Paper I Part A Q10 · 2 marks
Logarithms IG.C.E. O/L 2025
Antilog of a decimal using a given logarithm
🔒 Answer & steps: try it free in Gedara Sir→Short notes › Grade 10 · සිංහලෙන්
10 ශ්රේණිය · 18 පරිච්ඡේදය · ලඝුගණක I
69%
of Grade 10 term tests include it (37 of 54)
174
marks in past papers (50 questions)
11 min
full lesson with quick checks
aⁿ = b ⟺ logₐ b = n (a > 0, a ≠ 1)
✅ (i) lg 1000 = 3, log₅ (1/25) = −2 (ii) 2⁶ = 64
logₐ (xy) = logₐ x + logₐ y logₐ (x/y) = logₐ x − logₐ y
✅ 2
| aⁿ = b in logarithm form? | logₐ b = n |
| What does lg mean? | Logarithm to base 10. lg 1000 = 3. |
| logₐ 1 = ? | 0, because a⁰ = 1. |
| logₐ a = ? | 1, because a¹ = a. |
| Product law | logₐ (xy) = logₐ x + logₐ y |
| Quotient law | logₐ (x/y) = logₐ x − logₐ y |
| log₃ 81 = ? | 4, because 3⁴ = 81. |
| lg 5 in terms of lg 2? | lg 5 = lg 10 − lg 2 = 1 − lg 2. |
Now test yourself: 6 quick checks and 2 more worked examples
The full lesson checks your answers as you go, and Gedara Sir tracks which parts you've got.
G.C.E. O/L 2025
Antilog of a decimal using a given logarithm
🔒 Answer & steps: try it free in Gedara Sir→G.C.E. O/L 2024
Logarithm of a number from a given index form
🔒 Answer & steps: try it free in Gedara Sir→G.C.E. O/L 2019
Convert a base-2 logarithm to index form
🔒 Answer & steps: try it free in Gedara Sir→G.C.E. O/L 2018
Read a logarithm from an index statement
🔒 Answer & steps: try it free in Gedara Sir→