Logarithms: recover an exponent and respect the domain

A logarithm answers an exponent question

For a positive base b other than 1, log_b(a) asks what exponent on b produces a. The statement log_b(a) = c is equivalent to b^c = a. In real-number work the argument a must be positive. For example, log₂(8) = 3 because 2³ = 8. The number 3 is an exponent, not a multiplier to attach to 8. Writing the equivalent exponential statement is a dependable first step when the notation feels unfamiliar. The base belongs to the operation and cannot be silently changed.

Explain the rules through products of powers

If two positive numbers are powers of the same base, multiplying them adds their exponents. That is why the logarithm of a product becomes a sum of logarithms. Dividing positive quantities subtracts the corresponding exponents. Raising a positive quantity to a power multiplies its logarithm by that power. These rules require their original real logarithms to be defined. They do not turn the logarithm of a sum into a sum of logarithms. Before applying a memorized rule, identify whether the original operation is multiplication, division, exponentiation or addition.

Worked example 1: translate before solving

Solve log₂(x - 1) = 3 over the real numbers. The logarithm imposes x - 1 > 0, so any final answer must exceed 1. Translate the logarithmic equation into an exponential equation before moving terms. This keeps the base and exponent in their proper roles. The check should use the original expression rather than only a rearranged equation.

• Interpret the statement as 2³ = x - 1.
• Evaluate 2³ = 8, so x - 1 = 8 and x = 9.
• Check the domain: 9 - 1 is positive.
• Substitute into the original: log₂(9 - 1) = log₂(8) = 3. The answer satisfies both the numerical equation and its input restriction.

Worked example 2: a combined logarithm creates candidates

Solve log₁₀(x) + log₁₀(x - 3) = 1. Here both original logarithms must exist. Their combined restriction is x > 3. Combining them gives a quadratic equation, whose roots are only candidates until this restriction is checked. The problem illustrates why a correct factorization can still leave an invalid proposed solution.

• Use the product rule: log₁₀[x(x - 3)] = 1.
• Translate to x(x - 3) = 10, giving x² - 3x - 10 = 0.
• Factor as (x - 5)(x + 2) = 0. The algebraic candidates are 5 and -2.
• Keep only 5 because x must exceed 3. Although (-2)(-5) is positive, the original separate logarithms of -2 and -5 are not real.
• Check x = 5: the two positive arguments multiply to 10, whose base-10 logarithm is 1.

Use logarithms to compare multiplicative changes

On a base-10 logarithmic scale, a difference of 3 corresponds to a ratio of 10³, or 1,000, in the underlying positive quantity. Equal steps on the logarithmic scale therefore represent equal multipliers, not equal additions. To see the distinction, compare quantities 2, 20 and 200: each is ten times the previous one, so their base-10 logarithms differ by 1. Their ordinary differences, 18 and 180, are not equal. Always check which quantity a graph or statement actually reports before interpreting a change in its displayed numbers.

Misconception: logarithms distribute over addition

The expression log₁₀(10 + 90) equals log₁₀(100), which is 2. Splitting it as log₁₀(10) + log₁₀(90) would instead describe log₁₀(900). These expressions are not equal. Another trap is accepting an algebraic root without checking every original argument. Record restrictions before combining or exponentiating so they remain visible at the end. A calculator can approximate a logarithm, but it cannot decide which model or domain the question intended.

Sources & further reading