Logarithms and logarithmic functions

2ˣ = 8 is easy: x = 3. But what about 2ˣ = 5? No whole number works: 2² = 4 is too small and 2³ = 8 too large. The exact answer is a new number, log₂ 5 ≈ 2.32, «the power to which 2 must be raised to give 5». Logarithms belong to the syllabus line «basic elementary functions», and CSCA items test them in a few fixed ways: evaluating an expression, comparing three numbers, finding a domain, and solving an equation or an inequality. Everything here builds on the lesson «Power and exponential functions»: a logarithm is simply an exponent read backwards.

What a logarithm is

Laws of logarithms and change of base

Every law of logarithms is a law of powers read backwards. If M = aᵐ and N = aⁿ, then MN = aᵐ⁺ⁿ, so logₐ(MN) = m + n: multiplication inside the logarithm becomes addition outside, division becomes subtraction, and a power becomes a factor.

The logarithmic function y = logₐ x

Two quick consequences. Domain: in y = logₐ f(x) the argument must be positive, f(x) > 0; if x also appears in the base, the base must be positive and ≠ 1. Fixed point (定点): because logₐ 1 = 0 for every a, the graph of y = logₐ(kx + m) + c passes through the point where kx + m = 1, whatever the base.

Monotonicity of y = logₐ f(x). Work only inside the domain. With a > 1 the logarithm keeps the direction of f; with 0 < a < 1 it reverses it. Chinese students remember 同增异减: «same directions give increasing, different directions give decreasing».

Comparing without a calculator. Monotonicity is also the key to comparing logarithms. Try these four moves in order:

1. 1
   Same base

   Use monotonicity: log₃ 5 < log₃ 7 (base > 1), but log₀.₅ 5 > log₀.₅ 7 (base < 1).

2. 2
   Same argument

   If the argument and the bases are all above 1, the larger base gives the smaller value: log₂ 7 > log₃ 7 > log₅ 7 (by change of base, logₐ 7 = 1 / log₇ a).

3. 3
   Benchmarks 0 and 1

   Find the sign first, then compare with 1 = logₐ a. log₃ 2 lies between 0 and 1, log₂ 3 is above 1, log₀.₅ 3 is negative.

4. 4
   An intermediate number

   If both numbers lie between the same benchmarks, insert a number such as 1/2 or 3/2 and compare powers: log₂ 3 > 3/2 because 2^(3/2) = 2√2 ≈ 2.83 < 3.

Logarithmic equations and inequalities

1. 1
   Write the domain

   Every argument > 0; a base containing x must be > 0 and ≠ 1. Do this before any transformation (定义域优先, «domain first»).

2. 2
   Bring to one base

   Combine with the laws, change the base, or substitute t = logₐ x when the equation is quadratic in the logarithm.

3. 3
   Remove the logarithm

   logₐ f = b ⇔ f = aᵇ; logₐ f = logₐ g ⇔ f = g. In an inequality keep the sign for a > 1 and reverse it for 0 < a < 1.

4. 4
   Intersect with the domain

   Reject extraneous roots (增根) and intersect the solution set of an inequality with the domain.

How CSCA asks about this

• Evaluate an expression: combine with the laws, use the domino rule and a^(logₐ N) = N; the answer is usually a small integer or a simple fraction, so if you get something like lg 7, look for a mistake.
• Order a, b, c: signs first, then the benchmarks 0 and 1, then an intermediate number (1/2, 2/3, 3/2, 2); mixed items put an exponential value such as 2^0.3 next to the logarithms (see «Power and exponential functions»).
• Domain: a logarithm often comes together with a square root or a fraction; all conditions go into one system.
• Graph facts: the fixed point, which graph belongs to which base, and the monotonic intervals of logₐ f(x).
• Equations and inequalities: domain first; the options often contain the extraneous root or the answer without the domain restriction.
• Time-savers (75 s per item): substitute the options into an equation instead of solving it; for «which interval» items test one point from each option; when an item gives lg 2 ≈ 0.301, it expects an estimate, not an exact value.