Indices [Exponents]
ICSE Class 9 · Mathematics · Chapter 7
What this chapter covers
- Powers, bases and indices
- The product law
- The quotient law
- The power law and powers of products and quotients
- The zero index
- Negative indices
- Fractional indices
- The laws of indices together
- Simplifying expressions involving various indices
- Finding an unknown index
Key points
- In an, a is the base and n is the index (or exponent); for a positive integer n, an is the product of n factors of a.
- Product law: am × an = am+n, for the same base only.
- Quotient law: am ÷ an = am-n, with a ≠ 0.
- Power law: (am)n = amn; also (ab)n = anbn and ((a)/(b))n = (an)/(bn).
- Zero index: a0 = 1 for every non-zero a.
- Negative index: a-n = (1)/(an); a negative index means reciprocal, never a negative value; ((a)/(b))-n = ((b)/(a))n.
- Fractional index: a(1)/(n) = sqrt[n]a and a(m)/(n) = (sqrt[n]a)m, for a > 0; the denominator gives the root, the numerator the power.
- The laws act on products and quotients of powers, never on sums: (a+b)n ≠ an + bn.
- To simplify, write every number as a power of a prime, apply the laws to each base, and give the answer with positive indices.
- If ax = ay with a > 0 and a ≠ 1, then x = y; this finds an unknown index.
Questions answered
Define the index of a power, and name the base and the index in 74.
The index of a power is the number that shows how many times the base is used as a factor. In 74 the base is 7 and the index is 4.
Evaluate (-2)5 and -24.
(-2)5 = -32, because the index is odd. -24 = -(24) = -16, because the base is 2 and the minus sign stays outside.
Simplify 2a3b2 × 5a-1b4.
10a3 + (-1)b2+4 = 10a2b6.
Evaluate 125-(2)/(3).
125-(2)/(3) = dfrac1(sqrt[3]125)2 = (1)/(52) = (1)/(25).
Evaluate ((16)/(81))-(1)/(4).
Invert the fraction: ((81)/(16))(1)/(4) = (3)/(2), since 34 = 81 and 24 = 16.
Explain why a0 is taken to be 1 for a ≠ 0.
By the quotient law, an ÷ an = an-n = a0. But any non-zero number divided by itself is 1. For the law to stay true, a0 must equal 1.
Evaluate 9(3)/(2) - 3 × 50 - ((1)/(81))-(1)/(2).
9(3)/(2) = 33 = 27, 3 × 50 = 3 and ((1)/(81))-(1)/(2) = 81(1)/(2) = 9. So the value is 27 - 3 - 9 = 15.
Simplify (x-1 + y-1)-1.
x-1 + y-1 = (1)/(x) + (1)/(y) = (x + y)/(xy). Its reciprocal is (xy)/(x+y). It is not x + y, because the negative index cannot be moved across a sum.
Past board exam questions
ICSE 2019 sample paper · 3 marks
Simplify: ((8)/(27))-(1)/(3) × ((25)/(4))(1)/(2) × ((4)/(9))0 + ((125)/(64))(1)/(3)
Show answer
((8)/(27))-(1)/(3) = ((27)/(8))(1)/(3) = (3)/(2), ((25)/(4))(1)/(2) = (5)/(2), ((4)/(9))0 = 1, ((125)/(64))(1)/(3) = (5)/(4)
Expression = (3)/(2) × (5)/(2) × 1 + (5)/(4) = (15)/(4) + (5)/(4) = (20)/(4) = 5
ICSE 2019 sample paper · 3 marks
Evaluate for x: (√((5)/(3)))x-8 = ((27)/(125))2x-3
Show answer
((5)/(3))(x-8)/(2) = ((3)/(5))3(2x-3) = ((5)/(3))-3(2x-3)
(x-8)/(2) = -6x + 9 ⇒ x - 8 = -12x + 18 ⇒ 13x = 26
x = 2