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Logarithms Worksheet Answer Key – Algebra 2

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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 a logarithms worksheet answer key? Then this answer key gives every step of the Algebra 2 worksheet on Logarithms: The Inverse of Exponents. Each solution below shows the key step, usually the conversion to an exponent, so that mistakes are easy to trace. Compare every step with the work, not only the final answer, and redo any problem whose method differs.

Standards: F-LE.A.4, F-BF.B.4a, F-IF.C.7e · Level: Grade 11 Algebra 2

Problem 1 Exponential to logarithmic form

The base 2 stays the base and the exponent 9 becomes the value of the log. log2 512 = 9.

Problem 2 Logarithmic to exponential form

The base is 6 and the log value 3 is the exponent. 63 = 216.

Problem 3 Evaluate a basic logarithm

35 = 243, so log3 243 = 5.

Problem 4 Logarithm of a power of 10

0.001 = 10−3, so log 0.001 = −3.

Problem 5 Find the base from a graph

The point (4, 1) means logb 4 = 1, so b1 = 4 and b = 4. The other points agree: 42 = 16 and 4−1 = 0.25. Since 43 = 64, b = 4 and log4 64 = 3.

Problem 6 Logarithm with a fractional base

Let log1/2 8 = y. Then (1/2)y = 8. Since 8 = 23 = (1/2)−3, y = −3.

Problem 7 Logarithm giving a fraction

25y = 5 means (52)y = 51, so 2y = 1. y = 12.

Problem 8 Natural logarithm of a power of e

ln is the log base e, so ln e4 = 4. Also 1e = e−1, so ln 1e = −1. 4 and −1.

Problem 9 Expand a logarithm

log3 9 + log3 x − log3 y2 = 2 + log3 x − 2log3 y. 2 + log3 x − 2 log3 y.

Problem 10 Condense a logarithmic expression

Power rule: log5 m2 + log5 4 − log5 n. Product and quotient: log5(4m2n).

Problem 11 Change of base

log3 50 = ln 50 ÷ ln 3 ≈ 3.9120 ÷ 1.0986 ≈ 3.5609. ≈ 3.56.

Problem 12 Domain and asymptote

The argument must be positive: x − 3 > 0, so x > 3. The asymptote is where the argument is 0. For the intercept, log2(x − 3) = 0 gives x − 3 = 1. Domain x > 3; asymptote x = 3; x-intercept (4, 0).

−1123456789101112−3−2−11234xyx-intercept (4, 0)
Asymptote x = 3 and intercept (4, 0).

Problem 13 Locate a log on a number line

1000 = 103 and 10,000 = 104, and 1000 < 5000 < 10,000. So the log lies between 3 and 4. A calculator gives log 5000 ≈ 3.699. Between 3 and 4; ≈ 3.7.

Problem 14 Reflect points to find an inverse

Reflecting over y = x swaps the coordinates: (1, 0), (3, 1) and (9, 2). These satisfy y = log3 x. (1, 0), (3, 1), (9, 2); inverse y = log3 x.

−1123456789−1123456789xy(1, 0)(3, 1)(9, 2)
The inverse points sit on the orange curve.

Problem 15 Earthquake energy comparison

A magnitude difference of 7.1 − 5.9 = 1.2 means the amplitudes differ by a factor of 101.2. 101.2 ≈ 15.85. About 15.85 times larger.

Problem 16 Decibel level of a vacuum cleaner

I/I0 = 3.2 × 10−5 ÷ 10−12 = 3.2 × 107. Then log(3.2 × 107) = log 3.2 + 7 ≈ 0.5051 + 7 = 7.5051. L = 75.051. ≈ 75.1 dB.

Problem 17 pH of a soil sample

pH = −log(2.5 × 10−6) = −(log 2.5 − 6) = 6 − 0.3979 = 5.6021. pH ≈ 5.60, which is acidic (below 7).

Problem 18 Inverse of a logarithmic function

Write y = 2 log5 x + 1, swap: x = 2 log5 y + 1. Then log5 y = (x − 1)/2, so y = 5(x−1)/2. Check: f(25) = 2·2 + 1 = 5 and f−1(5) = 52 = 25. f−1(x) = 5(x−1)/2.

Problem 19 Using given log values

18 = 2 · 32, so logb 18 = 0.43 + 2(0.68) = 1.79. Also 8/9 = 23/32, so logb(8/9) = 3(0.43) − 2(0.68) = 1.29 − 1.36 = −0.07. 1.79 and −0.07.

Problem 20 Justify a change-of-base consequence

log8 32 = log2 32log2 8 by change of base. Since 32 = 25 and 8 = 23, the quotient is 5/3. Check: 85/3 = (23)5/3 = 25 = 32. log8 32 = 5/3.

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