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Exponential and Logarithmic Equations Test Answer Key with Scoring – Algebra 2

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

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

Looking for an exponential and logarithmic equations test answer key? Then this answer key gives the solutions and the scoring guide of the Algebra 2 test on Using Logs to Solve Exponent Equations. Each answer shows the work and a scoring guide. Give partial credit for a correct setup with an arithmetic slip, and use the total out of 20 to judge readiness.

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

Question 2 Exponential equation (2 points)

x = ln 50 ÷ ln 6 = 3.9120 ÷ 1.7918 = 2.1833. x ≈ 2.18.

Scoring: 1 point for the logarithm setup, 1 point for the value.

Question 3 Isolate and solve with e (2 points)

e0.5x = 7.5, so 0.5x = ln 7.5 ≈ 2.0149 and x ≈ 4.030. x ≈ 4.03.

Scoring: 1 point for isolating, 1 point for the value.

Question 4 Logarithmic equation (2 points)

2x + 3 = 72 = 49, so 2x = 46. x = 23.

Scoring: 1 point for exponential form, 1 point for the solution.

Question 5 Equation with two logarithms (3 points)

log5(x² + 4x) = 1, so x² + 4x = 5 and (x + 5)(x − 1) = 0. The value −5 makes log5 x undefined. x = 1. Check: log5 1 + log5 5 = 0 + 1.

Scoring: 1 point for combining, 1 point for factoring, 1 point for rejecting −5.

Question 6 Decay model (3 points)

  1. The curve is at 40 a little after 4 hours.
  2. 100(0.8)x = 40, so 0.8x = 0.4 and x = ln 0.4 ÷ ln 0.8 = 4.106. ≈ 4.11 hours.
0246810020406080100xyx ≈ 4.11
The curve reaches 40 at about 4.11 hours.

Scoring: 1 point for the estimate, 1 point for the logarithm setup, 1 point for the value.

Question 7 Compound interest (2 points)

2500(1 + 0.038/12)12t = 4000, so (1.0031667)12t = 1.6. Then 12t = ln 1.6 ÷ ln 1.0031667 = 0.4700 ÷ 0.00316, so t ≈ 12.388. ≈ 12.4 years.

Scoring: 1 point for the setup, 1 point for the value.

Question 8 Half-life (2 points)

200(1/2)t/5 = 25, so (1/2)t/5 = 1/8 = (1/2)3. Then t/5 = 3. t = 15 hours.

051015202504080120160200xy15 h
The amount reaches 25 mg at 15 hours.

Scoring: 1 point for the equation, 1 point for the value.

Question 9 Extraneous solution (2 points)

x² + 9x = 10, so (x + 10)(x − 1) = 0. The value −10 is rejected because log(−10) is undefined. x = 1. Check: log 1 + log 10 = 0 + 1 = 1.

Scoring: 1 point for the solutions of the quadratic, 1 point for the rejection and answer.