Exponents and logarithms (applications)

Exponents and logarithms are inverse operations fundamental to modeling exponential growth, decay, and scale transformations.

Key Definitions:

Exponential Form: a^x = b where a is the base (a > 0, a ≠ 1), x is the exponent, and b is the result.

Logarithmic Form: log_a(b) = x reads as "logarithm of b to base a equals x". This answers: "To what power must a be raised to obtain b?"

Critical Laws of Logarithms:

• Product Rule: log_a(xy) = log_a(x) + log_a(y)
• Quotient Rule: log_a(x/y) = log_a(x) - log_a(y)
• Power Rule: log_a(x^n) = n·log_a(x)
• Change of Base: log_a(x) = log_b(x)/log_b(a) = ln(x)/ln(a)
• Special Values: log_a(1) = 0, log_a(a) = 1, a^(log_a(x)) = x

Essential Applications:

1. Exponential Growth/Decay: N(t) = N₀e^(kt) where k > 0 (growth) or k < 0 (decay)
2. Compound Interest: A = P(1 + r/n)^(nt)
3. Half-life/Doubling Time: t_(1/2) = ln(2)/|k|
4. pH Scale: pH = -log₁₀[H⁺]
5. Richter Scale: M = log₁₀(I/I₀)_

Mnemonic: "Logs Undo Exponents" — LUE reminds you they're inverses.

Natural Logarithm (ln) uses base e ≈ 2.718, appearing naturally in continuous growth models. Common logarithm (log) uses base 10, prevalent in scientific scales.

Why Logarithms Matter:

Imagine you're tracking a viral video's views. It gains 50,000 views on day 1, doubling daily. After t days: V(t) = 50,000 × 2^t. To find when it reaches 10 million views, you need logarithms: 2^t = 200, so t = log₂(200) ≈ 7.64 days.

Real-World Applications:

1. Carbon Dating (Archaeology): Carbon-14 decays with half-life 5,730 years. If a fossil retains 35% of original C-14: N(t) = N₀e^(-kt), where k = ln(2)/5730. Solving 0.35N₀ = N₀e^(-kt) gives t = -ln(0.35)/k ≈ 8,680 years.

2. Sound Intensity (Decibels): Sound level L = 10log₁₀(I/I₀) where I₀ = 10^(-12) W/m². A concert at 110 dB has intensity I = I₀ × 10^11 = 0.1 W/m². Doubling intensity adds ~3 dB, not double dB — logarithmic scales compress large ranges.

3. Investment Growth: You invest £5,000 at 4.5% annual interest compounded quarterly. To reach £8,000: 8000 = 5000(1.01125)^(4t). Taking logarithms: 4t·log(1.01125) = log(1.6), giving t ≈ 10.6 years.

4. Earthquake Magnitude: An earthquake measuring 7.0 releases 10^(7.0-6.0) = 10 times the energy of a 6.0 quake. This logarithmic compression makes huge variations manageable.

Analogy: Think of logarithms as a "volume control" for numbers. Exponentials expand rapidly; logarithms compress them back to human-scale values, perfect for phenomena spanning many orders of magnitude.

Example 1: The population of bacteria doubles every 3 hours. If initially 500 bacteria are present, find when the population reaches 50,000.

Solution:

• Model: N(t) = N₀ × 2^(t/3) where t is in hours
• Setup: 50,000 = 500 × 2^(t/3)
• Simplify: 100 = 2^(t/3)
• Apply logarithms: log₂(100) = t/3
• Change base: t/3 = ln(100)/ln(2) = 4.605/0.693 ≈ 6.644
• Answer: t ≈ 19.9 hours

Examiner Note: Always define variables clearly. Show conversion steps explicitly for full marks.

Example 2: Solve for x: 3^(2x-1) = 7^(x+2)

Solution:

• Take ln of both sides: ln(3^(2x-1)) = ln(7^(x+2))
• Apply power rule: (2x-1)ln(3) = (x+2)ln(7)
• Expand: 2x·ln(3) - ln(3) = x·ln(7) + 2ln(7)
• Collect x terms: x[2ln(3) - ln(7)] = ln(3) + 2ln(7)
• Solve: x = [ln(3) + 2ln(7)]/[2ln(3) - ln(7)]
• Calculate: x = [1.099 + 3.891]/[2.197 - 1.946] ≈ 19.88

Examiner Note: Keep exact forms until final step. Show all algebraic manipulation.

Example 3: A radioactive substance decays to 40% of its original mass in 8 years. Find the half-life.

Solution:

• Model: N(t) = N₀e^(-kt)
• From given: 0.4N₀ = N₀e^(-8k), so k = -ln(0.4)/8 ≈ 0.1145
• Half-life: t_(1/2) = ln(2)/k = 0.693/0.1145 ≈ 6.05 years_