 0 → 2x = 6 ⇒ x=3, so (3, 0). y-intercept: set x = 0 → −3y = 6 ⇒ y = −2, so (0, −2).

• f is said to have a local or relative maximum at c if there exists an interval (a, b) containing c ($c \in (a, b)$) such that f(c) ≥ f(x) $∀ x \in (a, b) ∩ D$. f is said to have a local or relative minimum at c if there exists an interval (a, b) containing c ($c \in (a, b)$) such that f(c) ≤ f(x) $∀ x \in (a, b) ∩ D$.💡Local extrema can only occur where the function stops rising or falling —either because the derivative is zero, the derivative doesn't exist, or you're at the edge of the domain.

• First Derivative Test. Let f be differentiable on an open interval containing c, except possibly at c itself, and let c be a critical point (so f′(c) = 0 or f′ is undefined). Then:

1. If f′(x) > 0 for x just to the left of c and f′(x) < 0 for x just to the right of c, then f(c) is a local maximum.
2. If f′(x) < 0 for x just to the left of c and f′(x) > 0 for x just to the right of c, then f(c) is a local minimum.
3. If f′(x) does not change sign at c (stays positive or stays negative), then f(c) is not a local extremum.

• Second Derivative Test. If the first derivate of a function at a point c is zero (f'(c) = 0) and the second derivate at this point exists, then: (i) If f''(c) > 0, then f(c) is a local minimum. (ii) If f''(c) < 0, then f(c) is a local maximum. (iii) If f''(c) = 0, the test is inconclusive.
• Vertical asymptotes are vertical lines (perpendicular to the x-axis) of the form x = a (where a is a constant) near which the function grows without bound when x approaches a from at least one side. The line x = a is a vertical asymptote of f if at least one of the one-sided limits is infinite: $\lim_{x \to a^{-}}f(x)=\pm\infty$ or $\lim_{x \to a^{+}}f(x)=\pm\infty$.
• Horizontal asymptotes are horizontal lines (y = c, parallel to the x-axis) that the graph of the function approaches as x grows very large (positively or negatively). In other words, the function values get arbitrarily close to c as long as x is sufficiently large or more formally: $\lim_{x \to \infty}f(x)=c$ and/or $\lim_{x \to -\infty}f(x)=c$.
• A line y = m+b is an oblique (slant) asymptote as $x \to \infty$ if $\lim_{x\to\infty} [f(x) - (mx+b)] = 0.$ This means that the curve of f(x) gets more and more closer to the line y = mx + b as x grows large. For an asymptote y = mx + b as $x \to \infty, m = \lim_{x\to\infty} \frac{f(x)}{x}, b = \lim_{x\to\infty} \bigl(f(x) - m x\bigr)$, provided both limits exist.

Definition. An exponential function is a mathematical function of the form f(x) = b·a^x, where the independent variable is the exponent, and a and b ($b\neq 0$, we want only nontrivial functions) are constants. a is called the base of the function and it should be a positive real number (a > 0).

For a = 1, we could treat it separately as a degenerate exponential, which is just the constant function f(x) = b. We normally want “exponential” to mean “genuinely growing or decaying” rather than just a constant function.

Examples are 2^x, 7^x, and (¹⁄₄)^x. There are three kinds of exponential functions depending on whether a > 1, a = 1, or 0 < a < 1 -Figure 1.a.- The more important example is the natural exponential function, f(x) = e^x where e is the Eulers Number, a mathematical constant approximately equal to 2.71828.

$\frac{d}{dx}e^x=e^x$ It’s the unique (up to scaling) function equal to its own derivative. $e=\lim _{n\rightarrow \infty }\left( 1+\frac{1}{n}\right) ^n$. Furthermore, $a^x=e^{x\ln (a)}$, so every exponential function is just a rescaled version of e^x.

1. Case a > 1. It can be observed that as the exponent increases, the curve get steeper. As x increases, f(x) heads to infinity, $\lim_{x \to ∞} a^x = ∞$. As x decreases, f(x) heads to zero, $\lim_{x \to -∞} a^x = 0$. It is strictly increasing ((e^x)’ = e^x > 0, (a^x)’ = a^x·ln(a) > 0) and has a horizontal asymptote along the x-axis.

   It is important to realize that as x approaches negative infinity, the results become very small but never actually attain zero, e.g., 2^-5 ≈ 0.03125, 2^-15 ≈ 0.00003052. Besides, the base of an exponential function determines the rate of growth or decay. For a > 1, the larger the base, the faster the function grows (Figure iii).

    
2. Case 0 < a < 1. As x increases, f(x) heads to zero, $\lim_{x \to ∞} a^x = 0$. As x decreases, f(x) heads to ∞, $\lim_{x \to -∞} a^x = ∞$. It is strictly decreasing and has a horizontal asymptote along the x-axis. $a^x=\left( \frac{1}{b}\right) ^x=b^{-x}, \text{ where } b = \frac{1}{a}>1$, decay is just growth reflected in the y-axis.
3. When a = 1, the graph is a horizontal line, y = 1. This is technically an exponential function, but it doesn’t model growth or decay, so we usually exclude a = 1 when talking about exponential growth/decay.  

Loosely speaking, the inverse function of a function f is a function that undoes the operation of the original function. If a function f maps each element x in its domain to a unique element y in its codomain, then the inverse function, denoted by $f^{-1}$, maps each such element y back to its corresponding x.

Formally, if f(x) = y, then $f^{-1}(y) = x$. This relationship can be summarized by the identities $f^{-1}(f(x)) = x, \forall x \in \mathrm{Dom}(f),$ and $f(f^{-1}(y)) = y, \forall y \in \mathrm{Dom}(f^{-1}).$

Logarithm function

In mathematics, the logarithm is the inverse operation of exponentiation. We call the inverse of a^x the logarithmic function with base a, that is, log_ax = y ↔ a^y = x. In words, the logarithm of a number x to the base a, log_ax, is the exponent to which a must be raised to produce x, e.g., log₄(64) = 3 because 4³ = 64, log₂(16) = 4 because 2⁴ = 16, log₈(512) = 3 because 8³ = 512, but log₂(-3) is undefined in the real numbers (logs require positive arguments).

The logarithm function is a fundamental mathematical tool that serves as the inverse of exponential growth, transforming multiplicative relationships into additive ones. Its applications span virtually every scientific discipline: in finance, logarithms calculate compound interest; in biology, they describe population dynamics and drug elimination from the body; in physics, they quantify radioactive decay and measure earthquake magnitudes. This remarkable versatility makes logarithms essential for solving real-world problems where quantities change at rates proportional to their current values.

Although the base of a logarithm can be any number, the most commonly used bases are 10 and e:

• The logarithm with base e is the natural logarithm. The natural logarithm is the inverse of e^x, that is, ln(x) = y ↔ e^y=x where e is the Eulers Number, a mathematical constant approximately equal to 2.71828.
• The logarithm with base 10 is the common logarithm. $log_{10}(x)$ often written simply log(x) in some contexts (engineering / log tables).

Domain, range, intercepts and asymptotes

• Domain: $(0, \infty)$. The logarithm’s domain consists of real positive numbers.
• Range: $(-\infty,\infty)$. Its range is ℝ.
• x-intercept: (1, 0) because $\log_a(1) = 0$.
• y-intercept: none, logarithm is not defined at x = 0.
• Vertical asymptote. It has a vertical asymptote along the y-axis at x = 0, i.e., the graph “blows up” as $x\to0^+$ (figure i and ii).

Monotonicity and limits

Let a > 0 with $a \neq 1$.

1. If a > 1, e.g. a = 2, e, 10. $\log_a(x)$ is strictly increasing. As x increases, f(x) heads to infinity, $\lim_{x \to ∞} log_a(x) = \infty$. $\lim_{x \to 0+} log_a(x) = -\infty$.

   The bigger the logarithm base, the graph approaches the asymptote x = 0 quicker (Figure iii and iv).

2. If 0 < a < 1, $\log_a(x)$ is strictly decreasing. As x increases, f(x) heads to minus infinity, $\lim_{x \to ∞} log_a(x) = -\infty$. As x decreases, f(x) heads to infinite, $\lim_{x \to 0+} log_a(x) =\infty$.

Basic identities

1. Product Rule. log_a(xy)=log_a(x) + log_a(y).

   Proof: Let $u = \log_a(x), v = \log_a(y)$. Then, $a^{u} = x, a^{v} = y.$ So $xy = a^{u}a^{v} = a^{u+v}$. Taking $\log_a$ gives $\log_a(xy) = u + v = \log_a(x)+\log_a(y)$.

2. Quotient Rule. log_a(^x⁄_y) = log_a(x) - log_a(y).
3. Power Rule. log_a(x^r) = r·log_a(x) ($r \in \mathbb{R}$).
4. Log of 1 and base. log_a(1) = 0 ↭ a⁰ = 1. log_a(a) = 1 ↭ a¹ = a.
5. Change of base formula. $log_b(x)=\frac{log_a(x)}{log_a(b)}$. In particular, $\log_{b}(x) = \frac{\ln(x)}{\ln(b)}$.

   Let y = log_b(x) ↭ b^y = x and z = log_a(x) ↭ a^z = x. Then, z = log_a(x) = $log_a(b^y) = y·log_a(b) ⇒ log_b(x) = y = \frac{z}{log_a(b)} = \frac{log_a(x)}{log_a(b)}$

6. Inversion identities / inverse property. $a^{\log_a(x)} = x, \log_a(a^{t}) = t.$

Differentiation, integration, and series expansion

Derivative of the Natural Logarithm

For x > 0, let y = ln(x) ⇒[By definition] e^y = x ⇒[Differentiating both sides of this equation with respect to x. Recall (e^x)’=e^x] $\frac{d}{dx}(e^y) = \frac{d}{dx}(x) \implies[\text{Applying the chain rule}] e^y\frac{dy}{dx} = 1 \implies[\text{Sove for } \frac{dy}{dx}] \frac{dy}{dx} = \frac{1}{e^y} = \frac{1}{x}.$ Thus, $\boxed{\frac{d}{dx}ln(x) = \frac{1}{x}}, \forall x \gt 0$

Derivative of the Logarithm with Base a

For x > 0, a > 0, and a ≠ 1, let y = log_a(x) ⇒[By definition] a^y = x ⇒ ln(a^y) = ln(x) ⇒[Power Rule. log_a(xⁿ) = n·log_a(x).] y·ln(a) = ln(x) ⇒ y = $\frac{ln(x)}{ln(a)}⇒$[¹⁄_ln(a) is a constant] $\frac{dy}{dx} = \frac{1}{ln(a)}\cdot \frac{1}{x} = \frac{1}{xln(a)}$. Thus, $\boxed{\frac{d}{dx}log_a(x) = \frac{1}{xln(a)}}, \forall x \gt 0$

Integrals and series expansion

• Natural logarithm, $\boxed{\int \ln(x) dx = x\ln(x) - x + C}$. Differentiate xln(x) - x to verify: $\frac{d}{dx}(xln(x) -x) = ln(x) + x\cdot \frac{1}{x} -1 = ln(x).$
• Logarithm with base a (a > 0, $a \ne 1$): $\boxed{\int \log_a(x) dx = \frac{x\ln(x) - x}{\ln(a)} + C}$. $\int \log_a(x) dx = \frac{1}{ln(a)} \int ln(x)dx = \frac{xln(x)-x}{ln(a)} + C$.
• For |t| < 1, the Taylor series of ln(1 + t) about t = 0 is $\boxed{\ln(1+t)=t-\frac{t^2}{2}+\frac{t^3}{3}-\frac{t^4}{4}+\cdots;}$ This gives good approximations for $\ln$ near 1, radius of convergence 1, excluding -1.

For |u| < 1, we have the geometric series: $\frac{1}{1+u} = 1 - u + u^2 -u^3 \cdots$. This is just the usual geometric series applied with x = -u.

$ln(1+t) = \int_{0}^{t} \frac{du}{1+u}$. This follows from these statements:

1. Derivative property: $\frac{d}{dt} \ln(1+t)=\frac{1}{1+t}$ (by the chain rule, since the derivative of ln(x) is 1/x).
2. Initial condition: $\ln(1 + 0) = \ln(1) = 0$.
3. By the Fundamental Theorem of Calculus any function F(t) satisfying $F'(t) = \frac{1}{1+t}$ and F(0) = 0 must be given by the integral $\int_{0}^{t} \frac{du}{1+u}$. Hence, ln(1 + t) is precisely such a function.

Since the geometric series converges uniformly on |u| < 1, we are allowed to integrate term-by-term: $\int_{0}^{t}\frac{du}{1+u} = \int_{0}^{t} (1 - u + u^2 -u^3)du = t-\frac{t^2}{2}+\frac{t^3}{3}-\frac{t^4}{4}+\cdots$

The geometric series converges only for |u| < 1. Since the integration is done over the interval from 0 to t, we need the entire path to stay inside the disk |u| < 1. That requires: |t| < 1. The endpoint t = -1 is excluded because $\ln(1+t)$ has a singularity at t =-1. Thus, the radius of convergence is exactly 1.

Examples:

• f(x) = $ln(\frac{x}{x+1})$

f(x) = $ln(\frac{x}{x+1}) = ln(x)-ln(x+1)$

f’(x) = $\frac{1}{x}-\frac{1}{x+1} = \frac{x+1}{x(x+1)} -\frac{x}{x(x+1)} = \frac{1}{x(x+1)}$

• f(x) = $ln(\frac{x^2·sin(x)}{2x+1})$

f(x) = $ln(\frac{x^2·sin(x)}{2x+1}) = ln(x^2·sin(x))-ln(2x+1) = ln(x^2)+ln(sin(x))-ln(2x+1)= 2ln(x) +ln(sin(x))-ln(2x+1)$

f’(x) = $\frac{2}{x} + \frac{cos(x)}{sin(x)} -\frac{2}{2x+1} = \frac{2}{x} + cot(x) -\frac{2}{2x+1}$

• f(x) = x^x.

y = x^x ⇒ ln(y) = ln(x^x) ⇒ ln(y) = xln(x) ⇒ $\frac{d}{dx} ln(y) = \frac{d}{dx} xln(x) ⇒ \frac{y'}{y} = ln(x)+1 ⇒ y' = y(ln(x)+1) = x^x(ln(x)+1).$

Examples:

• Let’s convert $\log_2(9)$ to base e (natural logarithm), $log_2(9)=\frac{ln(9)}{ln(2)}$= [$ln(9) \approx 2.1972, ln(2) \approx 0.6931$] ≈ 3.1699 ≈ 3.17.

• $log_3(81)=\frac{log_{10}(81)}{log_{10}(3)}$= [$log_{10}(81) \approx 1.9082, log_{10}(3) \approx 0.4771$] ≈ 4.

Solved examples

• Simplify $log_8(\frac{1}{12})+log_8(\frac{12}{5})$

$log_8(\frac{1}{12})+log_8(\frac{12}{5})$ =[Product Rule. log_a(xy)=log_a(x) + log_a(y)] $log_8(\frac{1}{12}·\frac{12}{5}) = log_8(\frac{1}{5}).$

• Solve ln(4x-3) = 7.

ln(4x -3) = 7 ⇒[Apply e^x to both sides] $e^{ln(4x-3)} = e^7 ⇒ 4x -3 = e^7 ⇒ x = \frac{e^7+3}{4}.$

• Solve log₄(x²-2x) = log₄(5x-12).

log₄(x²-2x) = log₄(5x-12) ⇒[a = 4 > 1, logarithm is strictly increasing⇒ one-to-one] x²-2x = 5x-12 ⇒ x² -7x + 12 = 0 ⇒ (x-3)(x-4) = 0 ⇒ x = 3 or 4.

It is essential to make sure to plug these numbers into the original equation so we don’t end up taking logarithms of zero or negative numbers.

log₄(3²-2·3) = log₄(3) = log₄(5·3-12) and log₄(4²-2·4) = log₄(8) = log₄(5·4-12).

• Solve the logarithmic equation $log_3(\sqrt{-7x+1})-7 = -4$

$log_3(\sqrt{-7x+1})-7 = -4$⇒[Isolate the Logarithmic Term] $log_3(\sqrt{-7x+1}) = 3$⇒[Convert to Exponential Form] $3^{log_3(\sqrt{-7x+1})}=3^3 = 27 ⇒ \sqrt{-7x+1} = 27$ ⇒[Square both sides of the equation to solve for x] -7x+1 = 27² = 729 ⇒ -7x = 728 ⇒ $x = \frac{-728}{7}$ = -104.

Besides, $log_3(\sqrt{-7·-104+1})-7 = -4↭log_3(\sqrt{729})-7 = -4 ↭ log_3(27)-7 = -4 ↭ 3-7 = -4$ and obviously $log_3(\sqrt{729})$ is a valid number.

• Solve the logarithmic equation $log_3(2x+1)+log_3(x+8)=3$

$log_3(2x+1)+log_3(x+8)=3$⇒[Product Rule. log_a(xy)=log_a(x) + log_a(y)] $log_3((2x+1)(x+8)) = 3 ⇒$[By definition of logarithms] (2x+1)(x+8) = 3³ = 27.

Expand and rearrange the equation, 2x² +17x +8 = 27. It has two solutions x = -¹⁹⁄₂, x = 1.

It is essential to make sure to plug these numbers into the original equation so we don’t end up taking logarithms of zero or negative numbers.

1. $log_3(2\frac{-19}{2}+1)= log_3(-18)$ ⊥ Since logarithms of negative numbers are undefined, this is not a valid solution.
2. $log_3(2·1+1)+log_3(1+8)=3↭log_3(3)+log_3(9)=3↭1+2=3$, so 1 is a valid solution.

Bibliography

1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Calculus. Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
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5. Michael Penn, Andrew Misseldine, blackpenredpen, and MathMajor, YouTube’s channels.
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