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Transcript of LAWS OFLAWS OF EXPONENTS - · PDF fileLaws of Indices for -ve integral indices: ... base...

  • LAWS OFLAWS OF EXPONENTSEXPONENTS

  • We know that,,8 = 2x2x2 625 = 5x5x5x5 -27 = (-3)x(-3)x(-3)

    The same factor is repeated for several times So the aboveseveral times. So the above numbers can be denoted in a h t t shortcut way

  • 8 = 2x2x2 625 5 5 5 5625 = 5x5x5x5 -27 = (-3)x(-3)x(-3) In 8, the factor 2 is repeated 3 times

    Therefore, 8 = 23 In 625, the factor 5 is repeated 4 times

    Therefore, 625 = 54I 27 the factor 3 is repeated 3 times In -27, the factor -3 is repeated 3 timesTherefore, -27 = (-3)3

  • This way of writing a number is called y gThe Exponential form or The Base Index form.

    In general, the product axaxa......(to n g p (factor) is denoted as an ,where a is called the base and n is the exponent por index of the number.

  • 1)The exponential form provides a shortcut notation to represent theshortcut notation to represent the product of repeated factors

    2)The product of the repeated factors is also known as the power of theis also known as the power of the factor

    3)an is also called as the nth power of a or a is raised to the power na or a is raised to the power n

  • Exponents :Exponents : exponent

    35power 5basebase

    Example: 125 = 53 means that 53 is the exponential form of 125the exponential form of 125

  • There are 5 laws of exponents

    First Law of Exponents:

    Consider the following examples:1)33x32 = (3x3x3) x (3x3)1)33x32 = (3x3x3) x (3x3)

    = 3x3x3x3x3= 35 = 33+2= 35 = 33 2

    2)a3xa4 =(axaxa) x (axaxaxa)=axaxaxaxaxaxa=axaxaxaxaxaxa= a7 = a 3+4

  • From the above examples, we can li th l f llgeneralize the law as follows:

    If i l b d 0If a is any real number and a0, m and n are +ve integers, then am x an = a m+nEx : 1) 314 x 312 = 314+12 = 326

    2) 23 x 28 x 26 = 23+8+6 = 2173) a10 x a12 = a 10+12 = a22

  • Second law of exponents:

    Case 1:F ll l l f ( 0) dFor all real values of a (a0) and

    m and n are +ve integers, then when m>n nm

    n

    m

    aaa =

    Example : 1) a10 = a 10-3 = a 7a3

    2) x24 = x 24-20 = x 4x20

  • Case-2:

    am = 1 when n>mn n man a n-m

    E ample 1) 45 = 1 = 1Example : 1) 45 = 1 = 1 48 48-5 43

    2) x12 = 1 = 12) x12 = 1 = 1 x20 x20-12 x8

  • Definition of a 0 :

    We know that when m and n are +ve integers. nmnm

    aaa =

    Now if m=n

    a

    Now if m=n, the above result becomes nnn

    n

    aaa =

    ie 1= a 0 ' a 0 =1 where a0ie 1= a . . a =1, where a0

  • Definition of a -n :Definition of a :We know that when a0, m and n are +ve integers

    N i th lt f 0

    nmn

    m

    aaa =

    Now assuming the result for m=0 we get n

    n aaa = 0

    0

    ie 1 = a-n ( ' .' a0 =1)

    a

    ( )an

    ' a-n = 1 when a0. . a-n = 1 when a0an

  • Third law of exponents:Consider the following example:Consider the following example:

    Find 32x32x32Find 32x32x32

    By first law of exponentsBy first law of exponents, 32x32x32 = 32+2+2 = 36we can write this also aswe can write this also as(32)3 = 36 =32x3

    '.'. (32)3= 32x3

  • From the above example, we can generalisethe relationship asthe relationship as

    (am)n = amn

    for all real values of a (a0 ), and for all +ve integers m and n.g

    Example: 1) (54)3 = 54x3 = 5122) (33)6 = 33x6 = 318

  • Fourth law of exponents:

    Consider the following example(3x5)2 = (3x5) (3x5) = (3x3)(5x5) = 32x52( ) ( ) ( ) ( )( )

    So from the above example, we can li th l ti higeneralise the relationship as

    (ab)m = am bmfor all real values of a and b a0 b0 andfor all real values of a and b a0, b0, andfor all +ve integer m.

  • Fifth law of exponents:Consider the below example:

    5

    55

    34

    3333344444

    34

    34

    34

    34

    34

    34

    ==

    =

    xxxx

    5333333333333 xxxx

    So from the above example, we can li th l ti higeneralise the relationship as

    mm a=a

    for all real values of a & b, a0, b0, for all

    mb=

    b

    , , ,+ve integer m

  • Laws of Indices for -ve integral indices:With definition of n 1With definition of

    we can extend the five laws of exponents for any integral indices

    nn

    aa =

    any integral indices.

    Definition of rational index: a be any +vereal number, n be any +ve integer and m is any integer , then we define

    ( )mnn mnm

    aa ==aif m=1, then nn aa =

    1

  • Laws of exponents for rational indices:n m

    m

    By using the definition ofall the five laws can be extended to rational i di f ll

    n mn aa =

    indices as followsLet a>0 be a real number and p and q be rational numbers then we haverational numbers, then we have

    .2.1 aaaaxa qpqp

    qpqp == +

    ( ) ( ) )0(.4.3 realisbbaabaaa

    ppppqqp

    q

    >==

    )0(.5 realisbba

    ba

    p

    pp

    >=

  • Exercise 1:

    Express81inexponential or

    Answer:

    exponentialorbaseindexform. 81 = (3x3x3x3) = 3

    4

    Express 216 in exponential or base

    Answer:216exponential or base

    index form216==(2x2x2)(3x3x3)23 X 33=23 X33

  • Exercise 2:

    Filluptheblankusingfirstlawof

    Answer:

    indices415 x 411 =

    4264 x4 =_____

    Answer:

    (2/3)5x(2/3)3 = ____(2/3)8

  • Exercise 3:

    Fill up the blank using laws of exponents

    Answer:

    1052 =103_____1010)1 2

    5

    =

    Answer:

    10

    2) 76 x 74 73 =___ 77

  • Exercise 3:

    Fill up the blank using laws of exponents

    Answer:

    z15_____zz)3 52

    715

    =zxzx

  • Exercise 4:

    Simplify by using the law of exponents

    Answer:

    k40( ) ____85 =k

    Whichisgreater Answer:

    ( ) 2323 22 or 232

  • Exercise 5:

    Fill up the blank using laws of

    Answer:

    indices1) (3x7)15 =

    315 X715) ( ) ____

    Answer:

    ____)28

    2 =

    cba

    816

    8

    cba

  • Exercise 6:

    Express the following as a rational number.

    Answer:1) 64/49

    1) (7/8)-2) /2) 4/15

    1

    453)2

    x

    Expressthefollowingi t f

    Answer:inrootform:

    1)23/2 2 8)

    2)53/44 125

  • Exercise 7:

    Find

    1) 161/4Answer:1)21) 161/4

    2) (-125)1/3

    )

    2)5

    Expressthefollowingi b i d f

    Answer:inbaseindexform:

    3 17)1 4 3/1)2

    41

    3)2 qp

    31

    7)1)

    a qp yx)33)2

    aq

    ap

    yx)3

  • Exercise 8:

    Simplify using laws of exponents

    Answer:

    45x10y7572

    23

    35)1

    yxyx

    Answer:52

    2/814352

    3.33.3.2)2

  • Indices leading to logarithm:g g

    The laws of indices made an important b k th h i th d l t fbreak through in the development of logarithms.

    In the first law of exponents, the product of the numbers is replaced by the sum of p ythe exponents and in the second law, the quotient of two numbers is replaced by the q p ydifference of Indices.

  • Indices leading to logarithm:g g

    So this concept of product replacing to d ti t l i t diffsum and quotient replacing to difference

    has led to the development of logarithms hich in t rn ill help in doing thewhich in turn will help in doing the

    calculations in Physics & Engineering.

    But after the invention of calculators theBut after the invention of calculators, the importance of logarithm tables has gone down !

  • CDCD

    Concentration

    and

    Dedication

    Take it as aTake it as a challenge!!!!

    THANK YOU