EXPONENTIALAND

LOGARITHMICFUNCTIONS

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 BASEab - is the INDEX or EXPONENT

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

Try 37 = 2187 ( < 10 000)

38 = 6561 ( < 10 000)

39 = 19683 ( > 10 000)

n = 9

Graphs of ax – Exponential Functions

For y = ax if a > 1

The graph:

•Is always positive

•Never crosses the x-axis

•Is increasing

•Passes through (0,1)

•Is called a growth function

x

y

(0,1)

y = ax

For y = ax if 0 < a < 1

The graph:

•Is always positive

•Never crosses the x-axis

•Is decreasing

•Passes through (0,1)

•Is called a decay function

Notes:

y = ax - is called an exponential function

- has base a

- a 0

x

y

(0,1)

y = ax

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 = £100

After 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

ii.

Try n = 9 : y9 = (1.15)9 × 100 = £351.7876292

Try 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?

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: r0

After year 1: r1 = r0 × 1.12

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

Formula : rn = (1.12)n × r0

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 )

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).

x

y

x

y

If we plot f (x) and f `(x) for f (x) = 2.7x:

f (x)f `(x)

(0,1)

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

If we plot f (x) and f `(x) for f (x) = 2.8x:

f (x)f `(x)

(0,1)

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

x

y

x

y

From these graphs we can see that:

•If f(x) is an exponential function then f `(x) is also an exponential function

•f `(x) = k f (x)

•A value exists between 2.7 and 2.8 for which f ` (x) = f (x)

This value is: 2.718…..

This value is denoted by: e

If f (x) = ex then f `(x) = ex

f (x) = 2.718x = ex

is called the exponential function to the base e

f (x) = f `(x) = ex

(0,1)

x

y

Examples

a) Evaluate e5.

e5 = 148.4131591

= 148.4 (1dp)

b) Solve for x, ex = 5.

x = 1.609437912

= 1.61 (2dp)

LogarithmsFor the function ( ) which is an exponential function,xf x e

1it's inverse ( ) exists. The inverse is a reflection in the line f x y x

y

x

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 of x.

If then logxay a x y

If log then yay x x a

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) log2 1/128 = -71/128 = 2-7

log4 64 3= log2 32 5=

7 1 1(check: 2

2 2 2 2 2 2 2 128

Laws of Logarithms

Law A: log a a n = n

Evaluate:

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

2 8n 3n

3 1n 0n

12

16n

4n

Law B:

Law C:

Law D:

log ( ) log loga a axy x y

log log loga a a

xx y

y

log logna ax n x

2 2 3 3 3

Evaluate:

1(a) log 8 log 4 ( ) log 27 log 3 ( ) log 81

4b c

2log 8 4

2log 325

3

27log

3

3log 9

2

14

3log 81

3log 3

1

Logarithmic Equations

Rules learned can now be used to solve logarithmic equations.

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

log 3 x + log 3 x3 – log 3 8 - log 3 x = 0

log 3 x3 – log 3 8 = 0

log 3 x3 = log 3 8 x3 = 8 x = 2

33 3 31. Solve for 0, log log log 8 0x x x x

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

log 4 x + log 4 32 = 0

log 4 x = - log 4 32

log 4 x = 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

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

Natural LogarithmsLogarithms to the base e are called natural logarithms.

log e x = ln x i.e. log 2.718 x = ln x

Examples

a)

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

ln x = 5

log e x = 5

x = e 5

x = 148.41316

x = 148.4 (1dp)

b) e x = 7

ln e x = ln 7

xloge e = ln 7

(Take natural logs of each side)

x = ln 7

x = 1.9459101

x = 1.9 (1dp)

(log 1)e e

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) = ½ P(0)?

i) P(0) = 50 e –2(0)

= 50 e 0

= 50

ii) ½ P(0) = 25

50 e -2t = 25

e -2t = ½

ln e -2t = ln ½ Remember: ln e = loge e

-2t = -0.6931471

t = 0.3465735

t = 0.3 (1dp)

Formula from experimental data

•If data from an experiment is analysed, say x and y, and plotted, and it is found to form a straight line then x and y are related by the formula:

y = mx + c

•If data from an experiment is analysed, say x and y, and plotted, and it is found to form an exponential growth curve then x and y are related by the formula:

y = k x n

(k and n are constants)

x

y

x

y

Using logarithms y = k x n can be expressed as a straight line.

y = k x n

log 10 y = log 10 k x n

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

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

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

Y = mX + C

If y = k x n

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

Where: Y = log 10 y

X = log 10 x

C = log 10 k

n = gradient

When / why do we use this?

•If we are given two variables of data (x and y) 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 = k x n).

•If we are given two variables of data in logarithmic form (log 10 x and log 10 y) 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 given formula can be used to calculate k and n and hence find the formula relating them (y = k x n).

Example

Experimental data are given in the table:

xy

2.4 5.62 18.2 31.6 129

8.3 26.3 115 209 1100

i) Show that x and y are related in some way.

ii) Find the values of k and n (to 1dp) and state the formula that connects x and y.

i) log10 x

log10 y

0.38 0.75 1.26 1.50 2.11

0.92 1.42 2.06 2.32 3.04

As a straight line is produced when log 10 x is plotted against log 10 y then the formula relating x and y is of the form y = k x n.

x

y

ii) Take 2 points on best fitting straight line:(0.38,0.92) & (2.11,3.04)

Put them into Y = nX + C, where Y=log10y, X=log10x, C=log10k.

0.92 = 0.38n + C3.04 = 2.11n + C

12

2 - 1

2.12 = 1.73n

n = 1.225433526

n = 1.2 (1dp)

Substitute 1.225… into 1

0.92 = 0.38(1.225…) + C0.92 = 0.46566… + C

C = 0.4543352601

C = log 10 k

log 10 k = 0.45433…

k = 10 0.45433… k = 2.846657779 k = 2.8 (1dp)

Formula relating x and y is: y = k x n

y = 2.8 x 1.2

Steps:

i) Make a table of log10x and log10y.

ii) Plot log10x against log10y.

iii) If a straight line is produced then state that the form is y = k x n.

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 = k x n.

Steps: Version 2

i) Make a table of log10x and log10y.

ii) Plot log10x against log10y.

iii) If a straight line is produced then state that the form is y = k x n.

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

v) Use these two points to find the gradient. This will give you n. vi) Use one of the points and y – b = m(x – a) to find the equation of the line.

vii) You now have the value of C.

viii) Use the value for C to calculate k (i.e. C = log10k k =10C).

ix) State complete formula : y = k x n.

Related GraphsThe rules previously learned for graph manipulation can be applied to logarithmic and exponential curves.

As ex and ln x are opposite operations (inverse) then the graph of y = ln x is a reflection of y = ex about y = x.

y = xy = ex

y = ln x

x

y

x

yy = ex

y = ex+a

a Shift up ‘a’

x

y

y = ln xy = ln (x+a)

1-a

Shift left ‘a’

x

y

y = exy = e-x

Reflection in y-axis

x

y

y = -ln x

y = ln x

Reflection in x-axis

x

yy = ex

y = e(x-b)

(b,1) Shift right ‘b’

x

y

y = ln x

y = a ln x

Stretch by ‘a’ about the x-axis

Examples

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

i) On the same diagram sketch y = log 2 1/x .

y = log 2 x

x = 2 y

If y = 0, x = 1

If y = 1, x = 2

( 1 , 0 ) , ( 2 , 1 )

y = log 2 1/x

y = log 2 x -1

y = -log 2 x

Graph is a reflection of y = log 2 x about the x-axis.

x

y y = log2x

y = log21

/x

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

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

y = log 2 2x

y = log 2 2 + log 2 x

y = 1 + log 2 x

Graph of y = log 2 x moves up 1 place.

Crosses x-axis when y=0

1 + log 2 x = 0

log 2 x = -1 x = 2 -1

x = ½.

( ½ , 0)

x

y

y = log2x

y = log22x

Further Formula from Experimental Data

•If data from an experiment is analysed, say x and y, and plotted, and it is found to form an exponential growth curve then x and y are related by the formula:

y = a b x

(a and b are constants)

x

y

Using logarithms y = a b x can be expressed as a straight line.

y = a b x

log 10 y = log 10 a b x

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

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

When / why do we use this?

• If we are given two variables of data (x and y) 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 10 y) 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 given formula can be used to calculate a and b and hence find the formula relating them (y = a b x).

Example

Experimental data are given in the table:

xy

2.15 2.13 2.00 1.98 1.95

83.33 79.93 64.89 62.24 59.70

i) Show that x and y are related by the formula y = a b x.

ii) Find the values of a and b (to 1dp) and state the formula that connects x and y.

i) x

log10 y

2.15

1.921 1.903 1.812 1.794 1.776

1.93

57.26

2.13 2.00 1.98 1.95 1.93

1.758

As a straight line is produced when x is plotted against log 10 y then the formula relating x and y is of the form y = a b x.

x

y

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

13

0.163 = 0.22m

m = 0.740909

Substitute m = 0.740909 into 1

1.921 = 2.15(0.740909) + c1.921 = 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)

c = log 10 a

log 10 a = 0.3280454

a = 10 0.3280454

a = 2.1283618 a= 2.1 (1dp)

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

y = 2.1 (5.5) x

Steps:

i) Make a table of x and log10y.

ii) Plot x against log10y.

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