What is the model for exponential growth and decay?

N = N₀e^(kt), where N₀ is the initial amount, N is the amount at time t, and k is positive for growth and negative for decay.

How are logarithms used in these models?

They bring the unknown down from the exponent, so you can find a time or a rate. For example, 2 = e^(0.04t) gives t = ln 2 / 0.04 ≈ 17.33.

What is the decibel formula?

L = 10 log(I/I₀), where I is the sound intensity and I₀ = 10⁻¹² W/m² is the threshold of hearing.

How many times more intense is a 90 dB sound than a 60 dB sound?

10³ = 1000 times. Each 10 dB step multiplies intensity by 10.

Which board sets my paper if I study in Gujrat?

Gujrat district is listed under the Board of Intermediate and Secondary Education, Gujranwala. PECTAA writes the books. Your board sets the paper and the dates, so follow its notices.

Growth, Decay and Decibels: Applications of Logarithms: PECTAA Class 12 Maths, Punjab Board

A population that keeps doubling, a radioactive sample that keeps shrinking, a sound that is a thousand times stronger yet only "30 decibels louder": all three are exponential or logarithmic at heart. Section 1.11 of the book puts the laws of logarithms to work on them.

Unit 1 of the PECTAA Mathematics 12 book is Graphical Representation of Functions. After solving exponential and logarithmic equations and inequalities (section 1.10), section 1.11 applies these functions to real-life situations. This note covers growth and decay and sound intensity in decibels. This note is for FSc Part 2 (2nd year, Class 12) students in Lahore, Faisalabad, Sargodha, Rawalpindi, Multan, Gujranwala and the other Punjab boards. Gujrat district comes under the Gujranwala board. PECTAA writes the books. Your own board sets your paper.

What to know

Why exponentials and logarithms. Exponential models describe quantities that change over time; logarithms find unknowns inside them, such as a time or a rate. The book takes the growth and decay formula as a given model (it is derived later, in the chapter on differential equations) and the decibel formula as a standard result.

Growth and decay. Processes such as population growth and radioactive decay change at a rate proportional to the current amount. The model is

N = N₀e^(kt)

N₀: the initial amount; N: the amount at time t.
k > 0: growth (N increases with time).
k < 0: decay (N decreases with time).

To find t or k: isolate the exponential, take ln of both sides, then solve.

Sound intensity. Intensity is the sound energy passing through unit area per unit time, measured in W/m². It varies over a huge range, so a logarithmic scale is used, the decibel (dB) scale:

L = 10 log(I/I₀)

I: intensity of the sound.
I₀ = 10⁻¹² W/m²: the threshold of hearing, the smallest intensity a normal ear detects.

Because the scale is logarithmic, a small rise in decibels means a large rise in intensity.

Doubling and halving times. The steps of Example 1 work for any growth model: setting N = 2N₀ gives 2 = e^(kt), so the doubling time is t = ln 2 / k, whatever the starting amount. For decay, setting N = N₀/2 gives t = ln 2 / |k|, the time to fall to half. This is why the answer in Example 1 does not depend on the 200 at the start.

Reading the graphs. The graph of N = N₀e^(kt) passes through (0, N₀). For k > 0 it rises ever more steeply; for k < 0 it falls towards the t-axis without touching it. The decibel scale works the other way round: it squeezes intensities from 10⁻¹² W/m² upward into a short, manageable range of numbers.

Key points

In N = N₀e^(kt), the sign of k tells growth from decay.
Isolate e^(kt) before taking ln.
L = 10 log(I/I₀); I₀ = 10⁻¹² W/m².
Each +10 dB multiplies intensity by 10; +30 dB multiplies it by 1000.

Worked examples

Example 1 (book Example 22, growth). A population is N = 200e^(0.04t), t in hours. When does it reach 400?

400 = 200e^(0.04t), so 2 = e^(0.04t).
ln 2 = 0.04t.
t = ln 2 / 0.04 = 0.6931 / 0.04 ≈ 17.33 hours.

Example 2 (book Example 23, decay). A substance decays as N = 500e^(kt). After 20 hours, N = 300. Find k.

300 = 500e^(20k), so e^(20k) = 3/5 = 0.6.
20k = ln 0.6 ≈ −0.5108.
k ≈ −0.5108 / 20 ≈ −0.0255 per hour. The book prints −0.0256; ln 0.6 / 20 = −0.02554, which rounds to −0.0255. k is negative, as decay requires.

Example 3 (book Example 24, finding sound level). I = 10⁻⁴ W/m², I₀ = 10⁻¹². Find L.

I/I₀ = 10⁻⁴ / 10⁻¹² = 10⁸.
L = 10 log 10⁸ = 10 × 8 = 80 dB.

Example 4 (book Example 25, finding intensity). L = 60 dB, I₀ = 10⁻¹². Find I.

60 = 10 log(I/10⁻¹²), so 6 = log(I/10⁻¹²).
Exponential form: I/10⁻¹² = 10⁶.
I = 10⁶ × 10⁻¹² = 10⁻⁶ W/m².

Example 5 (book Example 26, comparing levels). L₁ = 90 dB and L₂ = 60 dB. Find I₁/I₂.

9 = log(I₁/I₀), so I₁/I₀ = 10⁹.
6 = log(I₂/I₀), so I₂/I₀ = 10⁶.
Divide: I₁/I₂ = 10⁹ / 10⁶ = 10³ = 1000.

Common mistakes in the exam

Taking ln before isolating the exponential. First divide by N₀.
Dropping the minus sign of k in decay.
Using ln in the decibel formula. The decibel formula uses log base 10.
Forgetting the factor 10 in L = 10 log(I/I₀).
Subtracting intensities instead of dividing them when comparing levels.

Exam technique

Write the model, substitute, isolate, take logs, solve: one step per line.
Keep ln 2 ≈ 0.6931 and ln 0.6 ≈ −0.5108 to four decimals, and round at the end.
State units: hours for t, per hour for k, dB for L, W/m² for I.
Check your board's notice for the sections in your paper.

Practice questions

N = 100e^(0.05t). Find t when N = 200. Answer: t = ln 2 / 0.05 ≈ 13.86.
Is N = 80e^(−0.1t) growth or decay? Answer: Decay, because k = −0.1 < 0.
Find L for I = 10⁻² W/m². Answer: 10 log 10¹⁰ = 100 dB.
Find I for L = 40 dB. Answer: 10⁴ × 10⁻¹² = 10⁻⁸ W/m².
How many times more intense is 70 dB than 50 dB? Answer: 10² = 100 times.

Ready to practise this? Continue in 2nd Year Mathematics.