Logarithm (Base a)

Deep Breakdown

What it tells me (Situation Awareness): The derivative of $\log_a x$ (logarithm base $a$) is $1/(x \ln a)$. This is the generalization of $d/dx(\ln x) = 1/x$: the natural log $\ln x = \log_e x$ has $\ln e = 1$, so the factor $1/\ln a$ disappears. For base $a = 2$ (bits): $d/dx(\log_2 x) = 1/(x \ln 2) = 1/(0.693x)$, about $1.44$ times larger than the natural log derivative at the same $x$. For base $a = 10$: $d/dx(\log_{10} x) = 1/(x \ln 10) \approx 1/(2.303 x)$, smaller than the natural log derivative. In DA/DS: base-$2$ logarithms appear in information theory (entropy in bits), base-$10$ in decibels and order-of-magnitude analyses, and this derivative is needed when differentiating information-theoretic loss functions written in non-natural-log bases.

Flowchart

Real-world use

Where it is used

Differentiating Shannon entropy $H = -\sum_i p_i \log_2 p_i$ with respect to $p_i$: $dH/dp_i = -(1 + \log_2 p_i) = -(1 + \ln p_i/\ln 2)$, using $d/dp(\log_2 p) = 1/(p \ln 2)$. Differentiating mutual information expressions. Gradient of log-likelihood when the model uses $\log_{10}$ for interpretability. Change-of-base in information-theoretic optimization problems.

Problem it solves

When a loss function or metric is expressed in base-2 or base-10 logarithms (e.g., bits of entropy, nats vs bits conversion, decibel-scale regression), the gradient must be computed for optimization. Using the wrong base in the derivative introduces a constant-factor error equal to $\ln a$.

How it solves the problem

Change-of-base formula: $\log_a x = \ln x / \ln a$. Since $\ln a$ is a constant (not depending on $x$), $d/dx(\log_a x) = (d/dx(\ln x)) / \ln a = (1/x) / \ln a = 1/(x \ln a)$.

Example in a business

A DS computes the gradient of binary cross-entropy loss expressed in bits: $L = -[y \log_2 p + (1-y) \log_2(1-p)]$. The gradient: $dL/dp = -[y/(p \ln 2) - (1-y)/((1-p)\ln 2)] = -[y(1-p) - p(1-y)]/(p(1-p)\ln 2)$. Compared to the natural-log version, it differs by exactly $1/\ln 2 \approx 1.4427$. This constant factor affects the effective learning rate but not the direction of optimization.

The formula, piece by piece

• $\log_a x$: logarithm base $a$; defined for $x > 0$, $a > 0$, $a \ne 1$
• $\ln a$: natural log of the base; constant w.r.t. $x$; always positive for $a > 1$ (negative for $0 < a < 1$)
• $1/(x \ln a)$: the derivative; has the same $1/x$ structure as the natural log, scaled by $1/\ln a$
• Change of base: $\log_a x = \ln x / \ln a \Rightarrow d/dx(\log_a x) = (1/x)/\ln a$
• Special cases: $a = e \Rightarrow \ln e = 1 \Rightarrow d/dx(\ln x) = 1/x$; $a = 2 \Rightarrow d/dx(\log_2 x) = 1/(x \ln 2) \approx 1.4427/x$; $a = 10 \Rightarrow d/dx(\log_{10} x) = 1/(x \ln 10) \approx 0.4343/x$
• Chain rule: $d/dx(\log_a f(x)) = f'(x)/(f(x) \ln a)$
• Integral: $\int 1/(x \ln a) dx = \log_a x + C$

The formula in words

"The derivative of $\log_a x$ is $1$ over $x$ times the natural log of the base. The factor $1/\ln a$ converts from natural log units to base-$a$ units. When the base is $e$, this factor is $1$ and you get the natural log derivative. For other bases, it's a constant rescaling." Tiny examples: $d/dx(\log_2 x)|_{x=1} = 1/(1 \cdot \ln 2) \approx 1.4427$. $d/dx(\log_{10} x)|_{x=1} = 1/(1 \cdot \ln 10) \approx 0.4343$. $d/dx(\ln x)|_{x=1} = 1/1 = 1$. $d/dx(\log_2(3x))|_{x=2} = 3/((3 \cdot 2) \cdot \ln 2) = 3/(6 \ln 2) \approx 0.7213$.

Critical Concepts

• The base matters for magnitude, not direction: all log functions increase with $x$ (for $a > 1$), so all derivatives are positive. But the magnitude scales inversely with $\ln a$: larger base $\Rightarrow$ smaller derivative at the same $x$.
• Optimization is base-independent: gradient descent with a base-$a$ log loss will converge to the same minimum as with the natural log, because the derivative differs only by a constant $1/\ln a$, which just rescales the effective learning rate. For optimization, always use $\ln$ internally and apply the base-$a$ formula only when the output needs to be in specific units.
• Entropy gradient: $dH/dp_i = -(\log_a p_i + 1/\ln a)$ for Shannon entropy in base $a$. The $1/\ln a$ term comes from $d/dp(p_i \log_a p_i) = \log_a p_i + 1/\ln a$ (product rule).
• Conversion factor: $\log_a x = \log_b x / \log_b a$ for any base $b$. The derivative of any log base $a$ is the natural log derivative scaled by $1/\ln a$. This factor is: $1/\ln 2 \approx 1.4427$ (nats to bits), $1/\ln 10 \approx 0.4343$ (nats to decibels).
• Negative $\ln a$ for $0 < a < 1$: when $0 < a < 1$, $\ln a < 0$ and the derivative $1/(x \ln a) < 0$ for $x > 0$.

This is correct: $\log_a x$ is a decreasing function when $0 < a < 1$.

Step-by-step calculation (tiny example)

Differentiate $f(p) = -p \log_2 p$ (contribution to binary entropy) and evaluate at $p = 0.25$.

Step 1: Product rule. $f'(p) = -[\log_2 p + p \cdot d/dp(\log_2 p)]$.

Step 2: Apply $d/dp(\log_2 p) = 1/(p \ln 2)$. $f'(p) = -[\log_2 p + p \cdot 1/(p \ln 2)] = -[\log_2 p + 1/\ln 2]$.

Step 3: Evaluate at $p = 0.25$. $\log_2(0.25) = \log_2(2^{-2}) = -2$. $1/\ln 2 \approx 1.4427$. $f'(0.25) = -[(-2) + 1.4427] = -[-0.5573] = 0.5573$. Interpretation: the entropy contribution from $p_i = 0.25$ increases if $p_i$ increases slightly (slope is positive at $p = 0.25$).

Code examples

Beginner example (Python)

import numpy as np

def log_base_a_deriv(a: float, x: float) -> float:
    """d/dx(log_a(x)) = 1 / (x * ln(a)). Requires x > 0, a > 0, a != 1."""
    if x <= 0:
        raise ValueError(f"x = {x} <= 0: log undefined.")
    if a <= 0 or a == 1:
        raise ValueError(f"a = {a} invalid: need a > 0, a != 1.")
    return 1.0 / (x * np.log(a))   # Compare derivatives for different bases at
x = 1
print("d/dx(log_a(x)) at x=1 for different bases:")
print(f"{'Base a':>8} {'ln(a)':>10} {'Derivative at x=1':>20}")
for a in [2.0, np.e, 10.0, 100.0]:
    deriv = log_base_a_deriv(a, 1.0)
print(f"{a:>8.4f} {np.log(a):>10.6f} {deriv:>20.6f}")   # Shannon entropy gradient

def entropy_gradient(p: np.ndarray, base: float = 2.0) -> np.ndarray:
    """d/dp(-p * log_a(p)) = -(log_a(p) + 1/ln(a))."""
    return -(np.log(p) / np.log(base) + 1.0 / np.log(base))
print("\nShannon entropy gradient d/dp(-p*log2(p)):")
for p in [0.1, 0.25, 0.5, 0.75, 0.9]:
    grad = entropy_gradient(np.array([p]))[0]
print(f" p={p:.2f}: d/dp(-p*log2(p)) = {grad:.4f}")

Beginner example (R)

# d/dx(log_a(x)) = 1 / (x * log(a)) [log() in R is natural log]
log_base_a_deriv <- function(a, x) 1 / (x * log(a))
cat("d/dx(log_a(x)) at x=1 for different bases:\n")
cat(sprintf("%-8s %-10s %-20s\n", "Base a", "ln(a)", "Derivative at x=1"))
for (a in c(2, exp(1), 10, 100)) { cat(sprintf("%-8.4f %-10.6f %-20.6f\n", a, log(a), log_base_a_deriv(a, 1)))
}  # Binary entropy and its gradient
binary_entropy_grad <- function(p) -(log(p)/log(2) + 1/log(2))
cat("\nBinary entropy gradient d/dp(-p*log2(p)):\n")
for (p in c(0.1, 0.25, 0.5, 0.75, 0.9)) { cat(sprintf(" p=%.2f: %.4f\n", p, binary_entropy_grad(p)))
}

Professional example (Python)

import numpy as np
from scipy.special import entr
np.random.seed(42)   # -- Information-theoretic loss: cross-entropy in bits vs nats --
# Use case: a DA computes KL divergence in bits and needs to verify that
# optimizing in nats vs bits gives the same argmin but different gradient magnitudes.

def cross_entropy_bits(p_true: np.ndarray, p_pred: np.ndarray) -> float:
    """H(P, Q) = -sum_i p_i * log_2(q_i). Cross-entropy in bits."""
    p_pred_safe = np.clip(p_pred, 1e-10, 1.0)
    return float(-np.sum(p_true * np.log2(p_pred_safe)))

def cross_entropy_nats(p_true: np.ndarray, p_pred: np.ndarray) -> float:
    """H(P, Q) = -sum_i p_i * ln(q_i). Cross-entropy in nats."""
    p_pred_safe = np.clip(p_pred, 1e-10, 1.0)
    return float(-np.sum(p_true * np.log(p_pred_safe)))

def cross_entropy_bits_gradient(p_true: np.ndarray, p_pred: np.ndarray) -> np.ndarray:
    """d/d(q_i)(-p_i * log_2(q_i)) = -p_i / (q_i * ln(2)). Uses d/dx(log_2(x)) = 1/(x * ln(2)). """
    p_pred_safe = np.clip(p_pred, 1e-10, 1.0)
    return -p_true / (p_pred_safe * np.log(2))

def cross_entropy_nats_gradient(p_true: np.ndarray, p_pred: np.ndarray) -> np.ndarray:
    """d/d(q_i)(-p_i * ln(q_i)) = -p_i / q_i."""
    p_pred_safe = np.clip(p_pred, 1e-10, 1.0)
    return -p_true / p_pred_safe   # Softmax probabilities from a small 3-class distribution
K = 5
p_true = np.array([0.4, 0.3, 0.15, 0.1, 0.05])   # target distribution
p_pred = np.array([0.3, 0.25, 0.2, 0.15, 0.1])   # predicted distribution (unnormalized after softmax)
grad_bits = cross_entropy_bits_gradient(p_true, p_pred)
grad_nats = cross_entropy_nats_gradient(p_true, p_pred)
ce_bits = cross_entropy_bits(p_true, p_pred)
ce_nats = cross_entropy_nats(p_true, p_pred)
print("Cross-Entropy in Bits vs Nats: d/dq(-p*log_a(q)) = -p/(q * ln(a))")
print(f"\nCross-entropy: {ce_bits:.6f} bits = {ce_nats:.6f} nats")
print(f"Conversion: {ce_bits:.6f} / {ce_nats:.6f} = {ce_bits/ce_nats:.6f} (should = log2(e) = {np.log2(np.e):.6f})")
print(f"\n{'i':>3} {'p_true':>8} {'p_pred':>8} {'grad (bits)':>14} {'grad (nats)':>14} {'ratio':>8}")
for i in range(K):
    ratio = grad_bits[i] / grad_nats[i]
print(f"{i+1:>3} {p_true[i]:>8.3f} {p_pred[i]:>8.3f} {grad_bits[i]:>14.6f} {grad_nats[i]:>14.6f} {ratio:>8.5f}")
print(f"\nGradient ratio (bits/nats) = 1/ln(2) = {1/np.log(2):.6f} (constant, as expected)")
print(f"Key: optimizing bits cross-entropy = optimizing nats cross-entropy (same argmin, scaled gradient)")

Source: domain knowledge

#calculus#differential-calculus#table-of-standard-derivatives