Chapter 1.6 — Exercise 1.5 — Logarithms

Exponential Form Meets Logarithmic Form

If a is a positive number other than 1, and aˣ = N, then x is called the logarithm of N to the base a, written x = log_a N. These two statements describe exactly the same relationship between three numbers — only which one is the "subject" of the sentence changes.

For example, 2³ = 8 means log₂8 = 3, and 10² = 100 means log₁₀100 = 2. Comfort moving back and forth between these two forms is the single most useful skill for the rest of this exercise.

Common Logarithms and Natural Logarithms

Two particular bases show up so often that they get their own shorthand:

In this exercise, whenever "log N" appears with no base written, read it as log₁₀N — the common logarithm. Quantities that vary across an enormous range — earthquake energy, sound intensity, acidity — are often measured on a logarithmic scale precisely because logarithms compress huge ranges into small, manageable numbers: a magnitude-6 earthquake releases far more than twice the energy of a magnitude-3 one, and each whole step on the pH scale represents a tenfold change in acidity.

The Six Defining Properties

Every problem that follows is solved with one of these six properties, or a combination of them. Each follows directly from the laws of exponents.

Proof: let logₐm = x and logₐn = y, so m = aˣ and n = aʸ. Then mn = aˣ·aʸ = aˣ⁺ʸ, so logₐ(mn) = x + y = logₐm + logₐn.

Proof: with the same m = aˣ, n = aʸ, dividing gives m/n = aˣ⁻ʸ, so logₐ(m/n) = x − y = logₐm − logₐn.

Proof: let m = aˣ, so mⁿ = aⁿˣ, which means logₐ(mⁿ) = nx = n·logₐm. This is the property used most often in this exercise — it pulls an exponent out in front of a logarithm.

• logₐa = 1 — because a¹ = a, so by definition the log of a to its own base is 1.
• logₐ1 = 0 — because a⁰ = 1 for any valid base a, so the log of 1 in any base is always 0.
• a^logₐm = m — if logₐm = x, then aˣ = m by definition; substituting back gives a^logₐm = m directly.

Determining a Value: Nine Quick Examples

Question 1 asks you to evaluate nine logarithmic expressions. The strategy repeats throughout: rewrite the number inside the log as a power of the base — using roots, negative exponents, or straightforward factorisation — then apply Property 3 followed by logₐa = 1.

The last row mixes ordinary exponent rules with Property 6 — 2^2+log₂3 splits into 2² · 2^log₂3 using aᵏ⁺ˡ = aᵏ·aˡ, and then 2^log₂3 collapses straight to 3 by definition, leaving 4 × 3 = 12.

Combining Logs Into a Single Expression

Question 2 runs Properties 1, 2, and 3 in reverse — collapsing several logarithms into one "log N" and evaluating it where possible.

• log 2 + log 5 = log(2×5) = log 10 = 1
• log₂16 − log₂2 = log₂(16/2) = log₂8 = log₂2³ = 3
• 3 log₆₄4 = log₆₄4³ = log₆₄64 = 1
• 2 log 3 − 3 log 2 = log 3² − log 2³ = log 9 − log 8 = log(9/8)
• log 10 + 2 log 3 − log 2 = log 10 + log 9 − log 2 = log(10×9/2) = log 45

Working in Terms of x and y

Question 3 gives you x = log₂3 and y = log₂5, and asks you to express other base-2 logarithms using only x and y. The trick is to factorise the number inside the log into 2s, 3s, and 5s, then turn the product into a sum with Property 1.

For log₂6750, the exponents 3 on both the 3 and the 5 come straight out as coefficients on x and y — logₐ(mⁿ) = n·logₐm turns 3³ and 5³ into 3x and 3y respectively, which is exactly why Property 3 is the property to reach for whenever a number inside a log is itself a power.

Expanding a Single Logarithm

Question 4 works in the opposite direction from Question 2 — breaking one logarithm into a sum or difference of simpler ones, using Properties 1, 2, and 3 together.

• log 1000 — 1000 = 2³ × 5³, so log 1000 = 3 log 2 + 3 log 5
• log(128/625) — 128 = 2⁷ and 625 = 5⁴, so this becomes 7 log 2 − 4 log 5
• log(x²y³z⁴) = 2 log x + 3 log y + 4 log z
• log(p²q³/r⁴) = 2 log p + 3 log q − 4 log r
• log√(x³/y²) = (1/2)(log x³ − log y²) = (3/2) log x − log y

When expanding a quotient inside a log, every term in the denominator picks up a minus sign — easy to lose track of once three or more terms are involved, as in the p²q³/r⁴ example above.

An Algebraic Proof Using Logs

Question 5 gives x² + y² = 25xy and asks you to prove that 2 log(x+y) = 3 log 3 + log x + log y — a proof, not a plain evaluation, so it works from both sides toward the middle.

Adding 2xy to both sides of the given equation turns the left side into a perfect square, (x+y)². Taking the logarithm of both sides of (x+y)² = 27xy then gives 2 log(x+y) = log 27 + log x + log y, and since 27 = 3³, log 27 becomes 3 log 3 — exactly matching the right-hand side that needed to be proved.

Two More Worked Equations

The last two number problems in this exercise ask you to solve for an unknown expression rather than evaluate a fixed quantity.

• Question 6 — given log[(x+y)/3] = ½(log x + log y), find x/y + y/x. Doubling both sides and applying Property 3 gives [(x+y)/3]² = xy, which expands to x² + y² + 2xy = 9xy, so x² + y² = 7xy. Dividing every term by xy gives x/y + y/x = 7.
• Question 7 — given 2.3ˣ = 0.23ʸ = 1000, find 1/x − 1/y. Each equation converts to 1/x = log₁₀₀₀2.3 and 1/y = log₁₀₀₀0.23, so 1/x − 1/y = log₁₀₀₀(2.3/0.23) = log₁₀₀₀10. Since 10 = 1000^(1/3), this equals 1/3.

Solving an Exponential Equation With Logs

Question 8 asks for x in 2ˣ⁺¹ = 3¹⁻ˣ — an equation where the variable sits in the exponent on both sides, with two different bases. Taking the logarithm of both sides is the standard move whenever that happens:

Expanding both sides first (x log 2 + log 2 = log 3 − x log 3), then collecting every x-term on one side, is what turns an equation with x buried in two different exponents into one simple linear equation for x.

Is log 2 Rational? Is log 100?

The closing question asks you to classify two logarithms — and the two answers come out opposite, for revealing reasons.

log 2: assume it's rational, log 2 = p/q for integers p, q with 0 < p < q (since log 1 = 0 and log 10 = 1, and log 2 sits strictly between them). Then 2 = 10^(p/q), so 2^q = 10^p = 2^p·5^p, which rearranges to 2^(q−p) = 5^p. The left side is always even (since q > p), but the right side is always odd — an even number can never equal an odd one. The contradiction means log 2 is irrational.

log 100: this one needs no contradiction at all. log 100 = log 10² = 2 log 10 = 2 × 1 = 2, by Property 3 directly. Since 2 can be written as 2/1, log 100 is rational — a reminder that a logarithm isn't automatically irrational just because logarithms often are.

Closing the Chapter

Logarithms bring together everything covered earlier in Real Numbers — the rational/irrational classification from Exercise 1.4 resurfaces directly in Question 9 above, and the algebraic manipulation practised here carries forward into Chapter 2 and beyond. For the number-system foundations this exercise assumes, revisit the chapter introduction, or see prime factorisation applied a different way in Exercise 1.2.