Chapter 31 logarithms

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  • Logarithms

    31Contents:

    A Logarithms in base a [3.10]

    B The logarithmic function [3.10]

    C Rules for logarithms [3.10]

    D Logarithms in base 10 [3.10]

    E Exponential and logarithmic

    equations [3.10]

    In Chapter 28 we answered problems like the one above by graphing the exponential function and using

    technology to find when the investment is worth a particular amount.

    However, we can also solve these problems without a graph using logarithms.

    We have seen previously that y = x2 and y =px are inverse functions.

    For example, 52 = 25 andp25 = 5.

    If y = ax then we say x is the logarithm of y in base a, and write this as x = loga y.

    LOGARITHMS IN BASE a [3.10]A

    Opening problem#endboxedheading

    Tony invests $8500 for n years at 7:8% p.a. compounding annually. The interest rate is fixed for the

    duration of the investment. The value of the investment after n years is given by V = 8500 (1:078)ndollars.

    Things to think about:

    a How long will it take for Tonys investment to amount to $12 000?

    b How long will it take for his investment to double in value?

    Logarithms were created to be the .inverse of exponential functions

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  • For example, since 8 = 23 we can write 3 = log2 8. The two statements 2 to the power 3 equals 8

    and the logarithm of 8 in base 2 equals 3 are equivalent, and we write:

    23 = 8 , log2 8 = 3Further examples are: 103 = 1000 , log10 1000 = 3

    32 = 9 , log3 9 = 24

    1

    2 = 2 , log4 2 = 12

    The symbol ,is equivalent to

    In general, y = ax and x = loga y are equivalent statements

    and we write y = ax , x = loga y.

    Example 1 Self Tutor

    Write an equivalent:

    a logarithmic statement for 25 = 32 b exponential statement for log4 64 = 3:

    a 25 = 32 is equivalent to log2 32 = 5.

    So, 25 = 32 , log2 32 = 5.b log4 64 = 3 is equivalent to 4

    3 = 64.

    So, log4 64 = 3 , 43 = 64.

    Example 2 Self Tutor

    Find the value of log3 81:

    ) 3x = 81

    ) 3x = 34

    ) x = 4

    ) log3 81 = 4

    EXERCISE 31A

    1 Write an equivalent logarithmic statement for:

    a 22 = 4 b 42 = 16 c 32 = 9 d 53 = 125

    e 104 = 10000 f 71 = 17 g 33 = 127 h 27

    1

    3 = 3

    i 52 = 125 j 2 1

    2 = 1p2

    k 4p2 = 22:5 l 0:001 = 103

    2 Write an equivalent exponential statement for:

    a log2 8 = 3 b log2 1 = 0 c log212

    = 1 d log2

    p2 = 12

    e log2

    1p2

    = 12 f logp2 2 = 2 g logp3 9 = 4 h log9 3 = 12

    3 Without using a calculator, find the value of:

    a log10 100 b log2 8 c log3 3 d log4 1

    e log5 125 f log5(0:2) g log10 0:001 h log2 128

    i log212

    j log3

    19

    k log2(

    p2) l log2

    p8

    The logarithm of in baseis the exponent or powerof which gives .

    813

    3 81

    Let log3 81 = x

    626 Logarithms (Chapter 31)

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    Y:\HAESE\IGCSE01\IG01_31\626IGCSE01_31.CDR Monday, 27 October 2008 3:01:48 PM PETER

  • The logarithmic function is f(x) = loga x where a > 0, a 6= 1.

    Consider f(x) = log2 x which has graph y = log2 x.

    Since y = log2 x , x = 2y, we can obtain the table of values:

    y 3 2 1 0 1 2 3x 18

    14

    12 1 2 4 8

    Notice that:

    the graph of y = log2 x is asymptotic to the y-axis the domain of y = log2 x is fx j x > 0g the range of y = log2 x is fy j y 2 R g

    THE INVERSE FUNCTION OF f(x) = loga x

    Given the function y = loga x, the inverse is x = loga y finterchanging x and yg) y = ax

    So, f(x) = loga x , f1(x) = ax

    THE LOGARITHMIC FUNCTION [3.10]B

    x

    8642

    y

    O

    y xlogx

    627Logarithms (Chapter 31)

    m log7

    3p7

    n log2(4p2) o logp2 2 p log2

    1

    4p2

    q log10(0:01) r log

    p2 4 s log

    p3

    13

    t log3

    1

    9p3

    4 Rewrite as logarithmic equations:

    a y = 4x b y = 9x c y = ax d y = (p3)x

    e y = 2x+1 f y = 32n g y = 2x h y = 2 3a

    5 Rewrite as exponential equations:

    a y = log2 x b y = log3 x c y = loga x d y = logb n

    e y = logm b f T = log5a2

    g M = 12 log3 p h G = 5 logbm

    i P = logpbn

    6 Rewrite the following, making x the subject:

    a y = log7 x b y = 3x c y = (0:5)x d z = 5x

    e t = log2 x f y = 23x g y = 5

    x2 h w = log3(2x)

    i z = 12 3x j y = 15 4x k D = 110 2x l G = 3x+1

    7 Explain why, for all a > 0, a 6= 1: a loga 1 = 0 b loga a = 1

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  • Example 3 Self Tutor

    Find the inverse function f1(x) for: a f(x) = 5x b f(x) = 2 log3 x

    a y = 5x has inverse function x = 5y

    ) y = log5 x

    So, f1(x) = log5 x

    b y = 2 log3 x has inverse function

    x = 2 log3 y

    )x

    2= log3 y

    ) y = 3x2

    So, f1(x) = 3x2

    EXERCISE 31B

    1 Find the inverse function f1(x) for:

    a f(x) = 4x b f(x) = 10x c f(x) = 3x d f(x) = 2 3xe f(x) = log7 x f f(x) =

    12(5

    x) g f(x) = 3 log2 x h f(x) = 5 log3 x

    i f(x) = logp2 x

    2 a On the same set of axes graph y = 3x and y = log3 x.

    b State the domain and range of y = 3x.

    c State the domain and range of y = log3 x.

    3 Prove using algebra that if f(x) = ax then f1(x) = loga x.

    x

    y

    OO

    yx

    y xlogx

    y x

    If f(x) = g(x),

    graph y = f(x)

    and y = g(x)

    on the same set

    of axes.

    4 Use the logarithmic function log on your graphics calculator

    to solve the following equations correct to 3 significantfigures. You may need to use the instructions on page 15.

    628 Logarithms (Chapter 31)

    For example, if f(x) = log2 x then f1(x) = 2x.

    The inverse function y = log2 x is the reflection of y = 2x

    in the line y = x.

    a log10 x = 3 x b log10(x 2) = 2xc log10

    x4

    = x2 2 d log10 x = x 1

    e log10 x = 5x f log10 x = 3

    x 3

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    Y:\HAESE\IGCSE01\IG01_31\628IGCSE01_31.CDR Tuesday, 18 November 2008 11:12:59 AM PETER

  • Consider two positive numbers x and y. We can write them both with base a: x = ap and y = aq, forsome p and q.

    ) p = loga x and q = loga y ...... (*)

    Using exponent laws, we notice that: xy = apaq = ap+q

    x

    y=

    ap

    aq= apq

    xn = (ap)n = anp

    ) loga(xy) = p+ q = loga x+ loga y ffrom *gloga

    x

    y

    = p q = loga x loga yloga(x

    n) = np = n loga x

    loga(xy) = loga x + loga y

    loga

    x

    y

    = loga x loga y

    loga(xn) = n loga x

    Example 5 Self Tutor

    If log3 5 = p and log3 8 = q, write in terms of p and q:

    a log3 40 b log3 25 c log3

    64125

    a log3 40

    = log3(5 8)= log3 5 + log3 8

    = p+ q

    b log3 25

    = log3 52

    = 2 log3 5

    = 2p

    c log3

    64125

    = log3

    82

    53

    = log3 8

    2 log3 53= 2 log3 8 3 log3 5= 2q 3p

    RULES FOR LOGARITHMS [3.10]C

    629Logarithms (Chapter 31)

    Example 4 Self Tutor

    Simplify: a log2 7 12 log2 3 + log2 5 b 3 log2 5

    a log2 7 12 log2 3 + log2 5= log2 7 + log2 5 log2 3

    1

    2

    = log2(7 5) log2p3

    = log2

    35p3

    b 3 log2 5

    = log2 23 log2 5

    = log285

    = log2(1:6)

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    Y:\HAESE\IGCSE01\IG01_31\629IGCSE01_31.CDR Tuesday, 18 November 2008 11:13:47 AM PETER

  • EXERCISE 31C

    1 Write as a single logarithm:

    a log3 2 + log3 8 b log2 9 log2 3 c 3 log5 2 + 2 log5 3d log3 8 + log3 7 log3 4 e 1 + log3 4 f 2 + log3 5g 1 + log7 3 h 1 + 2 log4 3 3 log4 5 i 2 log3 m+ 7 log3 nj 5 log2 k 3 log2 n

    2 If log2 7 = p and log2 3 = q, write in terms of p and q:

    a log2 21 b log237

    c log2 49 d log2 27

    e log279

    f log2(63) g log2

    569

    h log2(5:25)

    3 Write y in terms of u and v if:

    a log2 y = 3 log2 u b log3 y = 3 log3 u log3 vc log5 y = 2 log5 u+ 3 log5 v d log2 y = u+ v

    e log2 y = u log2 v f log5 y = log5 ug log7 y = 1 + 2 log7 v h log2 y =

    12 log2 v 2 log2 u

    i log6 y = 2 13 log6 u j log3 y = 12 log3 u+ log3 v + 14 Without using a calculator, simplify:

    alog2 16

    log2 4b

    logp 16

    logp 4c

    log5 25

    log515

    d logm 25logm

    15

    Logarithms in base 10 are called common logarithms.

    y = log10 x is often written as just y = log x, and we assume the logarithm has base 10.

    Your calculator has a log key which is for base 10 logarithms.

    Discovery Logarithms#endboxedheading