MathematicsExam code: 0580
Algebraic Roots and Indices
Algebra and Sequences
Matching the bases
- Where an unknown appears in the index, rewrite both sides using the same base, then equate the indices
- This works because a single base raised to two powers is equal only when the powers match
- Recognising powers of 2, 3, 5 and 7 is what makes the rewriting possible
Worked example
Solving an index equation
Solve 2^(x + 3) = 8ˣ.
Solution:
- Write 8 as a power of 2: 8 = 2³
- The right-hand side becomes (2³)ˣ, which is 2^(3x)
- Both sides now share base 2, so the indices are equal: x + 3 = 3x
- Rearranging gives 3 = 2x, so x = 1.5
Worked example
Changing both sides to a shared base
Solve (1/4)ˣ = 8^(x + 1).
Solution:
- 1/4 and 8 are both powers of 2, so write each side with base 2
- One over a power is a negative power: 1/4 = 1/2² = 2⁻², so (1/4)ˣ = (2⁻²)ˣ = 2^(−2x)
- 8 = 2³, so 8^(x + 1) = (2³)^(x + 1) = 2^(3(x + 1))
- The 3 multiplies the whole index, giving 3x + 3, not 3x + 1
- The bases now match, so the indices are equal: −2x = 3x + 3
- Then −5x = 3, so x = −3/5
- Where the base carrying the unknown is the larger power, the answer is often a fraction
- 27ˣ = 3 becomes 3^(3x) = 3¹, so 3x = 1 and x = 1/3
- Logarithms are not required by this specification, so every such equation can be solved by matching bases, unless a question asks you to solve it from a graph (2ˣ = 5, say, has no shared base)