Logarithmic and Exponential Models

Real-world applications including compound interest, population growth, and radioactive decay

Logarithmic and Exponential Models

Exponential Growth Model

The general exponential growth model is:

A(t)=A0ektA(t) = A_0 e^{kt}

where:

  • A(t)A(t) = amount at time tt
  • A0A_0 = initial amount (at t=0t = 0)
  • kk = growth rate constant (k>0k > 0 for growth)
  • tt = time
  • e≈2.71828e \approx 2.71828 (Euler's number)

Finding the Growth Rate kk

If you know two data points, you can solve for kk:

  1. Substitute known values
  2. Divide by A0A_0
  3. Take natural log of both sides
  4. Solve for kk

Exponential Decay Model

Same formula, but k<0k < 0 (negative):

A(t)=A0e−ktA(t) = A_0 e^{-kt}

where k>0k > 0 represents the decay rate.

Half-Life Formula

The half-life is the time it takes for half the substance to decay.

A(t)=A0(12)t/hA(t) = A_0\left(\frac{1}{2}\right)^{t/h}

where hh is the half-life.

Relationship to kk: h=ln⁡(2)kh = \frac{\ln(2)}{k}

Compound Interest Models

Compound Interest (n times per year)

A=P(1+rn)ntA = P\left(1 + \frac{r}{n}\right)^{nt}

where:

  • PP = principal (initial investment)
  • rr = annual interest rate (as decimal)
  • nn = number of times compounded per year
  • tt = time in years

Continuous Compounding

A=PertA = Pe^{rt}

This is the limit as n→∞n \to \infty (compounding infinitely often).

Doubling Time

The time it takes for a quantity to double:

2A0=A0ekt2A_0 = A_0e^{kt} 2=ekt2 = e^{kt} ln⁡(2)=kt\ln(2) = kt t=ln⁡(2)kt = \frac{\ln(2)}{k}

Doubling time: td=ln⁡(2)k≈0.693kt_d = \frac{\ln(2)}{k} \approx \frac{0.693}{k}

Population Growth Models

Unlimited Growth (Malthusian)

P(t)=P0ektP(t) = P_0e^{kt}

Assumes unlimited resources (not realistic long-term).

Logistic Growth

P(t)=L1+Ce−ktP(t) = \frac{L}{1 + Ce^{-kt}}

where:

  • LL = carrying capacity (maximum sustainable population)
  • CC = constant based on initial conditions
  • kk = growth rate

The population approaches LL as t→∞t \to \infty.

Newton's Law of Cooling

Temperature of an object over time:

T(t)=Ts+(T0−Ts)e−ktT(t) = T_s + (T_0 - T_s)e^{-kt}

where:

  • T(t)T(t) = temperature at time tt
  • TsT_s = surrounding (ambient) temperature
  • T0T_0 = initial temperature
  • kk = cooling rate constant

Logarithmic Models

Some phenomena grow logarithmically:

y=a+bln⁡(x)y = a + b\ln(x)

Examples:

  • Earthquake intensity (Richter scale)
  • Sound intensity (decibels)
  • pH scale (acidity)

Richter Scale

Earthquake magnitude: M=log⁡(II0)M = \log\left(\frac{I}{I_0}\right)

where II is intensity and I0I_0 is reference intensity.

An increase of 1 on the Richter scale means 10 times more intense!

Solving Applied Problems

General Strategy:

  1. Identify the type of model needed
  2. Write the equation with known values
  3. Solve for the unknown (often kk first, then answer the question)
  4. Check if the answer makes sense in context

📚 Practice Problems

1Problem 1medium

❓ Question:

A population of bacteria grows from 100 to 500 in 3 hours. Assuming exponential growth P(t)=P0ektP(t) = P_0e^{kt}, find the growth rate kk and predict the population after 5 hours.

💡 Show Solution

Solution:

Part 1: Find kk

Given: P0=100P_0 = 100, P(3)=500P(3) = 500, t=3t = 3

Step 1: Write the equation. 500=100e3k500 = 100e^{3k}

Step 2: Divide by 100. 5=e3k5 = e^{3k}

Step 3: Take natural log. ln⁡(5)=3k\ln(5) = 3k

Step 4: Solve for kk. k=ln⁡(5)3≈1.6093≈0.536 per hourk = \frac{\ln(5)}{3} \approx \frac{1.609}{3} \approx 0.536 \text{ per hour}

Part 2: Find P(5)P(5)

P(5)=100e0.536(5)P(5) = 100e^{0.536(5)} =100e2.68= 100e^{2.68} ≈100(14.58)\approx 100(14.58) ≈1458\approx 1458

Answers:

  • Growth rate: k≈0.536k \approx 0.536 per hour
  • Population after 5 hours: approximately 1,458 bacteria

2Problem 2easy

❓ Question:

Invest $5,000 at 6% annual interest compounded monthly. How much will you have after 10 years?

💡 Show Solution

Solution:

Given:

  • P=5000P = 5000 (principal)
  • r=0.06r = 0.06 (6% as decimal)
  • n=12n = 12 (monthly compounding)
  • t=10t = 10 years

Step 1: Write the compound interest formula. A=P(1+rn)ntA = P\left(1 + \frac{r}{n}\right)^{nt}

Step 2: Substitute values. A=5000(1+0.0612)12(10)A = 5000\left(1 + \frac{0.06}{12}\right)^{12(10)}

=5000(1+0.005)120= 5000\left(1 + 0.005\right)^{120}

=5000(1.005)120= 5000(1.005)^{120}

Step 3: Calculate. (1.005)120≈1.8194(1.005)^{120} \approx 1.8194

A≈5000(1.8194)≈9097A \approx 5000(1.8194) \approx 9097

Answer: You will have approximately $9,097 after 10 years.

3Problem 3hard

❓ Question:

Carbon-14 has a half-life of 5,730 years. If a fossil contains 25% of its original carbon-14, how old is the fossil?

💡 Show Solution

Solution:

Given:

  • Half-life: h=5730h = 5730 years
  • Current amount: A(t)=0.25A0A(t) = 0.25A_0 (25% remaining)

Step 1: Write the half-life formula. A(t)=A0(12)t/hA(t) = A_0\left(\frac{1}{2}\right)^{t/h}

Step 2: Substitute known values. 0.25A0=A0(12)t/57300.25A_0 = A_0\left(\frac{1}{2}\right)^{t/5730}

Step 3: Divide by A0A_0. 0.25=(12)t/57300.25 = \left(\frac{1}{2}\right)^{t/5730}

Step 4: Recognize that 0.25=14=(12)20.25 = \frac{1}{4} = \left(\frac{1}{2}\right)^2. (12)2=(12)t/5730\left(\frac{1}{2}\right)^2 = \left(\frac{1}{2}\right)^{t/5730}

Step 5: Set exponents equal. 2=t57302 = \frac{t}{5730}

Step 6: Solve for tt. t=2(5730)=11,460 yearst = 2(5730) = 11,460 \text{ years}

Alternative method using logarithms:

ln⁡(0.25)=t5730ln⁡(0.5)\ln(0.25) = \frac{t}{5730}\ln(0.5)

t=5730ln⁡(0.25)ln⁡(0.5)=11,460 yearst = \frac{5730 \ln(0.25)}{\ln(0.5)} = 11,460 \text{ years}

Answer: The fossil is approximately 11,460 years old.

4Problem 4medium

❓ Question:

A bacteria population starts at 500 and doubles every 3 hours. Write an exponential model and find the population after 12 hours.

💡 Show Solution

Step 1: Use exponential growth model: P(t) = P₀ · 2^(t/d) where P₀ = initial amount, d = doubling time

Step 2: Identify values: P₀ = 500 d = 3 hours

Step 3: Write the model: P(t) = 500 · 2^(t/3)

Step 4: Find population at t = 12: P(12) = 500 · 2^(12/3) = 500 · 2⁴ = 500 · 16 = 8000

Answer: Model: P(t) = 500 · 2^(t/3); Population after 12 hours: 8000

5Problem 5hard

❓ Question:

The magnitude M of an earthquake is given by M = log(I/I₀), where I is the intensity and I₀ is a reference intensity. How many times more intense is an earthquake of magnitude 7 compared to one of magnitude 5?

💡 Show Solution

Step 1: Set up equations for both magnitudes: M₁ = 7 = log(I₁/I₀) M₂ = 5 = log(I₂/I₀)

Step 2: Convert to exponential form: 7 = log(I₁/I₀) → I₁/I₀ = 10⁷ 5 = log(I₂/I₀) → I₂/I₀ = 10⁵

Step 3: Find the ratio I₁/I₂: I₁ = I₀ · 10⁷ I₂ = I₀ · 10⁵

I₁/I₂ = (I₀ · 10⁷)/(I₀ · 10⁵) = 10⁷/10⁵ = 10² = 100

Step 4: Interpretation: An earthquake of magnitude 7 is 100 times more intense than an earthquake of magnitude 5.

Answer: 100 times more intense

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