Core Formula
log S = log C + Z log A
Equivalent to: S = C × A^Z (power function)
Variable Definitions
| Symbol | Name | Description |
|---|---|---|
| S | Species richness | Number of species in area A |
| A | Area | Geographic area (any unit, must be consistent) |
| Z | Regression coefficient | Slope of log-log line; range: 0.1–1.2 |
| C | Y-intercept | Species richness when A = 1 (same units) |
Z Value Reference Table
| Context | Z Range | Slope | Interpretation |
|---|---|---|---|
| Continental regions | 0.1–0.2 | Gentle | Slow species loss with area reduction |
| Oceanic islands | 0.6–1.2 | Steep | Rapid species loss with area reduction |
Ratio Method for Comparing Two Areas
$S_{2}$/$S_{1}$ = ($A_{2}$/$A_{1}$)^Z
Example: If $A_{2}$ = 0.5 × $A_{1}$ (50% area reduction):
- Continental (Z=0.15): $S_{2}$/$S_{1}$ = (0.5)^0.15 = 0.90 → 10% species loss
- Island (Z=0.85): $S_{2}$/$S_{1}$ = (0.5)^0.85 = 0.56 → 44% species loss
Key Numerical Problems
Problem 1: An island has 200 species on 1,000 . If Z = 0.8 and C = 5, how many species would a 100 island have?
log S = log 5 + 0.8 × log 100 = 0.699 + 0.8 × 2 = 0.699 + 1.6 = 2.299
S = 10^2.299 ≈ 199 — wait, A = 100 (not 1,000)
Ratio: $S_{2}$/$S_{1}$ = (100/1000)^0.8 = (0.1)^0.8 = 0.158
$S_{2}$ = 200 × 0.158 ≈ 32 species
Problem 2: Continental region, Z = 0.15. If habitat reduced from 10,000 ha to 1,000 ha (90% loss):
$S_{2}$/$S_{1}$ = (1000/10000)^0.15 = (0.1)^0.15 = 10^(-0.15) = 0.708 → 29% species loss
NEET Tips
- Always identify whether the question refers to islands (Z=0.6–1.2) or continents (Z=0.1–0.2)
- Log-log plot → straight line; arithmetic plot → power curve (concave up, flattening)
- Z is the SLOPE; C is the Y-INTERCEPT (not a species count directly)