Logarithms Worksheet Answer Key – Precalculus

Problem 1 Switch Between Forms

(a) log₆ 216 = 3.

(b) 9² = 81.

(c) log 0.01 = −2 (base 10).

Problem 2 Evaluate Logarithms Exactly

(a) 2⁶ = 64, so 6. (b) 5⁻² = 1/25, so −2. (c) 9^1/2 = 3, so 1/2. (d) The natural log undoes e⁴: 4.

Problem 3 Expand a Logarithm

log₃(27a²) − log₃ b = log₃ 27 + log₃ a² − log₃ b.

3 + 2 log₃ a − log₃ b.

Problem 4 Condense a Logarithmic Expression

2 ln 5 = ln 25. Then ln 25 − ln 10 = ln 2.5 and ln 2.5 + ln 4 = ln 10.

ln 10.

Problem 5 Match the Bases

(a) 32 = 2⁵, so x = 5.

(b) 81 = 3⁴, so x + 1 = 4 and x = 3.

Problem 6 Read a Logarithmic Graph

The point (4, 2) gives log₂ 4 = 2. The point (8, 3) gives log₂ x = 3 at x = 8.

The curve exists only for positive x: domain (0, ∞), vertical asymptote x = 0.

Problem 7 Change of Base

log₇ 50 = ln 50 ÷ ln 7 ≈ 3.9120 ÷ 1.9459 ≈ 2.010.

log₄ 0.3 = ln 0.3 ÷ ln 4 ≈ -1.2040 ÷ 1.3863 ≈ -0.868.

Problem 8 Exponential Equation with Logs

Take ln of both sides: (x − 2) ln 7 = ln 19, so x − 2 = ln 19ln 7 ≈ 1.5131.

x = 2 + ln 19ln 7 ≈ 3.513.

Problem 9 Isolate a Natural Exponential

Divide by 3: e^0.4x = 7. Take ln: 0.4x = ln 7.

x = ln 70.4 ≈ 4.865.

Problem 10 Logarithmic Equation with a Check

Combine: log₃[(x + 1)(x − 5)] = 3, so x² − 4x − 5 = 27 and x² − 4x − 32 = (x − 8)(x + 4) = 0.

For x = −4 the inputs −3 and −9 are negative: rejected. For x = 8: log₃ 9 + log₃ 3 = 2 + 1 = 3 ✓. x = 8.

Problem 11 Natural Log Equation

Rewrite as 2x − 1 = e³. Then x = e³ + 12.

x = e³ + 12 ≈ 10.543.

Problem 12 Inverse Pairs and Reflections

Swapping coordinates gives the point (9, 2) on g. Since 3³ = 27, g(27) = 3. The range of g is the domain of f: all real numbers.

Problem 13 Solve with a Fractional Exponent

Divide by 6: 2^x/3 = 16 = 2⁴. So x/3 = 4.

x = 12.

Problem 14 Doubling Culture

8,000 cells: 2^t/3 = 16, so t/3 = 4 and t = 12 hours (the marked point).

20,000 cells: 2^t/3 = 40, so t = 3 · ln 40ln 2 ≈ 15.97 hours.

Problem 15 Logistic Starting Value and Limit

L(0) = 500 ÷ (1 + 9) = 50. As t grows, e^−0.5t → 0, so the capacity is 500.

Set 500 ÷ (1 + 9e^−0.5t) = 250: 1 + 9e^−0.5t = 2, so e^−0.5t = 1/9 and t = 2 ln 9 ≈ 4.39 years.

Problem 16 Continuously Compounded Savings

After 8 years: 2500e^0.336 ≈ $3,498.35.

For $4,000: e^0.042t = 1.6, so t = ln 1.60.042 ≈ 11.19 years.

Problem 17 Medicine Half-Life

25 mg: (1/2)^t/6 = 1/8 = (1/2)³, so t/6 = 3 and t = 18 hours.

60 mg: (1/2)^t/6 = 0.3, so t = 6 · ln 0.3ln 0.5 ≈ 10.42 hours.

Problem 18 Comparing Earthquakes

The difference is 6.8 − 5.3 = 1.5, so the amplitude ratio is 10^1.5 ≈ 31.6 times.

M = log 5000 ≈ 3.70.

Problem 19 Sound Levels

The difference is 45 dB, so 10 log(ratio) = 45 and the ratio is 10^4.5 ≈ 31,623 times.

Two sources double the intensity: 10 log 2 ≈ 3.01 dB more.

Problem 20 Flu Spread in a School

400: 1 + 49e^−0.35t = 2, so e^−0.35t = 1/49 and t = ln 490.35 ≈ 11.1 days.

720: 1 + 49e^−0.35t = 10/9, so e^−0.35t = 1/441 and t = ln 4410.35 ≈ 17.4 days.