Chapter 15 Logs Exponentials

7/31/2019 Chapter 15 Logs Exponentials

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Revision : Indices

Examples

a) 34 = 3 3 3 3

= 81

b) Solve 2n = 1024

Try 28 = 256 (too low)

29

= 512 (too low)

210 = 1024 n = 10

ba - is called the BASEa

b - is the INDEX or EXPONENT


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c) Find the smallest integer value ofn such that 3n > 10 000.

Try 37

= 2187 ( < 10 000)

38 = 6561 ( < 10 000)

39 = 19683 ( > 10 000)

n = 9


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Graphs of ax ExponentialFunctions

For y = ax ifa > 1

The graph:Is always positive

Never crosses thex-axis

Is increasing

Passes through (0,1)

Is called a growth function4 -3 -2 -1 1 2 3 4

5

x

(0,1)

y = ax


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Examples

a) If you deposit 100 at the Northern Rock Bank, each year the

deposit will grow by 15% (compound interest).

i. Find a formula for the amount in the bank after n years.

ii. After how many years will the balance exceed 500?

i. After year 0: y0 = 100After year 1: y1 = y0 1.15 = 115

After year 2: y2 = y1 1.15 = (1.15)2 100 = 132.25

After year 3: y3 = y2 1.15 = (1.15)3 100 = 152.0875

Formula : yn = (1.15)n 100


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ii.

Try n = 9 : y9 = (1.15)9 100 = 351.7876292Try n = 10 : y10 = (1.15)

10 100 = 404.5557736

Try n = 11: y11 = (1.15)11 100 = 465.2391396

The balance will exceed 500 after 12 years.

Try n = 12: y12= (1.15)12

100 = 535.0250105

After how many years will the balance exceed 500?


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b) The rabbit population on an island increases by 12% per year.

i. Find a formula for the number of rabbits on the island after n years.

ii. How many years will it take for the population to at least double?

i. After year 0: r0After year 1: r1 = r0 1.12

After year 2: r2 = r1 1.12 = (1.12)2 r0

Formula : rn = (1.12)n r0


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For the population to at least double :

(1.12)n > 2

ii.

Try: (1.12)3 = 1.404928

Try : (1.12)4 = 1.57351936

Try : (1.12)

5

= 1.762341683

Population will double within 7 years.

Try : (1.12)6 = 1.973822685

Try : (1.12)7 = 2.2106841407

(too low)

(too low)

(too low)

(too low)

( > 2 )


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3. Exponential FunctionsIf we plot f (x) and

f `(x) for f (x) = 2x:

f (x)f `(x)

(0,1)

We can see that f `(x)

is below f (x).

If we plot f (x) and

f `(x) for f (x) = 3x:

f (x)f `(x)

(0,1)

We can see that f `(x)

is above f (x).

For some value between 2 and 3, f (x) = f `(x).

4 -3 -2 -1 1 2 3 4

5

x

4 -3 -2 -1 1 2 3 4

5

x


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f `(x) for f (x) = 2.7x

:f (x)

f `(x)

(0,1)

We can see that f `(x)is just below f (x).


f `(x) for f (x) = 2.8x

:

f (x)f `(x)

(0,1)

We can see that f `(x)is just above f (x).

4 -3 -2 -1 1 2 3 4

5

x

4 -3 -2 -1 1 2 3 4

5

x


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Logarithms

For the function ( ) which is an exponential function,x

f x e1

it's inverse ( ) exists. The inverse is a reflection in the linef x y x-

y

x


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If we consider the function ( ) 2xf x 1

2Then its inverse is the logarithmic function ( ) logf x x-

This is pronounced as log to the base 2 ofx.

If then logx

ay a x y

If log theny

ay x x a


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Examples

a) Write the following in logarithmic form:

i) 43 = 64 ii) 25 = 32 iii) p = 82

log8 p = 2

8= 23

(Check: 2 2 2 = 8)

b) Express the following in exponential form:

i) log28 = 3 i) log21/128 = -7

1/128 = 2-7

log4 64 3= log2 32 5=

7 1 1(check: 2

2 2 2 2 2 2 2 128

-


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Laws of Logarithms

Law A: log a an = n

Evaluate:

a) log 2 8 b) log 3 1 c) log 21/16

2 8n

3n 3 1

n 0n

12

16

n

4n -


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Law B:

Law C:

Law D:

log ( ) log loga a a

xy x y

log log loga a a

xx y

y

-

log log

n

a ax n x

2 2 3 3 3

Evaluate:

1(a)log 8 log 4 ( ) log 27 log 3 ( ) log 814

b c -

2log 8 4

2log 32

5

3

27log

3

3log 92

1

43

log 81

3log 31


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Logarithmic Equations

Rules learned can now be used to solve logarithmic equations.

log 3x + log 3x3(log 3 8 + log 3x) = 0

log 3x + log 3x3log 3 8 - log 3x = 0

log 3x3log 3 8 = 0

log 3x3

= log 3 8 x3 = 8

x = 2

3

3 3 31. Solve for 0, log log log 8 0x x x x -


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2) log 4x + 2 log 4 3 = 0

log 4x + log 4 32 = 0

log 4x = - log 4 32

log 4x = log 4 3-2

x = 3 -2

x = 1/32

x = 1/9

3) 32x-7 = 243

35 = 243

5 = 2x - 7

2x - 7 = 5

2x = 12

x = 6


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4. Solve for 0, log (2 1) log (3 10) log 11a a a

x x x x -

(2 1)(3 10) 11x x x -

26 3 20 10 11x x x x - -

26 28 10 0x x - -

2(3 1)( 5) 0x x -

15

3x or x -

since 0, 5x x


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Natural LogarithmsLogarithms to the base e are called natural logarithms.

log ex = lnx i.e. log 2.718x = lnx

Examples

a)

Solve the following, rounding correct to 1 d.p.:

lnx = 5

log ex = 5

x = e 5

x = 148.41316

x = 148.4 (1dp)

b) e x = 7

ln e x = ln 7

xlogee = ln7

(Take natural logs of each side)

x = ln 7

x = 1.9459101

x = 1.9 (1dp)

(log 1)e

e


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c) 37x+2 = 30

ln 3 7x+2 = ln 30

(7x+2) ln 3 = ln 30

7x+2 = ln 30/ln 3

7x+2 = 3.0959033

7x = 1.0959033

x = 0.1565576

x = 0.2 (1dp)

d) For the formula P(t) = 50 e -2t

i) Evaluate P(0).

ii) For what value of t is P(t) = P(0)?

i) P(0) = 50 e 2(0)

= 50 e 0

= 50

ii) P(0) = 25

50 e -2t= 25

e -2t=

ln e -2t= ln

Remember:

ln e = loge e

-2t = -0.6931471

t = 0.3465735

t = 0.3 (1dp)


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Using logarithmsy = kxn can be expressed as a straight line.

y = kxn

log 10 y = log 10 kxn

log 10 y = log 10 k + log 10xn

log 10 y = log 10 k + n log 10x

log 10 y = n log 10x + log 10 k

Y = mX + C

If y = kxn

Then log 10 y = n log 10x + log 10 k

Where: Y = log 10 yX = log 10x

C = log 10 k

n = gradient


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When / why do we use this?

If we are given two variables of data (x andy) related in some

way and when plotted they produce an experimental growth curve,

through (0,0), then the previously given formula can be used to

work out k and n. This can then be used to decide the formula to

relate them (y = kxn).

If we are given two variables of data in logarithmic form (log 10 x

and log 10y) and when plotted a straight line is produced then it can

be said that the original data (x and y) would produce an exponential

growth curve. If it passes through (0,0) the previously givenformula can be used to calculate k and n and hence find the formula

relating them(y = kxn).


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Example

Experimental data are given in the table:

x

y

2.4 5.62 18.2 31.6 129

8.3 26.3 115 209 1100

i) Show that x andy are related in some way.

ii) Find the values of k and n (to 1dp) and state the formula thatconnectsx andy.

i) log10x

log10y

0.38 0.75 1.26 1.50 2.11

0.92 1.42 2.06 2.32 3.04


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As a straightline is produced

when log 10x is

plotted against

log 10y then the

formula relating

x andy is of the

formy = kxn.


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S


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Steps:

i) Make a table of log10x and log10y.

ii) Plot log10

x against log10

y.

iii) If a straight line is produced then state that the form isy = kxn.

iv) Choose 2 points from the best fit line (2 far apart are best).

v) Substitute 2 points into Y = nX + C to form 2 equations.

vi) Solve by simultaneous equations.

vii) Value for n can be used in final formula.

viii) Value for C is used to calculate k (i.e. C = log10k k =10C).

ix) State complete formula :y = kxn.


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Related GraphsThe rules previously learned for graph manipulation can be

applied to logarithmic and exponential curves.

As ex and lnx are opposite operations (inverse) then the graph of

y = lnx is a reflection ofy = ex abouty =x.

y =xy = ex

y = lnx

4 -2 2 4 6 8

-4

-2

2

4

6

x


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y = ex


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-4 -3 -2 -1 1 2 3 4

-1

1

2

3

4

x

y

y = e

y = e(x-b)

(b,1)

Shift

right b

-1 1 2 3 4 5 6

-4

-3

-2

-1

1

2

3

4

x

y

y = lnx

y = a lnx

Stretch by

a about

thex-axis


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Examples

a) Sketch the graph of y = log 2x.

i) On the same diagram sketch y = log 21/x .

y = log 2x

x = 2 y

Ify = 0,x = 1

Ify = 1,x = 2

( 1 , 0 ) , ( 2 , 1 )

y = log 21/x

y = log 2x -1

y = -log 2x

Graph is a

reflection ofy = log 2x

about thex-

axis.

1 1 2 3 4 5 6

-3

-2

-1

1

2

x

y = log2x

y = log21/x


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ii) Sketch the graph of y = log 2x again.

On the same diagram sketch y = log 2 2x.

y = log 2 2x

y = log 2 2 + log 2x

y = 1 + log 2x

Graph of y = log 2x

moves up 1 place.

Crosses x-axis when y=0

1 + log 2x = 0

log 2x = -1

x = 2 -1

x = .

( , 0)

5 10

4

3

2

1

1

2

3

4

x

y = log2x

y = log22x


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Further Formula fromExperimental Data

If data from an experiment is analysed, sayx andy, and plotted,

and it is found to form an exponential growth curve thenx andy

are related by the formula:

y = a b x

(a and b are constants) 0.5 1 1.5 2x


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Using logarithmsy = a b xcan be expressed as a straight line.

y = a b x

log10

y = log10

a b x

log 10 y = log 10 a + log 10 bx

log 10 y = log 10 a +x log 10 b

log 10 y =x log 10 b + log 10 a

Y = mx + c

If y = a b x

Then log 10 y =x log 10 b + log 10 a

Where: Y = log 10 y

x =x

c = log 10 a

m = log 10 a


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When / why do we use this?

If we are given two variables of data (x andy) related in some

way and when plotted they produce an experimental growth curve,

through (0,+?), then the previously given formula can be used to

work out a and b. This can then be used to decide the formula to

relate them (y = a b x).

If we are given two variables of data in logarithmic form (x and

log 10y) and when plotted a straight line is produced then it can be

said that the original data (x and y) would produce an exponential

growth curve. If it passes through (0,+?) the previously givenformula can be used to calculate a and b and hence find the formula

relating them(y = a b x).


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As a straightline is produced

whenx is

plotted against

log 10y then the

formula relating

x andy is of the

formy = a b x.


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ii) Take 2 points on best fitting straight line:(2.15,1.921) & (1.93,1.758)

Put them into Y = mx + c, where Y=log10y,x=x, m = log10b, c=log10a.

1.921 = 2.15m + c1.758 = 1.93m + c

12 -1

1.921 = 2.15m + c

-1.758 = -1.93m - c

1

3

0.163 = 0.22m m = 0.740909

Substitute m = 0.740909 into 1

1.921 = 2.15(0.740909) + c

1.921 = 1.5929545 + c

c = 0.3280454

m = log 10 b

log 10 b= 0.740909

b = 10 0.740909

b = 5.5069241

b= 5.5 (1dp)


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c = log 10 a

log 10 a = 0.3280454

a = 100.3280454

a = 2.1283618

a= 2.1 (1dp)

Formula relating x and y is: y = a b x

y = 2.1 (5.5) x

Steps:


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Steps:

i) Make a table ofx and log10y.

ii) Plotx against log10y.

iii) If a straight line is produced then state that the form isy = a b x.

iv) Choose 2 points from the best fit line (2 far apart are best).

v) Substitute 2 points into Y = mx + c to form 2 equations.

vi) Solve by simultaneous equations.

vii) Value for m is used to calculate b (i.e. m = log10b b = 10m).

viii) Value for c is used to calculate a (i.e. c = log10a a = 10c).

ix) State complete formula :y = a b x.

Chapter 15 Logs Exponentials

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