Exponential and Logarithmic Equations Worksheet Answer Key

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Last updated: October 22, 2026 · Written by the Math Made Easy math teachers

Algebra 2 Math Worksheet Answer Keys – Mathbot, the Math Made Easy robot coach, with free printable worksheet answer keys for Algebra 2

Looking for an exponential and logarithmic equations worksheet answer key? Then this answer key gives every step of the Algebra 2 worksheet on Using Logs to Solve Exponent Equations. Every solution shows the isolating step, the logarithm step and the check. Keep extra decimals until the final line, and rewrite any answer that disagrees with the rounding shown here.

Standards: F-LE.A.4, A-SSE.B.3c · Level: Grade 11 Algebra 2

Problem 2 Same base on both sides

64 = 4³, so x + 1 = 3 and x = 2.

Problem 3 Simplest logarithmic equation

Rewrite as x = 5³. x = 125.

Problem 4 Negative log value

x = 2⁻³ = 1/8. x = 1/8.

Problem 5 Isolate, then match bases

Divide by 3: 2^x = 16 = 2⁴. x = 4.

Problem 6 Exponential equation with logs

x ln 7 = ln 20, so x = 2.9957 ÷ 1.9459 = 1.5395. x ≈ 1.54.

Problem 7 Base e equation

Take ln: 2x = ln 15 ≈ 2.7081, so x ≈ 1.3540. x ≈ 1.35.

Problem 8 Logarithm with a shifted argument

Rewrite: x + 4 = 10² = 100. x = 96. Check: log 100 = 2.

Problem 9 Graph of an exponential equation

Divide by 3: 2^x = 15, so x = log₂ 15 = ln 15 ÷ ln 2 = 3.9069. The graph crosses the line just before x = 4. x = log₂ 15 ≈ 3.91.

The intersection is at x ≈ 3.91.

Problem 10 Natural log on a graph

ln x = 2 means x = e² ≈ 7.3891. x = e² ≈ 7.39.

Problem 11 Condense logs and solve

Combine: ln(x² + 2x) = ln 24, so x² + 2x − 24 = 0 and (x + 6)(x − 4) = 0. The value −6 makes ln x undefined. x = 4.

Problem 12 Different bases on each side

Take ln: (x + 3) ln 2 = x ln 5. Then 3 ln 2 = x(ln 5 − ln 2), so x = 2.0794 ÷ 0.9163 = 2.2694. x ≈ 2.27.

Problem 13 Logarithm equation leading to a quadratic

Combine: log(x² − 3x) = 1, so x² − 3x = 10 and (x − 5)(x + 2) = 0. Since x = −2 is extraneous, x = 5. Check: log 5 + log 2 = log 10 = 1.

Problem 14 Equation with a coefficient on the log

Subtract 5: 3 log₂(x − 1) = 9, so log₂(x − 1) = 3. Then x − 1 = 8. x = 9.

Problem 15 Savings goal with annual compounding

Solve 6000(1.045)^t = 9000, so 1.045^t = 1.5. Then t = ln 1.5 ÷ ln 1.045 = 0.40547 ÷ 0.04402 = 9.212. ≈ 9.21 years.

The balance reaches $9,000 at about 9.21 years.

Problem 16 Medicine half-life

Solve 80(1/2)^t/6 = 15, so (1/2)^t/6 = 3/16. Then t/6 = ln(3/16) ÷ ln(1/2) = log₂(16/3) = 2.4150. t = 14.490. ≈ 14.5 days.

Problem 17 Town population growth

Solve 1200e^0.035t = 2000, so e^0.035t = 5/3. Then 0.035t = ln(5/3) ≈ 0.5108 and t ≈ 14.60. That is during the 15th year after 2020. 2035.

Problem 18 Doubling time at 3.2%

Set Pe^0.032t = 2P. Dividing by P gives e^0.032t = 2, so t = ln 2 ÷ 0.032 = 21.661. ≈ 21.7 years. The amount P cancels, so the doubling time is the same for every deposit.

Problem 19 Hydrogen ion concentration

−log[H⁺] = 4.3, so log[H⁺] = −4.3 and [H⁺] = 10^−4.3 ≈ 5.0119 × 10⁻⁵. ≈ 5.01 × 10⁻⁵ moles per liter.

Problem 20 Comparing two sound levels

The levels differ by 24 dB. Then 10 log(I₁/I₂) = 24, so log(I₁/I₂) = 2.4 and the ratio is 10^2.4 = 251.19. ≈ 251 times.

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Exponential and Logarithmic Equations Worksheet Answer Key – Algebra 2

Problem 1 Match the bases