Indices and Surds
Senior 2 · Mathematics
Indices and Surds
Lesson 2 of 5
22 min
Roots and powers are two views of the same idea. Once you can write √9 as 912, every law from the last lesson works for roots too, and you no longer need separate rules for them.
Equations with the unknown in the index describe growth. A savings group (ikimina) in Huye that doubles its membership every year, or bacteria that double every hour, both lead to questions such as "after how many doublings do we reach 160?". You will solve these by making the bases equal.
What should 912 mean? Use the power-of-a-power law: (912)2 = 912 × 2 = 91 = 9. So 912 is the positive number whose square is 9, which is 3. In the same way, 813 is the number whose cube is 8, which is 2.
In general, for a > 0 and a positive whole number n:
a1/n = n√a (the n-th root of a).
A fraction m/n in the index means "take the n-th root, then raise to the power m":
am/n = (n√a)m = n√(am).
Worked example 1
Evaluate 2723.
The denominator 3 means cube root: ∛27 = 3. The numerator 2 means square: 32 = 9. So 2723 = 9.
Worked example 2
Evaluate 1634 and 32−25.
1634: the fourth root of 16 is 2, and 23 = 8.
32−25: the minus sign means "one over", so 32−25 = 1 ÷ 3225. The fifth root of 32 is 2 and 22 = 4, so the answer is 14.
If ax = ay with a > 0 and a ≠ 1, then x = y. So the method is: write both sides as powers of the same base, then equate the indices.
Worked example 3
Solve 3x + 1 = 81.
81 = 34, so 3x + 1 = 34. Equate the indices: x + 1 = 4, so x = 3. Check: 34 = 81 ✓.
Worked example 4
Solve 4x = 32.
Both numbers are powers of 2: 4x = (22)x = 22x and 32 = 25. So 2x = 5 and x = 52.
Worked example 5
(a Rwandan context). An ikimina starts with 5 members and the number of members doubles every year. After how many years will it have 160 members?
After n years there are 5 × 2n members. Solve 5 × 2n = 160. Divide by 5: 2n = 32 = 25. So n = 5 years.
Worked example 6
(unknown in the base). Solve x32 = 27 for x > 0.
Raise both sides to the reciprocal power 23: x = 2723 = 9. Check: 932 = (√9)3 = 33 = 27 ✓.
Tip
Memory hook. In am/n, the denominator is the root ("the root is at the bottom, like the roots of a tree") and the numerator is the power. Take the root first: 6423 = 42 = 16 is easier than working out 642 = 4096 first.
Evaluate 12513.
Solution: 5 × 5 × 5 = 125, so 12513 = 5.
Evaluate 6423.
Solution: ∛64 = 4, and 42 = 16.
Evaluate 81−34.
Solution: The fourth root of 81 is 3, and 33 = 27. The negative index gives 127.
Solve 2x − 1 = 16.
Solution: 16 = 24, so x − 1 = 4 and x = 5.
Solve 25x = 125.
Solution: 25x = 52x and 125 = 53, so 2x = 3 and x = 32.
Remember
Sources and credits
The Rwandan worked examples, the practice items and the exercises were written by Ganzaa.
