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Cambridge IGCSE Mathematics — 0580 Extended

Topic 2.4: Algebra and graphs — Indices II

The Number chapter covered numerical powers. This lesson uses the same laws with letters, then solves exponential equations by rewriting both sides with the same base. Logarithms are not required for 0580.

Laws with algebraic terms

Treat the number and each letter separately. Multiply coefficients; add or subtract the indices of matching letters.

LawIn letters
Multiply\(a^m \times a^n = a^{m+n}\)
Divide\(a^m \div a^n = a^{m-n}\)
Power of a power\((a^m)^n = a^{mn}\)
Negative index\(a^{-n} = \dfrac{1}{a^n}\)

Method

  1. Rewrite a negative power as a reciprocal before (or after) applying the other laws — both orders work if you are consistent.
  2. A product raised to a power: \((5x^3)^{-2} = 5^{-2}(x^3)^{-2}\).
  3. When dividing, subtract the index of the denominator, including if that index is already negative: subtracting a negative adds.

Simplify \((5x^3)^{-2}\).

Worked simplification of 5 x cubed to the power negative 2 equals 1 over 25 x to the 6
Negative power as a reciprocal, then square: \(\dfrac{1}{25x^6}\).

Simplify \(12a^5 \div 3a^{-2}\).

Worked division 12 a to the 5 over 3 a to the negative 2 equals 4 a to the 7
Coefficients \(12 \div 3 = 4\); indices \(5-(-2)=7\).

Simplify \(6x^7 y^4 \times 5x^{-5} y\).

Worked product 6 x to the 7 y to the 4 times 5 x to the negative 5 y equals 30 x squared y to the 5
\(30x^{7-5}y^{4+1} = 30x^2 y^5\).

Fractional and negative indices

\(a^{1/2} = \sqrt{a}\) and \(a^{1/n} = \sqrt[n]{a}\). A fractional index \(m/n\) means the \(n\)th root, then raise to the power \(m\): \(a^{m/n} = \left(\sqrt[n]{a}\right)^m\).

Method

  1. Rewrite integers such as \(8\), \(27\) and \(32\) as prime powers before you apply a fractional index.
  2. For a quotient, divide coefficients and subtract the indices of each letter.
  3. For \((8a^6)^{2/3}\), apply \(2/3\) to \(8\) and to \(a^6\) separately.

Simplify \(\dfrac{27x^{3/2}}{3x^{-1/2}}\).

Worked quotient 27 x to the 3 over 2 divided by 3 x to the negative 1 over 2 equals 9 x squared
\(27/3=9\) and \(\frac{3}{2}-(-\frac{1}{2})=2\): \(9x^2\).

Simplify \((8a^6)^{2/3}\).

Worked power 8 a to the 6 to the power 2 over 3 equals 4 a to the 4
\(8^{2/3}=(2^3)^{2/3}=4\) and \(6 \times \frac{2}{3}=4\): \(4a^4\).

Solving by the same base

If \(a^m = a^n\) with \(a > 0\) and \(a \neq 1\), then \(m = n\). The skill is rewriting every number as a power of the same prime.

Method

  1. Write \(32\), \(8\), \(25\), \(27\), \(9\), \(4\) as powers of \(2\), \(3\) or \(5\).
  2. Use \((a^m)^n = a^{mn}\) so that both sides have the identical base.
  3. Equate the indices and solve the ordinary linear equation.

Solve \(32^x = 2\).

Worked solution 32 to the x equals 2 giving x equals 1 over 5
\(32=2^5\), so \(5x=1\) and \(x=\dfrac{1}{5}\).

Solve \(5^{x+1} = 25^x\).

Worked solution 5 to the x plus 1 equals 25 to the x giving x equals 1
\(25=5^2\), so \(x+1=2x\) and \(x=1\).

Paper 2 (non-calculator)

Rewrite \(32=2^5\), \(8=2^3\), \(25=5^2\), \(27=3^3\), \(9=3^2\) and \(4=2^2\) before you touch the unknown. Trying to “bring the power down” without a common base is a logarithm method — not on this paper.

Try this

Solve \(8^x = 2^{x+4}\).

Show answer
Answer

  1. Write \(8\) as a power of \(2\)

    \[ (2^3)^x = 2^{x+4} \]

  2. Power of a power, then equate indices

    \[ 2^{3x} = 2^{x+4} \quad \Rightarrow \quad 3x = x + 4 \]

  3. Final answer

    \[ x = 2 \]

Exam Traps

  • \(a^{1/2}\) is \(\sqrt{a}\), not \(\dfrac{1}{2a}\). The index \(1/2\) is a root; a negative index is a reciprocal.
  • Dividing \(12a^5 \div 3a^{-2}\) subtracts indices: \(5-(-2)=7\). Treating \(a^{-2}\) as “already a denominator” and subtracting \(2\) instead of adding it gives \(a^3\), which is wrong.

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