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Indices, Logarithms and Surds

Mathematics · WAEC and JAMB · SS2 and SS3

This topic feeds three or four objective questions and often a full Paper 2 question. Examiners like it because one index law applied wrongly destroys the whole answer, so the marks separate candidates who know the rules from those who guess.

What you need to know

  • The three core index laws are a^m x a^n = a^(m+n), a^m / a^n = a^(m-n) and (a^m)^n = a^(mn). Everything else in the topic is built from these three.
  • Any non-zero number raised to power zero is 1, so 5^0 = 1 and (3x)^0 = 1. This is not zero, and candidates throw away marks on it every year.
  • A negative index means a reciprocal: a^-n = 1/a^n, so 2^-3 = 1/8 and (2/3)^-2 = (3/2)^2 = 9/4. Flipping the fraction is the quickest way to handle a negative power on a fraction.
  • A fractional index is a root: a^(1/n) = the nth root of a, and a^(m/n) = the nth root of a, all raised to power m. So 27^(2/3) = (cube root of 27)^2 = 3^2 = 9.
  • To solve an index equation, write both sides as powers of the SAME base, then equate the powers. 3^(2x+1) = 27^(x-1) becomes 3^(2x+1) = 3^(3x-3), so 2x + 1 = 3x - 3.
  • A logarithm is simply an index written the other way round: if a^x = b then log to base a of b = x. Reading it aloud as 'the power you raise a to, to get b' removes most of the confusion.
  • The log laws mirror the index laws: log(MN) = log M + log N, log(M/N) = log M - log N, and log(M^n) = n log M. Multiplication becomes addition, which is the whole point of logs.
  • Log of 1 is 0 in any base, and log of the base itself is 1, so log to base 10 of 10 = 1 and log to base 5 of 5 = 1.
  • A surd is a root that cannot be simplified to an exact value, such as sqrt(2) or sqrt(7). To simplify, pull out the largest perfect square factor: sqrt(72) = sqrt(36 x 2) = 6 sqrt(2).
  • You may only add or subtract surds that are alike: 3 sqrt(5) + 4 sqrt(5) = 7 sqrt(5), but sqrt(2) + sqrt(3) cannot be combined at all and must be left as it is.
  • To rationalise a single-term denominator, multiply top and bottom by that surd: 6/sqrt(3) = 6 sqrt(3)/3 = 2 sqrt(3).
  • To rationalise a two-term denominator, multiply by the conjugate, which is the same expression with the middle sign reversed. The conjugate of sqrt(3) - sqrt(2) is sqrt(3) + sqrt(2), and the product gives 3 - 2 = 1, a whole number.
  • Useful approximations for comparison questions: sqrt(2) is about 1.414, sqrt(3) is about 1.732 and sqrt(5) is about 2.236. They let you check whether a simplified surd answer is sensible.
  • In log tables and in theory questions, remember that log to base 10 of a number between 1 and 10 lies between 0 and 1, which is why a log of 1.6811 corresponds to a number in the tens.

Key terms

Index (plural indices)
The power to which a number, called the base, is raised.
Logarithm
The power to which a given base must be raised to produce a particular number.
Surd
An irrational root of a rational number that cannot be written exactly as a fraction or terminating decimal.
Rationalising the denominator
Removing a surd from the bottom of a fraction by multiplying the numerator and denominator by a suitable surd or conjugate.
Conjugate surd
The expression formed by reversing the sign between two surd terms, used because the product of conjugates is rational.
Characteristic
The whole-number part of a base-ten logarithm, which fixes the size of the number.

Formulae

  • a^m x a^n = a^(m+n)
  • a^m / a^n = a^(m-n)
  • (a^m)^n = a^(m*n)
  • a^0 = 1 for any a not equal to 0
  • a^-n = 1 / a^n
  • a^(m/n) = (nth root of a)^m
  • if a^x = b then log base a of b = x
  • log(MN) = log M + log N
  • log(M/N) = log M - log N
  • log(M^n) = n log M
  • sqrt(a) x sqrt(b) = sqrt(ab)
  • (sqrt(a) + sqrt(b))(sqrt(a) - sqrt(b)) = a - b

Worked examples

Solve for x: 3^(2x+1) = 27^(x-1).

  1. Write both sides to the same base. Since 27 = 3^3, the right side becomes (3^3)^(x-1).
  2. Apply the power-of-a-power law: (3^3)^(x-1) = 3^(3x-3).
  3. The equation is now 3^(2x+1) = 3^(3x-3), and the bases are equal.
  4. Equate the indices: 2x + 1 = 3x - 3.
  5. Collect like terms: 1 + 3 = 3x - 2x, so x = 4.
  6. Check: left side 3^(2(4)+1) = 3^9 = 19683; right side 27^(4-1) = 27^3 = 19683.

Simplify (sqrt(3) + sqrt(2)) / (sqrt(3) - sqrt(2)), leaving your answer in surd form.

  1. The denominator has two terms, so multiply top and bottom by its conjugate, sqrt(3) + sqrt(2).
  2. Denominator: (sqrt(3) - sqrt(2))(sqrt(3) + sqrt(2)) = 3 - 2 = 1.
  3. Numerator: (sqrt(3) + sqrt(2))^2 = 3 + 2 x sqrt(3) x sqrt(2) + 2.
  4. Since sqrt(3) x sqrt(2) = sqrt(6), the numerator is 3 + 2 sqrt(6) + 2 = 5 + 2 sqrt(6).
  5. Divide by the denominator 1.

Given that log 2 = 0.3010 and log 3 = 0.4771, evaluate log 48 without tables.

  1. Break 48 into factors whose logs you know: 48 = 16 x 3 = 2^4 x 3.
  2. Apply the product law: log 48 = log(2^4) + log 3.
  3. Apply the power law: log(2^4) = 4 log 2 = 4 x 0.3010 = 1.2040.
  4. Add log 3: 1.2040 + 0.4771 = 1.6811.

The mistake to avoid

Candidates add the bases instead of the indices, writing 2^3 x 2^4 as 4^7 instead of 2^7. The other killer is treating sqrt(a + b) as sqrt(a) + sqrt(b): sqrt(9 + 16) is sqrt(25) = 5, not 3 + 4 = 7.

In the exam

When an index equation looks awkward, look for a common base among 2, 3, 5 and 10 before anything else, because almost every WASSCE index equation is built from those. In surd questions, leave the answer in exact surd form unless the paper asks for a decimal, and state the number of decimal places it asks for. Show the conjugate step clearly, as it carries a method mark on its own.