Solutions — Applications of Logarithmic Functions

Q1 — Solving basic logarithmic equations

(a) log₂(x) = 6 → x = 2⁶ = 64. Domain: x > 0 ✓ (64 > 0).

(b) log₅(x − 3) = 2 → x − 3 = 5² = 25 → x = 28. Domain: x − 3 > 0 ✓ (25 > 0).

(c) log(x + 1) = 3 → x + 1 = 10³ = 1000 → x = 999. Domain: x + 1 > 0 ✓ (1000 > 0).

(d) ln(x) = 4 → x = e⁴ ≈ 54.598. Exact answer: x = e⁴. Domain: x > 0 ✓.

Q2 — Logs on both sides

(a) log₃(x + 5) = log₃(2x − 1): same base → equate arguments.
x + 5 = 2x − 1 → 6 = x → x = 6
Check: x+5=11>0 ✓ and 2(6)−1=11>0 ✓

(b) log(4x − 3) = log(2x + 7): equate arguments.
4x − 3 = 2x + 7 → 2x = 10 → x = 5
Check: 4(5)−3=17>0 ✓ and 2(5)+7=17>0 ✓

(c) ln(3x) = ln(x + 8): equate arguments.
3x = x + 8 → 2x = 8 → x = 4
Check: 3(4)=12>0 ✓ and 4+8=12>0 ✓

Q3 — pH calculations

(a) pH = −log(10⁻⁵) = −(−5) = 5

(b) pH = −log(2.5 × 10⁻³) = −[log(2.5) + log(10⁻³)] = −[0.3979 − 3] = −(−2.6021) ≈ 2.60

(c) pH = 8.3 → [H⁺] = 10^−8.3 = 10⁻⁹ × 10^0.7 = 10⁻⁹ × 5.012 ≈ 5.0 × 10⁻⁹ mol/L

Q4 — Using log laws first

(a) log₂(x) + log₂(x − 2) = 3
Product law: log₂(x(x − 2)) = 3
Convert: x(x − 2) = 2³ = 8
x² − 2x − 8 = 0 → (x − 4)(x + 2) = 0 → x = 4 or x = −2
Check domain: need x > 0 AND x − 2 > 0, so x > 2. x = 4: ✓ x = −2: fails.
x = 4

(b) log(x + 4) − log(x − 1) = 1
Quotient law: log((x + 4)/(x − 1)) = 1
Convert: (x + 4)/(x − 1) = 10
x + 4 = 10(x − 1) = 10x − 10 → 14 = 9x → x = 14/9
Check: x+4=14/9+36/9=50/9>0 ✓ and x−1=14/9−9/9=5/9>0 ✓
x = 14/9

Q5 — Richter scale

(a) M = 9.0 → I/I₀ = 10⁹ → I = 10⁹ I₀ (one billion times the reference intensity)

(b) Ratio of intensities = 10^{9.0 − 3.2} = 10^5.8 ≈ 630 957 ≈ 6.31 × 10⁵ times more intense

(c) I₁/I₀ = 10^M₁ and I₂/I₀ = 10^M₂.
Therefore I₁/I₂ = (10^M₁ I₀) / (10^M₂ I₀) = 10^{M₁ − M₂}. ✓

Q6 — Exponential equations using logarithms

(a) 5^x = 80
Take log of both sides: x log(5) = log(80)
x = log(80) / log(5) = 1.9031 / 0.6990 ≈ 2.723

(b) 3 × 2^x = 48 → 2^x = 16 = 2⁴ → x = 4

(c) e^{2x − 1} = 7
Take ln: 2x − 1 = ln(7)
2x = 1 + ln(7)
Exact: x = (1 + ln 7) / 2
Decimal: x = (1 + 1.9459) / 2 = 2.9459 / 2 ≈ 1.473

Q7 — Population growth model

(a) At t = 0: P = 12 000 e⁰ = 12 000. The constant 0.025 is the continuous growth rate — the population grows at 2.5% per year.

(b) t = 10: P = 12 000 e^{0.025 × 10} = 12 000 e^0.25 = 12 000 × 1.28403 ≈ 15 408 people

(c) Set P = 20 000:
12 000 e^0.025t = 20 000
e^0.025t = 20 000 / 12 000 = 5/3
0.025t = ln(5/3) = ln(5) − ln(3) = 1.6094 − 1.0986 = 0.5108
t = 0.5108 / 0.025 ≈ 20.43 years after 2020
The population first exceeds 20 000 in 2040 (early in the year).

Q8 — Equations requiring log laws and domain checks

(a) log₃(x + 1) + log₃(x + 3) = log₃(5)
Product law LHS: log₃((x+1)(x+3)) = log₃(5)
Equate arguments: (x+1)(x+3) = 5
x² + 4x + 3 = 5 → x² + 4x − 2 = 0
x = (−4 ± √(16 + 8)) / 2 = (−4 ± √24) / 2 = −2 ± √6
x = −2 + √6 ≈ 0.449 or x = −2 − √6 ≈ −4.449
Domain: need x+1>0 and x+3>0, so x>−1. Only x = −2+√6 ≈ 0.449 > −1 ✓
x = −2 + √6

(b) log₂(x − 1) − log₂(x + 2) = −2
Quotient law: log₂((x−1)/(x+2)) = −2
Convert: (x−1)/(x+2) = 2⁻² = 1/4
4(x−1) = x+2 → 4x−4 = x+2 → 3x = 6 → x = 2
Check: x−1=1>0 ✓ and x+2=4>0 ✓
x = 2

(c) 2log₅(x) − log₅(x − 4) = 2
Power law: log₅(x²) − log₅(x−4) = 2
Quotient law: log₅(x²/(x−4)) = 2
Convert: x²/(x−4) = 5² = 25
x² = 25(x−4) = 25x − 100
x² − 25x + 100 = 0 → (x−20)(x−5) = 0 → x = 20 or x = 5
Domain: need x>0 and x−4>0, so x>4. Both x=20 and x=5 satisfy x>4 ✓
x = 5 or x = 20

Q9 — Compound interest

(a) A = 8000(1.045)^t

(b) Double → A = 16 000:
8000(1.045)^t = 16 000 → (1.045)^t = 2
t log(1.045) = log(2)
t = log(2) / log(1.045) = 0.30103 / 0.01912 ≈ 15.747 years
0.747 × 12 ≈ 8.96 months ≈ 9 months
Approximately 15 years and 9 months

(c) Triple → A = 24 000:
(1.045)^t = 3 → t = log(3) / log(1.045) = 0.47712 / 0.01912 ≈ 24.9 ≈ 25 years

Q10 — Comparing two radioactive decay models

(a) At t = 0: M_A = 200 e⁰ = 200 g and M_B = 150 e⁰ = 150 g

(b) Set M_A = M_B:
200 e^−0.06t = 150 e^−0.04t
200/150 = e^−0.04t / e^−0.06t = e^{(−0.04 + 0.06)t} = e^0.02t
4/3 = e^0.02t
0.02t = ln(4/3) = ln(4) − ln(3) = 1.3863 − 1.0986 = 0.2877
t = 0.2877 / 0.02 = 14.38 years

(c) At t = 50:
M_A = 200 e^{−0.06 × 50} = 200 e⁻³ = 200 × 0.04979 ≈ 9.96 g
M_B = 150 e^{−0.04 × 50} = 150 e⁻² = 150 × 0.13534 ≈ 20.30 g
Since 20.30 > 9.96, substance B has more mass after 50 years. Even though B started lighter, it decays more slowly (smaller rate constant 0.04 vs 0.06).