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Exact value of a logarithm

Write it as a power

  • In 3 of 20 recent exams
  • Paper 1, no calculator
  • 1 to 3 marks
How it is set in the exam

Paper 1 (no calculator) only, 1–3 marks per part, usually several short parts together: part (a) of a log question, or evaluating ff or f−1f^{-1} early in a functions question. Typical (3 of 20 exam versions, 8 parts, all Paper 1 (no calculator)).

Spot it in the exam
Exam phrases to spot (3)

  • “find the exact value of log⁡b(…)\log_b(\ldots)”

  • “evaluate log⁡b(…)\log_b(\ldots) without a calculator”

  • “find f(a)f(a)”

In disguise (3)
  • the value of the inverse of an exponential function at a number, once f−1(x)=log⁡axf^{-1}(x) = \log_a x has been written down or found in an earlier part: f−1(b)=log⁡abf^{-1}(b) = \log_a b
  • a log function evaluated at a given xx where its argument is 11: the value is 00, with nothing to solve (finding where a log function is 00 is Solve a logarithmic equation)
  • an argument written as a decimal
Variants (11)
  • the argument is a reciprocal, a negative power of the base: a negative answer
  • the argument is a root: a fraction answer
  • the base and the argument are powers of the same prime, but neither is a power of the other
  • the base is a surd or a power of 1010: change of base, or log⁡anx=1nlog⁡ax\log_{a^n} x = \frac{1}{n}\log_a x
  • f−1f^{-1} of an exponential at a surd, once f−1(x)=log⁡axf^{-1}(x) = \log_a x has been written down; a log of a multiple of xx at one point, then at the point where its argument is 11

Not this type (check the first bullet first: are any log values given?):

  • a log value, or a letter for one, is given, or several logs must be combined first, even if a number is asked for: Log laws
  • the unknown is inside a log or is its base: Solve a logarithmic equation
  • the formula of f−1f^{-1} is asked for: Find the inverse function
  • f−1(b)f^{-1}(b) before any formula for f−1f^{-1} has been found (solve f(x)=bf(x) = b; for an exponential ff it is the same common-base working), or for an ff that is neither an exponential nor a log: Values of f, f∘g and f⁻¹ from a graph or a formula
  • a log-scale model in a context (pH, decibels): Exponential and log models
  • a decimal value of a log, needed on inside another question: Change of base (the exact quotient first)
Worked example

Three evaluations: a decimal, a root, a reciprocal

Paper 1 (no calculator)
Question

Find the exact value of
(a) log⁡50.04\log_5 0.04 [2]
(b) log⁡322\log_{32} \sqrt{2} [2]
(c) log⁡9127\log_9 \frac{1}{27} [2]

Try it yourself first, then compare.

Now try a question

Questions of this type are on their way. Try another one meanwhile: write your working on paper or on screen and see where each mark went. Free, with no account.

A handwritten page as Paper Zero marks it: ticks at the two lines that earned a mark, a cross on line 3 where +7 was copied as −7, and a short note beside that line.123456f′(x) = 4x − 124x − 12 = 0, so x = 3f(3) = 2(3)² − 12(3) − 7= 18 − 36 − 7 = −25minimum at (3, −25)M1 ✓A1 ✓A1 ✗11The +7 became −7: 18 − 36 + 7 = −11.
Which method?
  • The base and the argument are powers of one number (usually a prime)
    Method
    common base: write both as powers of the prime, then equate the exponents
    Example
    log⁡84\log_8 4: 23x=222^{3x} = 2^2, so 23\frac{2}{3}
  • The base is itself a power or a root (a power of 1010, a surd)
    Method
    change of base, or log⁡anx=1nlog⁡ax\log_{a^n} x = \frac{1}{n}\log_a x
    Example
    log⁡1001000=12log⁡1000=32\log_{100} 1000 = \frac{1}{2}\log 1000 = \frac{3}{2}
  • The argument is a reciprocal or a root of a power of the base
    Method
    a negative or fractional exponent (or the power law)
    Example
    log⁡717=log⁡77−12=−12\log_7 \frac{1}{\sqrt{7}} = \log_7 7^{-\frac{1}{2}} = -\frac{1}{2}
Steps

  1. Call the value xx

    and write the exponential form: log⁡ab=x  ⟺  ax=b\log_a b = x \iff a^x = b.

  2. Write both sides as powers of one number

    , usually a prime (or 1010, or the base itself, as in log⁡6136=−2\log_6 \frac{1}{36} = -2). Roots become fractional powers, reciprocals negative powers.

  3. Equate the exponents

    and solve the linear equation. The answer is exact: an integer or a fraction, often negative.

  4. A log function:

    substitute first and simplify the argument, then steps 1–3. Where the argument is 11, the value is 00.

  5. f−1(b)f^{-1}(b) once f−1(x)=log⁡axf^{-1}(x) = \log_a x has been written down

    (for f(x)=axf(x) = a^x): substitute, f−1(b)=log⁡abf^{-1}(b) = \log_a b; then steps 1–3 (call the value yy there, since xx is already the function's variable).

  6. Check:

    the base raised to your answer must give the argument.

Common mistakes

  • Dividing the numbers instead of the exponents

    log⁡84\log_8 4 is not 48\frac{4}{8}. As powers of 22, 23x=222^{3x} = 2^2, so x=23x = \frac{2}{3}.

  • Losing the sign of a reciprocal

    log⁡5125=−2\log_5 \frac{1}{25} = -2, not 22 and not 12\frac{1}{2}.

  • Swapping the base and the argument

    log⁡497=12\log_{49} 7 = \frac{1}{2}, but log⁡749=2\log_7 49 = 2. Say it as "49 to what power gives 7?".

More mistakes (5)

  • A surd base with a number in front

    22=2322\sqrt{2} = 2^{\frac{3}{2}}, not 2122^{\frac{1}{2}}.

  • A decimal answer on

    the markscheme wants the exact integer or fraction.

  • Calling log⁡a1\log_a 1 undefined

    it is 00. It is log⁡a0\log_a 0 that does not exist.

  • Reading f−1(b)f^{-1}(b) as 1f(b)\frac{1}{f(b)}

    for f(x)=25xf(x) = 25^x, f−1(b)=log⁡25bf^{-1}(b) = \log_{25} b, not 25−b25^{-b}.

  • A bare answer in a 1-mark part

    in Example 1.3, (d) has no method mark of its own; its follows the substitution in (c). Some markschemes award that one for a substitution seen in either part, so show it in (d) too: write g(15)=log⁡251g\left(\frac{1}{5}\right) = \log_{25} 1 before the 00.

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