ACT Math study guide · Chapter 18
Logarithms
Three properties let logarithms be combined or split apart. The product rule, log(a) + log(b) = log(ab), turns a sum of logs into the log of a product. The quotient rule, log(a) − log(b) = log(a/b), turns a difference into the log of a quotient. The power rule, log(xⁿ) = n·log(x), pulls an exponent out as a multiplying coefficient.
These rules apply only where the logarithms involved are defined; in the real-number setting used here, each logarithm's argument must be positive.
Solving a logarithmic equation means rewriting it in exponential form. log₇(x) = k is equivalent to x = 7ᵏ: the base stays the base, the logarithm's value becomes the exponent, and the logarithm's argument becomes the exponential result. More generally, for any base b, a logarithm base b of x equal to k is equivalent to x = bᵏ. When an equation has two log terms, combine them into a single logarithm first using the product or quotient rule, then convert to exponential form.
An exponential equation with a base that doesn't convert neatly, such as 2^x = 30, is solved by taking the logarithm of both sides: log(2^x) = log(30), so x·log(2) = log(30), giving x = log(30)/log(2). This works with any base of logarithm, as long as the same base is used on both sides of the division.
Where this appears as a model: a quantity measured on a logarithmic scale. A sound with 100 times the reference intensity has a decibel level of 10log(100)=20 decibels, using the defining relationship between decibels and intensity ratio.
▸ Combining logarithms to solve an equation
Worked Example Solve log(x) + log(5) = log(35).
Combine the left side using the product rule, then set the arguments equal.
log(5x) = log(35)
5x = 35
x = 7
▸ Solving an exponential equation with logarithms
Worked Example Solve 4^x = 50 using logarithms. Round to the nearest hundredth.
Take the log of both sides, then solve for x.
x·log(4) = log(50)
x = log(50)/log(4) ≈ 2.82
Worked Example Solve log₂(x) + log₂(x − 2) = 3.
Recognize. Two logarithms with the same base are being added. Combine them with the product rule before converting to exponential form.
Set up.
log₂(x(x − 2)) = 3
Solve.
x(x − 2) = 2³ = 8
x² − 2x − 8 = 0
(x − 4)(x + 2) = 0
x = 4 or x = −2
Check. x = −2 would require log₂(−2), which is undefined, so it is rejected. For x = 4: log₂(4) + log₂(2) = 2 + 1 = 3. ✓
Best route. Hand algebra after combining the logs. A calculator does not decide which candidate solutions are valid.
Common error. Reporting both x = 4 and x = −2 without checking that a logarithm's argument must be positive.
▸ Practice
Question 1
- AA. log(1.5)
- BB. log(2)
- CC. log(10)
- DD. log(24)
Question 2
- FF. log(5)
- GG. log(80)
- HH. log(24)
- JJ. log(16)
Question 3
- AA. (log x)⁵
- BB. 5 log(x)
- CC. log(x)/5
- DD. log(5x)
Question 4
- FF. ab
- GG. a/b
- HH. a + b
- JJ. a − b
Question 5
- AA. 2.408
- BB. 0.903
- CC. 0.602
- DD. 0.301
Question 6
- FF. 18
- GG. 7
- HH. 24
- JJ. 63
Question 7
- AA. 6
- BB. 8
- CC. −2
- DD. 0.5
Question 8
- FF. 7
- GG. 25
- HH. 32
- JJ. 10
Question 9
- AA. 100
- BB. 16
- CC. 10000
- DD. 40
Question 10
- FF. 1.431
- GG. 0.778
- HH. 0.1436
- JJ. 0.176
Question 11
- AA. 3.91
- BB. 0.20
- CC. 15.00
- DD. 4.91
Question 12
- FF. 16
- GG. 4
- HH. 6
- JJ. 7
Question 13
- AA. 2 log(x)
- BB. log(3x)
- CC. log(x)
- DD. log(x⁵)
Question 14
- FF. 64c
- GG. 2c
- HH. c/3
- JJ. 3c
Question 15
- AA. 2.86
- BB. 20.00
- CC. 0.35
- DD. 1.86
Question 16
- FF. 1.5
- GG. 50.5
- HH. 50
- JJ. 101
Question 17
- AA. log(a+b−c)
- BB. log(a−b/c)
- CC. log(ab/c)
- DD. log(ab−c)
Question 18
- FF. 1.69
- GG. 3.38
- HH. 0.845
- JJ. 1.267
Question 19
- AA. 12
- BB. 1
- CC. 4/3
- DD. 7
Question 20
- FF. 1.78
- GG. 3.56
- HH. 8.33
- JJ. 0.89
▸ Answers and Explanations
Show the answers and explanations
1: D. The product rule states log(a) + log(b) = log(ab): log(6) + log(4) = log(24). Choice C adds the arguments (6+4=10) instead of multiplying them.
2: F. The quotient rule states log(a) − log(b) = log(a/b): log(20) − log(4) = log(5). Choice J subtracts the arguments (20−4=16) instead of dividing them.
3: B. The power rule states log(xⁿ) = n·log(x): log(x⁵) = 5 log(x). Choice A treats the exponent as applying to log(x) itself instead of pulling it out as a coefficient.
4: H. Since 15 = 3×5, log(15) = log(3) + log(5) = a + b. Choice F multiplies a and b instead of adding them, mismatching the product rule for logs with ordinary multiplication.
5: B. Since 8 = 2³, log(8) = 3·log(2) = 3(0.301) = 0.903. Choice D reports log(2) itself, forgetting to multiply by the exponent 3.
6: G. Combine the left side: log(3x) = log(21), so 3x = 21, giving x = 7. Choice F subtracts 3 from 21 instead of setting up the equation 3x=21.
7: B. Combine the left side: log(x/4) = log(2), so x/4 = 2, giving x = 8. Choice D divides 2 by 4 instead of multiplying, misapplying the quotient rule in reverse.
8: H. Rewrite in exponential form: x = 2⁵ = 32. Choice G squares the base instead of raising it to the given exponent, computing 5² instead of 2⁵.
9: A. Rewrite in exponential form: x² = 10⁴ = 10,000, so x = 100 (taking the positive root since x > 0). Choice C reports x² itself, forgetting to take the square root to find x.
10: G. Since 6 = 2×3, log(6) = log(2) + log(3) = 0.301 + 0.477 = 0.778. Choice H multiplies the two log values instead of adding them, mismatching the product rule for logs with ordinary multiplication.
11: D. Taking the log of both sides: x·log(2) = log(30), so x = log(30)/log(2) ≈ 4.91. Choice C divides 30 by 2 instead of using the logarithm relationship.
12: J. Rewrite in exponential form: x + 2 = 3² = 9, so x = 7. Choice H computes 2³−2 instead of 3²−2, swapping the base and exponent.
13: C. 3 log(x) = log(x³), so log(x³) − log(x²) = log(x³/x²) = log(x). Choice D adds the exponents instead of subtracting them when combining the two logarithmic terms.
14: J. Since 64 = 4³, log(64) = 3·log(4) = 3c. Choice G uses the wrong exponent, treating 64 as 4² (which equals 16, not 64) instead of 4³.
15: A. Taking the log of both sides: x·log(5) = log(100), so x = log(100)/log(5) ≈ 2.86. Choice B divides 100 by 5 instead of using the logarithm relationship.
16: G. Rewrite in exponential form: 2x − 1 = 10² = 100, so 2x = 101, giving x = 50.5. Choice H drops the −1 term when isolating x, solving 2x=100 instead of 2x=101.
17: C. Combining the product and quotient rules: log(a) + log(b) − log(c) = log(ab/c). Choice A adds and subtracts the arguments directly instead of applying the product and quotient rules.
18: F. Since 49 = 7², log(49) = 2·log(7) = 2(0.845) = 1.69. Choice H reports log(7) itself, forgetting to multiply by the exponent 2.
19: A. Combine the left side: log₅(x/3) = log₅(4), so x/3 = 4, giving x = 12. Choice B subtracts 3 from 4 instead of setting up the equation x/3=4.
20: F. Taking the log of both sides: 2x·log(3) = log(50), so 2x = log(50)/log(3) ≈ 3.56, giving x ≈ 1.78. Choice G stops after solving for 2x, forgetting to divide by 2 to isolate x.