Look for this. A logarithm answers the question ‘What power of this base gives that positive number?’
Translate logᵦ A = c into bᶜ = A, where b > 0, b ≠ 1 and A > 0. Rewrite the base and argument using a common base for an exact exponent, or bracket them between powers.
Worked example
Determine log₉(1/27) exactly, and check its sign before calculating.
Since 9 > 1 and its target lies between 0 and 1, the exponent is negative. Write 9ˣ = 1/27 as 3²ˣ = 3⁻³. Then 2x = −3, giving x = −3/2.
Try it yourself
An image doubles in area at each zoom level. A fractional setting is allowed. Describe the zoom level that multiplies area by 6, and give two consecutive integer bounds.
Check your reasoning
At level z the factor is 2ᶻ. Solving 2ᶻ = 6 gives z = log₂ 6. Because 2² = 4 < 6 < 8 = 2³, the setting lies between levels 2 and 3 (about 2.58).
Outcome RF8
Combine logarithms while keeping their original domain
Look for this. Logs with the same base can combine when they are added or subtracted. A multiplier can become an exponent.
Record positive-argument restrictions first. Use k logᵦ A = logᵦ(Aᵏ), then turn a sum into a product or a difference into a quotient. Check the result with an allowed value.
Worked example
Write 2 log₃ x − log₃(x − 2) as one logarithm, with its domain. Check at x = 3.
The arguments give x > 0 and x − 2 > 0, together x > 2. The result is log₃(x²/(x − 2)). At 3 the argument is 9, so its log is 2, matching 2 log₃ 3 − log₃ 1.
Try it yourself
You know log₅ p = 2 and log₅ q = 3. Without finding p and q first, work out log₅(√q/p²). Explain each change.
Check your reasoning
The quotient gives log₅(√q) − log₅(p²). Bring down the powers to get ½ log₅ q − 2 log₅ p = 3/2 − 4 = −5/2. Both p and q are positive because their logarithms were given.
Outcome RF9
Sketch an exponential and its inverse from anchors
Look for this. A shifted exponential has a horizontal asymptote. Reading its axes in reverse gives a logarithmic graph.
Locate the asymptote and choose inputs producing exponents 0 and 1. Swap the coordinates to sketch the inverse. The horizontal asymptote becomes a vertical one.
Worked example
Sketch y = 2^(x − 1) + 3 and its inverse. Label two points on each, both asymptotes and the inverse domain.
The exponential approaches y = 3 from above and passes through (1, 4) and (2, 5). Swap coordinates to plot (4, 1) and (5, 2). From x = 2^(y − 1) + 3, the inverse is y = 1 + log₂(x − 3), with asymptote x = 3 and domain x > 3.
Try it yourself
A graphing screen shows y = 4 − log₂(x + 1). A friend expects it to rise. Use two points, an intercept and the asymptote to decide how to sketch it.
Check your reasoning
The domain is x > −1, so draw the vertical asymptote x = −1. When x = 0 and 1, the points are (0, 4) and (1, 3). The graph decreases as x increases. At y = 0, x + 1 = 2⁴, so the x-intercept is (15, 0). Its range is all real y.
Outcome RF10
Solve for a time or an unknown logarithm argument
Look for this. If time is in an exponent, isolate the power. If the unknown is inside two logarithms, check the arguments and use the log laws.
Equating exponents works for equal bases. For other positive bases, taking a common log of both sides brings the exponent down. Substitute final candidates in the original model.
Worked example
A lamp’s brightness follows B(t) = 80(0.8)^(t/5) lux after shutdown, with t in minutes. When does it reach 45 lux? If readings are made at whole minutes, when is the first reading below 45?
Divide by 80, then take common logs. This gives t = 5 log(45/80)/log(0.8) ≈ 12.89 min. The brightness decreases, so the first whole minute below 45 is 13. At minutes 12 and 13 the model gives 46.83 and 44.78 lux.
Try it yourself
Solve log₂(x − 1) + log₂(x − 3) = 3. Decide which algebraic candidates are allowed.
Check your reasoning
The original domain is x > 3. Combine the logs, then undo the logarithm to obtain (x − 1)(x − 3) = 8. Thus x² − 4x − 5 = (x − 5)(x + 1) = 0. Only 5 is allowed. It gives log₂ 4 + log₂ 2 = 2 + 1 = 3.
Worked notes and diagrams
1
The words, first
The idea: an exponential question is often only hard because one of these terms was never pinned down.
The exponential and the logarithm are mirror images in the line y = x. Each one's asymptote is the axis the other one crosses.
Word
What it means
Base
The number being multiplied by itself, in 2⁵ and in log₂ 8 alike.
Exponent
The power applied to a base. Whole positive exponents count factors, negative exponents take reciprocals and fractional exponents can represent roots.
Logarithm
The exponent that turns the base into a given number: log₂ 8 = 3 because 2³ = 8.
Argument
The number inside the logarithm. It must be positive.
Common logarithm
The base-10 logarithm, written log x. For example, log 100 = 2.
Exponential function
y = bˣ, where the variable is the exponent.
Growth or decay factor
What the amount multiplies by each period: 2 for doubling, ½ for halving, 1.06 for 6% growth.
Half-life
The time for half of a sample to decay — a period with a factor of ½.
Asymptote
A line approached by a graph near a boundary or far along its domain. A graph can cross a horizontal asymptote.
Extraneous root
A solution the algebra produces that the original equation rejects.
2
Exponential and log graphs
The idea: y = bˣ and y = logb x are inverses, so their graphs are reflections of each other in the line y = x.
Swapping x and y turns one into the other: (0, 1) on y = 2ˣ becomes (1, 0) on y = log₂ x.
y = bˣ
y = logb x
Domain
all real numbers
x > 0
Range
y > 0
all real numbers
Intercept
(0, 1)
(1, 0)
Asymptote
horizontal, y = 0
vertical, x = 0
When b > 1 the exponential grows; when 0 < b < 1 it decays. Transformations work the way they always do: y = log(x − 2) moves the asymptote to x = 2, so its domain is x > 2.
3
What a logarithm actually is
The idea: logb x = y means exactly the same thing as by = x. The log is the exponent.
Read log₂ 8 as "2 to what power gives 8?" The answer is 3. Every evaluation works this way:
log 1000 = 3, because 10³ = 1000 (no base written means base 10)
log₅ 1 = 0, because anything to the power 0 is 1
log₃(1/9) = −2, because 3⁻² = 1/9
logb b = 1, always
Watch out: you cannot take the log of zero or a negative number. No power of a positive base ever gives you one.
4
The laws of logarithms
The idea: logs turn multiplying into adding, dividing into subtracting, and powers into multipliers — because that is exactly what exponents do.
Law
Rule
Example
Product
log(MN) = log M + log N
log 25 + log 4 = log 100 = 2
Quotient
log(M/N) = log M − log N
log₂ 12 − log₂ 3 = log₂ 4 = 2
Power
log(Mⁿ) = n log M
2 log 5 = log 25
Change of base
logb x = log x ÷ log b
log₃ 20 = log 20 ÷ log 3 ≈ 2.73
Writing logs in terms of other logs
If log 2 = a and log 3 = b, then log 12 = log(2² × 3) = 2 log 2 + log 3 = 2a + b. Break the number into factors you know, then apply the laws.
Watch out: log(M + N) is not log M + log N, and log M ÷ log N is not log(M/N). The laws only work in the shapes shown above.
5
Solving equations and real-world models
The idea: if you can make the bases match, match the exponents. If you cannot, take the log of both sides.
Method 1: common base
4ˣ⁺¹ = 8ˣ. Both are powers of 2: 2²⁽ˣ⁺¹⁾ = 2³ˣ. Now the exponents must be equal: 2x + 2 = 3x, so x = 2.
Method 2: take logs
3ˣ = 20 has no common base. Take the log of both sides: x log 3 = log 20, so x = log 20 ÷ log 3 ≈ 2.73. Sanity check: 3² = 9 and 3³ = 27, so the answer must be between 2 and 3.
Log equations: always check
log₂ x + log₂(x − 2) = 3 combines to x(x − 2) = 2³ = 8, which gives x = 4 or x = −2. But log₂(−2) is undefined, so −2 is extraneous. The only solution is x = 4.
The models you will see
Growth and decay: A = A₀(b)t/p, where p is how long one doubling or halving takes. 80 g with a 5-year half-life after 15 years: 80(½)³ = 10 g.
Compound interest: A = P(1 + i)ⁿ. To double $1000 at 6%: 2 = 1.06ⁿ, so n = log 2 ÷ log 1.06 ≈ 11.9 years.
Richter magnitude: at the same recording distance, M₂ − M₁ = log(A₂/A₁) for measured wave amplitudes A. A magnitude difference of 2 gives an amplitude ratio of 10² = 100.
Sound intensity level: L₂ − L₁ = 10 log(I₂/I₁) in decibels. A rise of 10 dB means 10 times the intensity. A rise of 1 dB means a factor of 100.1 ≈ 1.26.
pH: in the concentration model pH = −log[H⁺], increasing pH from 4 to 5 decreases the hydrogen ion concentration from 10⁻⁴ to 10⁻⁵ mol/L, one tenth as much.
Watch out: on a log scale, a difference of 2 means 100 times, not twice. The number on the scale is the exponent.
Practice by topic
Work on paper, choose an answer, then compare your method with the explanation.
Exponential and log graphs · 3 questions
Practice 1
What is the horizontal asymptote of y = 2ˣ?
x = 0
y = 1
y = 2
y = 0
Answer and explanation
y = 0
As x gets very negative, 2ˣ shrinks toward 0 but never reaches it, so the graph hugs y = 0. y = 1 is the y-intercept.
Practice 2
y = log₃ x is the inverse of which function?
y = 3x
y = 3ˣ
y = x/3
y = x³
Answer and explanation
y = 3ˣ
A logarithm undoes an exponential with the same base, so log₃ x and 3ˣ are inverses — reflections of each other in y = x.
Practice 3
What is the domain of y = log(x − 2)?
all real numbers
x > 0
x > −2
x > 2
Answer and explanation
x > 2
You can only take the log of a positive number, so x − 2 > 0, meaning x > 2. The graph has a vertical asymptote at x = 2.
Laws of logarithms · 7 questions
Practice 1
Evaluate log₂ 8.
16
4
2
3
Answer and explanation
3
A logarithm asks for an exponent: 2 to what power is 8? 2³ = 8, so the answer is 3.
Practice 2
Evaluate log 1000.
4
2
3
100
Answer and explanation
3
With no base written, the base is 10. 10³ = 1000, so log 1000 = 3.
Practice 3
Evaluate log₅ 1.
5
undefined
0
1
Answer and explanation
0
Any base to the power 0 is 1, so log of 1 is always 0. It is log₅ 5 that equals 1.
Practice 4
Evaluate log₂ 12 − log₂ 3.
4
3
log₂ 9
2
Answer and explanation
2
Subtracting logs with the same base means dividing inside: log₂(12 ÷ 3) = log₂ 4 = 2. log₂ 9 wrongly subtracts the insides.
Practice 5
Evaluate log₃(1/9).
1/2
2
−3
−2
Answer and explanation
−2
1/9 = 3⁻², so the exponent is −2. A fraction less than 1 always gives a negative log when the base is bigger than 1.
Practice 6
Evaluate 2 log 5 + log 4.
log 14
100
10
2
Answer and explanation
2
The power law moves the 2 inside: log 25 + log 4. The product law combines them: log 100 = 2.
Practice 7
If log 2 = a and log 3 = b, which expression equals log 12?
a + b
2ab
a²b
2a + b
Answer and explanation
2a + b
12 = 2² × 3, so log 12 = log 2² + log 3 = 2a + b. Logs turn multiplying into adding and powers into multipliers.
Solving equations · 4 questions
Practice 1
Solve 2ˣ = 32.
5
16
6
4
Answer and explanation
5
Write 32 as a power of 2: 32 = 2⁵. Same base on both sides means the exponents match, so x = 5.
Practice 2
Solve 4ˣ⁺¹ = 8ˣ.
2
3
4
1
Answer and explanation
2
Rewrite both with base 2: 2²⁽ˣ⁺¹⁾ = 2³ˣ. Match exponents: 2x + 2 = 3x, so x = 2.
Practice 3
Solve log₂ x + log₂(x − 2) = 3.
x = 8
x = −2 only
x = 4 only
x = 4 and x = −2
Answer and explanation
x = 4 only
Combine: log₂[x(x − 2)] = 3, so x² − 2x = 8, giving x = 4 or x = −2. But log₂(−2) does not exist, so −2 is extraneous. Always check.
Practice 4
Solve 3ˣ = 20, to the nearest hundredth.
1.82
6.67
0.37
2.73
Answer and explanation
2.73
No common base, so take logs: x log 3 = log 20, giving x = log 20 ÷ log 3 ≈ 2.73. Check: 3² = 9 and 3³ = 27, so it must be between 2 and 3.
Growth, decay and scales · 4 questions
Practice 1
A sample of 80 g has a half-life of 5 years. How much remains after 15 years?
10 g
40 g
20 g
5 g
Answer and explanation
10 g
15 years is 3 half-lives: 80 → 40 → 20 → 10 g. In formula form, 80(1/2)^(15/5) = 10.
Practice 2
At equal recording distances, two Richter magnitudes satisfy M₂ − M₁ = log₁₀(A₂/A₁), where A is measured wave amplitude. How many times as large is the amplitude for magnitude 7 as for magnitude 5?
20
100
1000
2
Answer and explanation
100
The difference 7 − 5 = 2 equals log₁₀(A₂/A₁). The amplitude ratio is 10² = 100.
Practice 3
$1000 is invested at 6% interest, compounded annually. To the nearest tenth, how many years until it doubles?
12.5
33.3
16.7
11.9
Answer and explanation
11.9
Set 2000 = 1000(1.06)ⁿ, so 2 = 1.06ⁿ and n = log 2 ÷ log 1.06 ≈ 11.9. 16.7 comes from 100 ÷ 6, which ignores compounding.
Practice 4
A bacteria culture doubles every 3 hours. How long does it take to grow from 500 to 64 000 bacteria?
42 hours
21 hours
24 hours
7 hours
Answer and explanation
21 hours
64 000 ÷ 500 = 128 = 2⁷, so it doubles 7 times. 7 doublings × 3 hours each = 21 hours. 7 is the number of doublings, not the time.
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