~ Section Contents

3.1 Exponential equations and logarithms 3

3.11 Laws of indices

' ) 3.12 Simple exponential equations ' 3.13 Logarithms

3.14 Logarithms and the cak:ulator

3.2 Further logarithms 7 3.21 Logarithms to base e

• 3.22 Evaluation of logarithms 3.23 Laws of logarithms 3.24 Change of base

3.3 Logarithmic and exponential. equations 11 3.31 Logarithmic equations 3.32 Exponential equations 3.33 Change of subject

.···-. 3.4 Graphs 14 . .: 3.41 The logarithmic graph 3.42 The exponential graph 3.43 Non-linear empirical equations 3.44 Log-log graphs

3.S Exponential growth and decay 23

3.Sl Verbally formulated problems 3.52 The exponential graph

~ Review 25 .

.

/----...\ 11 i# II Answers 27

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3 - l

Logs and exponentials

At the completion of this section you should be able to:

CJ manipulate and simplify arithmetic and algebraic expressions using the laws of :Indices and logarithms

CJ skefclt the graphs of simple exponential and logarithmic .functions showing behaviour for large and small values

CJ solve exponential and simple logarithmic equations using indices, logarithms, calculator and graphical techniques

Cl .convert logarithms from base 10 to base 2 and base e and vice­versa

CJ transform non-linear functions (including exponential) to linear form and plot data

a draw curves of best fit, interpolate data and estimate constants in suggesb!d relationships

a interp1et verbally formulated problems inVolving growth and decay and solve.

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Logs and exponentials

Indices were introduced in Maths A. Below is a table summarising the basic laws of indices.

3.11 Review of the laws of indices

1. a"' x an ::: am+n. 2. a'"' +an.= a"'"""

3. (am}"=a""' 4. a•= I 5. a"""' .. 2-

a"' I

6. '!(ti= ail

Note ao!O

It would be useful at this point to revise writing numbers as powers of a base. For example:

.l. = 2 ... 16

.l. 3-4 '1 = *"' 4-4 oh-= s-4

1-2'"" .- -Jr=rs it= 4-l tlr"' 5-3

1-2-2 ~- t"' 3-2 it= 4-• Yi= 5-2

t= 2-1 t=3_, 1-4""' 4- t= 5-1

I= 2° 1..; 3° 1= 4° J=·s• 2=21 3=31 4= 41 5=51

4= 22 9=32 16= 42 25= 52

8= 25 27=33 64=4' 12s- s• 16= 24 81= 34' 256= 44 625= s• 32= 2• 243"' 3• 1024= 4• 3125= 5'

All numbers can be expressed to some base. The table above shows some of the integer powers of the bases Z 3, 4, and 5. These powers are called · exponents.

More di££icult examples of indices include terms such as (2.3)"12• These can be

evaluated using a calcuJator.

Using a Casio fx-100 caiculator: (2.3)"12 = 23.9. NOil!: ~~KB.,

3-3

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. ····-··"···----.. -.. : .......

·Logs and exponentlals

3.12 Simple exponential equations

So far the laws of indices have been used to rearrange simple arithmetic and algebraic expressions. The laws can also be used to solve more difficult exponential equations of the form a• = b, where a and bare constants and the exponentx is unknown.

3.13 Logarithms

3-4

In example 1 above, the terms can all be expressed with easily calculated .-·-) powers. This represents only a small proportion of equations where the unknown is an exponent or index. For example; the equation 10• = 17 :requires you to write 17 as a power of 10. This is impo5sible without resorting to logarithms.

A logarithm is a tool which allows you to write numbers with an index in terms of a base (in this case the base 10). For example:

We know 100=102 so that (The loglOO ln the base 10 is 2 log

10100 .. 2 becauselOO = 102

.)

The expression 100 =102is in exponential form (theex]ionentis2). The expression log,0100 = 2 is :in log .fonn. · Logarithms can be written to any base, but the most important ones in this course are bases 2,. e and 10. Some easy examples of logs in the bases 2 and 10 are listed on page 5. ·

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Logs and exponentials

Indices base 2 Lo11:sbase2 htdices base 10 Lo2Bbase10 f,;"' 2 .... Iog,r. .. -4 ... 10 .... log, .... -4

*"" 2-3 log2 t= -3 iioo-=10-3 log10 ni\;y = -3

t= 2-• Iog.t=-2 i!o= 10-• 10810 rlo = -2

.l.- 2-• 2,... log2 t=-l ilr = 10-• 1og1oro= -1

l= 2° loth I"' 0 l• 10° log10l= 0

2: 21 log2 2=1 10= 101 log10 10= 1 ··

4= 22 log,4= 2 100= 10• log,. 100 = 2

g .. 23 log, s., 3 1000= 103 loglO 1000 = 3

16= 24 log2 16= 4 10000= 104 loglO 10000 = 4

32 = 2' Jog.32= s 100000= JO' log10 100000 = 5

3.14 Logarithms and the calculator You can use your calculator to find the Jogs of any number. The lag button on your calculator represents the common log to base 10.

3.5

....... --···· ····-·····-··-·-·········""'"'' ....... ·-· ........ ·~·······--"'""""'·'"'

Logs and exponentials

3.15 Activity 1

(1) Simplify (In a-d leave your answer in exponential form. In e-f leave your answer as a rational number).

(b) 21 +23

(d) 2° x2'""

(2) Use a calculator to evaluate the follow.Ing (to 3 sig. figures).

(a)(2.3)3

(c) (6.3)-M

(b)(3.2S)u

(d) (21.5)33 x(0.45)'

(3) Solve the following exponential equations

(a)2' =8 (b)3' =27

(c)2-• =32 (d)2' -4=0

(e)2' +5•13 (fl4' •8

(g}3' =! {h)f.'=81

(I) g2.>t-I = 16 U> 44-• ·-k

(4) Use the log button on your calculator or the log table on page 5 to solve

{a) 2' =16 (b) 2-· =t

Cc) 2· -1 (d)2-· =16

(e)lO' =50 (/)10-· =0.234

(g) 10>+3 = 234;2 (h) 102-

3' • 19

(I) tOi. = (3.S)' • U)lO• .. so

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Logs and exponentials

3.21 Logarithms to base e

Logarithms can be expressed to any base but probably the most significant base is the letter e. On your calculator next to the log button, is the In button. The In button ls log,; that ls log to the base e. The letter e repre&ents a number between 2 and 3 (approximately 2.718) that is very important in measuring exponential change. Using your calculator:

log, 2 = ln2 = 0.69:!1

log, 3 = In:!= 1.0986

log, 2.718=In2.718=1

3.22 Evaluation of logarithms

or e0'°'»1 = 2

or e1.ll98t! = 3

or e1 =2.718

Your calculator allows you to evaluate logs, or exponential expressions where the indices used are not integers.

3-7

Logs and exponentials

3.23 Laws of logarithms

3-8

Since a log is an index, then the laws of logs are essentially the laws of indices. The three basic laws of logs are:

(I) log(AxB) = logA+logB (ir)log(A,rB) = logA-logB

(iii) log A"= nx log A

(tv)log. a c 1

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·····-~-·-·-·-·-·- .. --.. ... .. ... . . ·~· '""' ·--- ... --····-·--'·-···· .

Logs and exponentials

· .. ) 3.24 Change of base

Although logs can be written to any base the calculator displays only base e and base 10. You can use the caJculator to find logs to any base by noting:

You could also use the In button on your calculator by noting that lnA

log1 A=-lnb

3-9

Logs and exponentials

3.25 Activity 2 ) (1) Evaluate (to 3 sig. figures)

(a) log10 S.21 (b) lnS.21

101301 eo.1• _10o.u (c)log 200 (d) -0.U

10 e

(e)31og1&26-101•1 (f) e3.6 -2Se-u

(g) IO'"'" -4ln4 (h) ln23idn4+Iog10 8

(2) Use the laWB of logs to simplify

(a) loglO S + Jog!O 2 (b) log10 40-log10 4 !' '\ \.. .. )

(c) log10 40+ log1, S-log10 2 (d) log2 8 + log2 1.5-log2 48

(e)tlog2 100-loih 2.S (f) lne3 +Jn.[;

(g) lnf-log2f (h) log2 16-log2 8-log2 2

. (t)f1nx-21nx (j) 1n,J; + 21nt

(k) tlog10 100x-1og16 9,J; <l)mW-1ne (3) Evaluate

(a) log4 9.6S (b) log2 10

(c) 4log, e-4log2 10 (d) log3 0.034S-flog2 0.2 ()

(4) Given Jn 2mfl.693 and ln 3=1.10, Without a calculator, evaluate:

(a)ln6 (b)ln4-ln6

(c) lnl.S (d) ln12

(5) Given In x-5.5 and In y-4.2, evaluate:

(a)lnxy (b) ln(t)

(c)·hu:2 + lny (d)lnxxlny

(} 3-10

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Logs and exponentials

3.31 Logarithmic equations

Equations where the unknown is contained within the log require a translation to exponential form.

3.32 Exponential equations

Jn section 1.2, exponential equations were introduced where the exponents were easy powers of the given base. When the powers are not easily found, the equation must be solved using logs (either to base 10 or base e).

3·11

Logs and exponentlills

3.33 Change of subject

3-12

Solving or rearranging log and exponential equations generally requires you to switch between log and exponential forms. The same is true when changing the subject of these types of equations.

)

' ) 3.34 Activitjr 3

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(1) Solve for the unknown: (to 3 sig. figures).

(a) log10 3x ""2.S

(c) ln3x=2.S

(g) lnx"' -3.S

(2) Solve for the unknown:

(a)lO' =0.1

(c)4' =-3.4

(e)22•+> =s'

(g)3x2.,._, =2

(b) log2 3x = 2.S

(d}log2(t+3)=-l

(f>m(t)=4 (h) log2 (x + 3) = Jog10 65

(b)8' =128

(d) 3•-I = s

(h) 112 = l51e"+2

(3) Oumge the subjl!CI: to the term in parentheses:

(a) log10(.x + 2) = Y (.r) (b)ln(x+2)=y (x}

(c) log2 2+log2 t=21 (t) (d) ~~) = -0.035 .(P)

(e) Jog,0(-1-) = r

t-1 (t) (f) z = 276log,o( 2:) (D)

(g) lnP+NlnV• lnG (G) (h) lo!!10( __!._) = k

v-r (r)

(l)X=X0 A" (t) U)P=P,,e-lt (k)

(k)y=y0 +Ae" (t) (l) v = v0 e1-" (k)

3-13

Logs and exponentials

3.41 The log graph

3-14

Consider the table of values below relating to the functiony =log. x.

x 0 .25 .50 :15 l 2 3 4 5 6 7 8

v - -1.39 -.69 ·.29 0 .69 1.10 1.1'19 1.61 1.79 1.95 2.08

By plotting these values on axes and joining the points smoothly, we create a graph of y =Jog. x, fur values ofx between 0 and 8.

·The note above applies generally to log.functions in any base.

The graph of y =Jog, x appears below.

;s y a.

y=log.x

.., t' 'I x

Notethat: When x~ the function ill not defined arui the curve tends towards negative infinity; As x gets larger so does y (the curve tends towards positive infinity); We cannot take the log of a negative nul!lber arui so there are no negative values of x.

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Logs and exponentials

3.42 The exponential graph

The function y .. e' is the inverse of the function y = log, :x • As in the construction of the log graph, consider the table below.

I; ·2 0 1 2 l 2.718 7.389 0.135 0.000 0.368

The graph of y = e' appears below. y=ez

3 2009

3-15

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. ·- ...... ·-·-···--·--··- .................. _ ... ···--··· .. ········---·- ..... ~-·-.. ·--···--· .. ··-····-··-······ . ....... --.-.. ~-------···-·-··--· . -···· ......... . .. -···---.. ··~ .. ·-~--~---·- .............. .

Logs and exponenfials

The graph of y = e'is an example of an exponentlal curve. Note that:

As x gets smaller the value ofy tends towards O; As x get8 larger so does y (as x approaches infinity y approaches infinity); There are no negative values of y.

Olher examples of exponentlal curves are shown in example 12.

3.43 Non-linear empirical equations

3·16

Empirical data refers to information collected from experiments. When such data is collected, it is usually to establish a relationship between the terms being measured.

One of the simplest relationships is the linear equationy = mx + b. When we find values form (the gradient) and b(the vertklil axis intercept), we detail the particular link between the terms being measured.

On the following page is an exampie of a linear relationship.

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Logs and exponentia!Jl

The following table links the distance, d (in metres), an object travels with time, t(in seconds), with readings being taken eveey 10 seconds.

I 1~~9 10 30 0 0 5 16

The relationshipisoftheformy =mx+b. Jn this experiment the actual variables are d = mt+ b . '

From the table we can plot the values of tand d and draw a line of best fit. (See Maths A section 4). The equation of this line establishes the link between ti.me and distance and the gradient of the line measures the rate of change of dlstance With respect to time (the velocity) .

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The value of mis found by usmg the d . -·~ .. - f . ···1

! gradient formula m = y, - Yi . .... ..... ~·······~~ ... 2f ............ ,_ ·1 ••• . x, -Xi

~o _ ... ' Using the points (0,0) (50,24.6) ..

!~ •..... i 24.6-0 lO - m=

50-0 =0.S.

w ·-~ The place where the line crosses the lo .... ... ..., fir> t vertical. axis is b. That is h-0.

• Note: When calculating myou can use any two points on your line.

From the data above the value of .m-0.5 and b-0. The expression has equation d-0.5~ with the value 0.5 ~ting a constant velocity of O.Sm/s.

Most experiments generate relationships that are not shitple linear equations. While this leads to more complex equations, there are techniques for converting ium'llnear data into linear form. One such technique involves the use of algebra to transform the equation so that it becomes linear.

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3-18

When dealing with exponential functions, the general way to transpose equations to linear form is to take logs of both sides of the equation. If the exponential function has a base e then the relevant log base is e (that ls ln). Otherwise taking log to the b~ 10 transforms the equation to linear form.

If the equation ls a quadratic function (see example 13 part (a)), then logs are not used in the transfunnation.

Where an experiment involves the construction of a lable, the infonnation recorded in the table must be changed so that the relationship is linear.

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···-···· -·-·-··--·""""'""'-·'········

Logs and exponenllals

3.44 Simple log-log graphs

Jn example 14 above, a table was constructed using the log values of current and voltage. This table was necessary to make the equation lillfar.

Another method that allows non-linear equations to become linear, is to use graph paper with a non-lillfar scale. Log-log graph paper uses a scale where each axis is Jogaritlunk (base 10).

The axes are divided into cycles of 10 with the gaps between numbers uneven, as shown below:

1 2 3 4 5 6 7 8 9 10

Depending on the values in the experiment, the numbers on the axes could be: 1,2,3,4,5,6,7,8,9,10or10,20,30,40,50,60,70,80,90,100or100-1000 etc.

By using log-log graph paper we do not have to construct a new table, as in example 14. We can directly plot the original points to generate a straight line.

The experim,ental values in 'example 14 are used to plot points on log-log graph paper overleaf. The particular paper used is called 2x2 cycle log-log graph paper, because the valties in the table are spread over a range from 1-100. The first cycle represents values between 1 and 10; the second cycle represents values between 10 and 100.

3-19

Logs and exponenl.:!ais

Table from ~xainple 14:

1. v 1 5 13 I 40 90 )

1.8 6 12 27

Plotting V on the horizontal axis and I on the vertical axis shows: · ·

/ .~ ·~

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Logs and exponetllals

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3.45 Activity 4

1. Sketch graphs of the following functions:

(a)y= lnx (h)y=3lnx

(c)y=31og10 x (d)y=e1'

(e) y=-2% (f)y=e...,_I

2. Convert the following to linear form:

(a)y=2.!:2 -3x (b)y=~

(c)y=2e% (d)y=3e1'

3. The following table of values was recorded in an experiment

:JC 1 2 3 6 9 l2 () 11 2.7 7.3 ' 14.4 36 64 96

The equalion1lnking the two variables is of the formy >=Ax"', where A and m are constants. (a) By taking the Jog of both sides of the equation and Jetting logy-Y, logx-X

and logA•b, show lhatthe linear equation relating X and Y is Y = mX +h. (b) Construct a new table using the log values of each score and use this table

to draw a line of best fit (on linear graph paper). (c) Use your graph to find m and b (and hence the equation linking y and :x.)

4. An experiment gave the foJlowing results:

v 1.5 3.2 6 16 40 85

E 65 30 16 5.8 2.2 1.0

The equation linking the two variables is of the form E =AV", where A and n () · ....•. are constants. Using Jog-Jog graph paper sketch a liru:i of best fit

5. The results be1ow were recorded from an experiment

:JC 10 20 40 60 80 100 v 53 22 9.5 6 4 2.8

Construct a newtab1e of values by taking the log of each entry. From your new table plot the points on ordinary (line8r) graph paper and draw a line of best fit. Find the gradient (m) and vertical axis :Intercept (b) of your line and hence .find its equation in the form y=mx+b.

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Logs and exponentials

Population growth or decay usually occurs at a rate dependant on the original population. The rate of change is exponential for which the value e Is highly significant. A formula for exponential growth (or decay) is·:

P=JOekl,

where PiB the current population, P0 represents the original population, k is the rate of growth (if k is negative, the population is decreasing) and t represents the time period.

In section 3.51 are some examples of the type of problems that involve using the exponential formula above.

3.51 Verbally formulated problems

A population is more than a collection of individual$. It can represent the number of bacteria in a culture, the mass of an object, the amount of money in a bank account or any one of numerous other applications .

3-23

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Logs and el<pOllenliala

3.52 Activity 5

1. The population of a city rotals 200000peoplein1998. If the growth rate of the dty is a constant 1.3%, when will the population double? Use the formula

P=P,,1l1 •

2.The mass of an organism growing in a pond is given by M = M 0e"', where M0 is the initial mass of 0.3kg, k-0.03 and time tis measured m days. What will be the mass of the organism after:

(a) lOdays? (b) 1year?

3. A radioactive element decays aa:ordh1g to the £or.mu1a M = M 0eu, where the initial mass is 100g. If the element decays to 50g in 10 years, find the value of k (as a decimal) and the mass of the element after 20 years.

4. Atmospheric pressure P ldlopascals, at a height hkm above sea level. is . given by the formula P = 100e-o.1,.... Find the:

(a) Atmosphericpressureatsealevel (b) Atmosphericpressureat25000m. (c) Atwhatheightthepressure drops to 1kPa.

5. The fntenstty of fight passing through a transparent medium, such as water, diminishe5 according to the Jaw I= 10e'"".-i, where I 0 is the initial intensity and I is the intensity at a depth of d metres. At what depth will the intensity De:

(a) 50% of the initial value? (b) 1% of the initial value?

(c) wm the intensity ever be zero?

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Logs and exponenllals

1. Use your calculator to evaluate the following: (a)lo'~ (b)lO-u (c)eu

(e) 4.leu ( h) l 8eo.1'69><IOO

2. Evaluate without a calcu1ator:

{a)log2 32 (b)log28 . (c)lne

(e) log10 1000 ([)log,. 0.01 (g) 1os2l (h)log2 o.s .,.,

(i) log10 .J100 (f) loS10 JOI (k)Jnl (I) Inf

3. Write in exponential form and solve:

(a)log216=x (c) log2 l=v

([) log,0 x = -0.S (g) Int .. 3.389 (h) Jnx = -0.4979

4. Evaluate to 3 sig. figures: ·

(a) log10 94.3 (c)ln4 (d) Jn(2.4x103)

(e) SOlog10 50 ([) log10 34.S- ln34.S (g} ln(e -1) (h)lnx=-0.4979

. 5. Use the laws of logs to simplify:

(a)log10 so+Jog10 20 (b)log2 l6-log2 8 (c)log .. 5-log10 o.s (d) log2 2+log2 o.s

(e) log2 20 +log2 1.6 (}) log10 l.2S+Iog1.s (g) log2 64 log2 8

(}) 2log2 48-2log2 6-log2 32

(k)2lne-lnf-3 (/) elne' +tln.Je-elog10 100

Logs and .exponentials

6. Use the Jaws of logs to simplify:

(a)Inxy-lnx (b}flnx-2lny (c}ln ~

e (d} 2lnx+3lny-lnx

7. ExPand using the Jaws of logs:

(a) lnxy (d) ln'ifi x (c) In .JY

8. Evaluate the following:

(a)log2 10 (b)log! 16 (c)log16 28 ( d) log6 0.035

9. Solve (answer to 3 sig. figures):

(a) log10 x=1.5 (b)log2 x=0.5 (c)lnt=0.1398 (d) 5.lnt = -3.65

(e)ln(t-1)=1.l (f) ln(f)=e (g) 10»= S.4 (h)e' .. 5,4

(1)5,. =10 U>2>l .. s (k) 4.5'e' = 4.S (l) e21-• "' 0.564

10. Change the subject of the following to the term in brackets:

(a) l' = 100.e' (t) (b)y• lnx• (x) (c) Y •Yo In(2-t) (t) (d) P = P0.e'°.ou (t)

11. The following results were recorded in an experiment.

v 1 4 10 30 60 100 I 5 . 10 17 so

(a) Plot the points on log-log graph paper and draw a line best fit. (b) Construct a new table of values by hiking the log of each reading. (c) Using the values in the new tllble, plot the poi:nlB on linear graph paper and draw

a line of best fit. · (d) Use your graph to find the equation of the linem the form y=mx+b.

. .

12. The formula T = 400ep8 is used to measure the maximum tension in a belt pulley, before slippage. Tension Tis measured in newtons, 0 is the angle of lap in radians and µ is the coefficient of friclion between the belt and the pulley. Calculate the: (a) tension T, whenµ= 0.5 and 0 = 3radians,

(b) coe.ff.identoffrlction, when T =1000N and 0•100°.

3-26

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l.ogll and expone.nflals

(c) y

-1

{e) y

2 (a)Letf=Y

Y=2x-3

x

x

(d)

, x

{f)

~ (1,e) ....

x

(b) log10 Y = 2nlog10 x or Y=2nX

(c) lny = ln2 + Jne• (d) lny= InJ+lne:i.

.s. (b)

=ln2+x Y= x+0,693

log10X 0

log10Y 0.431

0.301

0.863

= l.10+2x Y= 2.r+l.10

0.4'!'1 o.m 0.954 1.08

1.16 1.56 1.81 1.98

(c) Using tbe p0mtii (0.778,1.56) and (0,0.461), m-1.41, b=0.461 and y = 2. 89x"'2• .

5. I0810X 1 1.3 1.6 1.78 1.90 2.

log10Y 1.72 1.34 0.9'78 0.778 0.602. 0.447

Using the points (1,1.72} and (2.39,j)) m--1.2 and b=i.92

Adivfty5 1. t-53.3 yrs

4. (a) P=l.00 kPa (b) P-4.39 kPa (c) h-36.8 km

2. (a} M-0.405 kg (b) M=17086 kg

5. (a) d-1 m (b)da6.65m

s. k--0.0693 M"'25kg

5. (c) As depth increases, intensity tends Inwards Dim.

)

.. ' ' )

/. \ _,

,,.·--. \ ( }

.. . .. . . ·····• ········•··--··· ··········-········ . ········-············--··· . ' . ·-·······

.... ... .. ~--·-· ·-· ... . .. ··········--·········· ..

Logs and exponenlials

( \ Review )

1. (a)20 (b) o.os (c)3.67 (d) 0.082 (e) 6.76 (f)36.8

i' (g) 1.65 (h)17861 1!

i:~ 2. (a)S (b)S (c)l (d)-3 (e)S (f)-2 ·w (g)O (h)-1 (i) 1 (j)-0.5 (k)O (1)-1 j~ L· 3. (a)4 (b)-2 (c)O (d)-1 (e)3 (f) 0.316

r (g)29.6 (h) 0.608

I 4. (a) 1.97 (b)-0.367 (c) 1.39 (d) 7.78 (e)84.9 (f)-2

I (g)0.541 (h)0.608

5. (a)3 (b)l (c)l (d)O (e)5 (f) 1 (g)2 (h)0.25 (i) 1 (j)2 (k)O (I) -!-

6. (a)lny (b)7 (c)-2.% (d)xy' ,,.. ;

' } 'l. (a)lnx+Iny (b)lny (c)Jnx-tlny (d)}lnx

8. (a)3.32 (b) 1.72 (c}l.204 (d)-1.87

9. (a)31.6 (b) 0.25 (c)l.15 (d)0.482 (e)4 (f) 30.3 (g) 0.732 (h) 1.69 (i) 0.715 b)S.32 (k)O (1) 1.21

10. (a)ln(Wi) or y L (d)-SOln(f,)or lnV-lnlOO

(b)e2 (c)2-e"' - So{lnP- ln.fo)

........ ( ·, ;_ ,1 .,_

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