An exponential relationship such as y = abˣ is a curve, but taking logarithms turns it into a straight line. Once it is a line, the gradient and the vertical intercept give you the constants a and b.
This lesson uses the laws from combining logarithms and leads into the straight-line work in straight lines and linearisation. It sits within exponential and logarithmic reasoning.
How does the straight line appear?
Start with y = abˣ and take lg of both sides.
lg y = lg (abˣ) = lg a + lg (bˣ) = lg a + x lg b.
Rewrite as lg y = (lg b) x + lg a. Compare this with Y = mX + c:
| Part | Meaning |
|---|---|
| Y | lg y |
| X | x |
| gradient m | lg b |
| intercept c | lg a |
For a power relationship y = axⁿ the same idea gives lg y = n lg x + lg a. Here the vertical axis is lg y, the horizontal axis is lg x, the gradient is n and the intercept is lg a.
Steps for reading constants from a log graph
- Identify the axes. What is plotted against what?
- Write the straight-line equation in the form Y = mX + c using the graph’s variables.
- Find the gradient from two points, and the intercept from the line.
- Match to the theory: gradient = lg b, intercept = lg a.
- Reverse the log: b = 10gradient and a = 10intercept.
- Write the final relationship and test it with one of the given points.
Worked example
A graph of lg y against x is a straight line through (1, 1.2) and (5, 2.8). The variables are related by y = abˣ. Find a and b.
Step 1, gradient: m = (2.8 − 1.2) ÷ (5 − 1) = 1.6 ÷ 4 = 0.4.
Step 2, intercept: using lg y = 0.4x + c at (1, 1.2): 1.2 = 0.4 + c, so c = 0.8.
Step 3, match to theory: lg b = 0.4 and lg a = 0.8.
Step 4, reverse the logs: b = 100.4 = 2.51 and a = 100.8 = 6.31, both to 3 significant figures.
Answer: y ≈ 6.31 × 2.51ˣ.
Check: at x = 5, lg y = 0.8 + 0.4 × 5 = 2.8, so y = 102.8 ≈ 631. The rounded constants give 6.31 × 2.51⁵ ≈ 629, which agrees to within rounding. ✓
The mistake to watch for
The usual slip is to read the intercept as a itself, instead of lg a.
Mistaken answer: a = 0.8 and b = 0.4.
The student copied the gradient and intercept straight into the constants.
The correction is that the graph’s numbers are logs of the constants. The intercept 0.8 is lg a, so a = 100.8 = 6.31.
The gradient 0.4 is lg b, so b = 100.4 = 2.51. A test with x = 0 shows the issue: the original equation gives y = a, so if a were 0.8, lg y would be lg 0.8, which is negative. The graph shows lg y = 0.8 at x = 0, so y = 6.31.
Check yourself
1. For y = 2 × 5ˣ, a graph of lg y against x is drawn. State the gradient and the intercept to 3 decimal places.
Show answer
lg y = lg 2 + x lg 5. The gradient is lg 5 = 0.699 and the intercept is lg 2 = 0.301.
2. A graph of lg y against lg x is a straight line through (0, 1) and (2, 4). Find the relationship in the form y = axⁿ.
Show answer
Gradient = (4 − 1) ÷ (2 − 0) = 1.5, so n = 1.5. Intercept = 1, so lg a = 1 and a = 10.
So y = 10x1.5. Check: lg x = 2 gives x = 100, and 10 × 1001.5 = 10 × 1000 = 10 000, which has lg = 4. ✓
3. A graph of ln y against x has gradient 0.3 and intercept 2. The relationship is y = Aekx. Find A and k.
Show answer
ln y = ln A + kx, so k = 0.3 and ln A = 2. Then A = e² = 7.389. So y = 7.39e0.3x (A to 3 significant figures).
Where this leads next
Exponential growth models come from exactly this kind of equation, so the next lesson separates a growth constant from a percentage growth. When you are ready, try the mixed practice set.
If you understand the algebra but hesitate over which number is which when you read a graph, a teacher in online one-to-one Additional Mathematics tuition can practise that step with your own sketches.