Most tested MM5.2

Laws of Logarithms

Logarithms are the inverse of exponential functions, used to find the power to which a base must be raised to produce a given number. Mastery of the logarithm laws is crucial for solving equations involving unknown powers and for simplifying expressions without a calculator.

Part of the ESAT Mathematics 2 syllabus — revision for the Engineering and Science Admissions Test (ESAT), the UAT-UK admissions test for Cambridge, Imperial, Oxford and UCL.

Key points

  • The statement 'loga(c) = b' is equivalent to 'a^b = c'. This is the fundamental link between logarithms and exponentials.
  • Logarithms can only be taken of positive numbers, and the base must also be positive and not equal to 1.
  • Adding logs with the same base corresponds to multiplying their arguments: loga(x) + loga(y) = loga(xy).
  • Subtracting logs with the same base corresponds to dividing their arguments: loga(x) - loga(y) = loga(x/y).
  • A coefficient in front of a log can be moved to become a power on the argument: k × loga(x) = loga(x^k).
  • Special cases to remember: loga(a) = 1 and loga(1) = 0.
    since a1 = a
    since a0 = 1

Formulae

a^b = c ≤> b = loga(c)

To convert between exponential and logarithmic forms.

loga(x) + loga(y) = loga(xy)

To combine two logarithms of the same base that are being added.

loga(x) - loga(y) = loga(x/y)

To combine two logarithms of the same base that are being subtracted.

k × loga(x) = loga(x^k)

To handle a coefficient multiplying a logarithm.

loga(1/x) = -loga(x)

A special case of the subtraction or power rule, useful for arguments that are fractions.

loga(a) = 1

To simplify a logarithm where the base and argument are identical.

Definitions

Logarithm
The power to which a specified number (the base) must be raised to obtain a given value (the argument).
Base
In the expression loga(x), 'a' is the base. It is the number that is raised to a power.
Argument
In the expression loga(x), 'x' is the argument. It is the number you are finding the logarithm of.

Worked example

Given that log2(y) - 2 × log2(x) = 3, express y in terms of x.

  1. 1

    Use the power law on the second term:

    2 × log2(x) becomes log2(x2).

    The equation is now log2(y) - log2(x2) = 3.

  2. 2

    Use the subtraction law to combine the logarithms:

    log2(y / x2) = 3.

  3. 3

    Convert the logarithmic equation into exponential form, using the definition loga(c) = b ≤> a^b = c.

  4. 4

    This gives:

    y / x2 = 23
  5. 5

    Calculate the power of 2:

    23 = 8
  6. 6

    Rearrange the equation to make y the subject:

    y = 8 × x2

Answer: y = 8x2

Common mistakes

  • ×When solving equations, failing to check that the arguments of the original logarithms are positive. This can lead to including extraneous solutions.
  • ×Making sign errors, particularly when dealing with the subtraction law or negative coefficients. For example, incorrectly simplifying -(log(A) - log(B)) to -log(A) - log(B).
  • ×Arithmetic errors when evaluating the powers after converting from log form, for example calculating 34 as 12 instead of 81.

No-calculator tips

  • ✓To evaluate an expression like log3(81), ask yourself '3 to what power is 81?'. Mentally calculate powers: 31=3, 32=9, 33=27, 34=81. The answer is 4.
  • ✓For logs where the argument is not a simple integer power of the base, like log4(32), try to express both numbers as powers of a common base. Here, 4=22 and 32=25. So, if log4(32)=p, then 4^p=32 ⇒ (22)^p=25 ⇒ 2^(2p)=25 ⇒ 2p=5 ⇒ p=5/2.

Read this topic in the official UAT-UK ESAT guide →

All Mathematics 2 topics