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Chapter 10: Exponents and Powers

    Introduction

    Powers with Negative Exponents

    Laws of Exponents

    Use of Exponents to Express Small Numbers in Standard Form

    Comparing very large and very small numbers

    Exercise 10.1

    Exercise 10.2

    What Have We Discussed?

    Enhanced Curriculum Support

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Chapter 10: Exponents and Powers > Laws of Exponents

Laws of Exponents

We have learnt that for any non-zero integer a , am×an=am+n, where m and n are natural numbers.

Does this law also hold if the exponents are negative?

(i) Let us explore. 2−3 and 2−2

2−3and2−2

  • We know that 2−3 = and 2−2 =
  • Numerators and denominators are same, so add the powers
  • We get the result as
  • If we bring denominator to numerator then the power will be

(ii) Take −3−4×−3−3

−3−4×−3−3

  • We know that −3−4 = and −3−3 =
  • Numerators and denominators are same so, add the powers
  • We get the result as
  • If we bring denominator to numerator then the power will be

(iii) Now consider 5−2×54

5−2×54

  • we know that 5−2 =
  • Divide the exponents
  • Numerators and denominators are same so, subtract the powers
  • We get the result
  • If we bring denominator to numerator then the power will be

In general, we can say that for any non-zero integer am×an=am+n, where m and n are integers.

Try these.

Match the below exponents

Instructions

(–2)^(–3) × (–2)^(–4)
p^3 × p^(–10)
3^2 × 3^(-5) × 3^(6)
3^3
-2^(-7)
p^(-7)

On the same lines you can verify the following laws of exponents, where a and b are non zero integers and m, n are any integers.

(i) aman=am−n(ii) amn=amn(iii) am×bm=abm
(iv) ambm=abm(v) a0=1

Examples on Laws OF Exponents

Let us solve some examples using the above Laws of Exponents.

Example 1: Find the value of:

Instructions

(i) 2−3 = 123 =
(ii) 13−2 = = × =

Example 2: Simplify:

Instructions

(i) −45×−4−10
We know that am×an =
−45×−4−10 = −45−10 = =
(ii) 25÷2−6
We know that: am÷an =
25÷2−6 = 25−−6 =

Note :an= 1 only if n = 0.

This will work for any a.

For a = 1, 11= 12= 13 = 1−2... = 1 or 1n= 1 for infinitely many n.

For a = –1, −10= −12= −14= −1−2 = ... 1 or −1p = 1 for any even integer p.

Example 3: Express 4−3 as a power with the base 2.

Solution:

We have, 4 = 2 × 2 =

Therefore, 4−3 = 2×2−3 = = =

Example 4: Simplify and write the answer in the exponential form.

(i) 25÷285×2−5

simply the exponential form

  • Here exponential form is 25÷285×2−5
  • 2 is the Base of all terms so, subtract the powers in the bracket. We get the result is
  • Bases are same so, multiply the powers
  • we get the result is
  • converting the exponent:

(ii) −4−3 × 5−3 × −5−3

= −4×5×−5−3 = 100−3 = 11003

(iii) 18×3−3 = 123× 3−3 = 2−3× 3−3 = 2×3−3 = 6−3 = 163

(iv) −34 × 534 = −1×34 × 5434 = −14×34 × 5434 = −14 × 54 = (as −14 = )

Example 5: Find m so that −3m+1×−35=−37

Find value of m

  • Here is the exponent is −3m+1×−35=−37
  • Bases are same then add the exponents,we get power is
  • On both the sides, we have the same base, so their exponents must be equal.
  • subtracting the values
  • Hence m is

Example 6: Find the value of 23−2

23−2=2−23−2=3222=

In general, ab−m=bam

Example 7: Simplify

(i) 13−2−12−3÷14−2

= 1−23−2−1−32−3÷1−24−2

= 1−23−2−1−32−3÷1−24−2

= 312−213÷412

= ( - ) ÷ =

(ii) 58−7×85−5

= 5−78−7×8−55−5

= 5−75−5×8−58−7

= 5−7−−5×8−5−−7

= 5−2 × 82

= 8252 =