Set 5

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1 SUMMARY NOTES

UNITS 3 AND 4 MAmSMETHODS (CAS) NOTES SET 5

ALGEBRA OF LOGARITHMS

INDICES LAW

• Logarithmic function is simply the inverse of the exponential function. Logarithmic and exponential functionsare mainly used in real life situations for natural growth and decay.

BASIC LOGARITHMIC EXPRESSIONS

• Iloga a = 1 1 where a = a positive constant

• Iloga 1= 01 where a = a positive constant

ego loglo 10 =1; log44 = 1

ego log 2 1= 0 ; 10glO1= 0 ; 10glO01 = = 0

(iv) Index law: Iloga b = c I is equivalent to l a C = b l !

Eg. Simplify: a. logs 125 c. 210gloi f J O

SOLUTION a. logs125 = logs 53 = 310gs 5 (see iii) = 3(1) = = 3

b. log2/6 = log2(;4) = log2T4 = -410g22 (seeili) = -4 (1 ) =-4

Eg. Simplify: a. log 42

d. loglOOO5 + logJOOO2g. log9 486 -log9 2

b. log1255

e. 41og52 + 310gs6

h. 210g3 5 - 310g3 2

c. log654 + log64

f.210g3(t)+310g3( t)

i. tlogS 2 - tlogs 27

J

SOLUTION a. log42 = log4.J4 = log442 = ~log44 = t(l) = +

b. log12S 5 = log]2S 3JliS = log12512St = tlog12s(l25) = = t C I ) = tc. log654 + log6-4-= log6<54-x-4)-~log6 216 = loU 63_~310g6 6 ="-3 (1 ) = 3

3 r;;:;;;;:; 1 1 ]d. 1 0 g lOOO 5 + logLOOO = lo glO OO (S x 2) =1 0 g lOOO( 1 0 ) = log ]oO O . " \ 11000 = ] og IOOo( J 000 )3 = 3(1) = 3

4 'e. 4 logs 2 + 3 logs 6 = logs 2 + logs 6J = logs (16 x 216) = log, (3456)

(Note: Prior to addition or subtraction of a logarithmic term, eliminate any coefficient in front of logarithmic)

f. 210g 3 ( t ) + 3 1 o g 3 ( t) = l og3 (+ )2 + logj ( t)3 '" lo g 3(1 S ) + l o g 3 U 4 ) = lo g3 (i5 x ; 4 ) = l o g 3 ( lO~O ) = J o g 3 1 O -3

= - 3 J o g 3 1O

Ig. log9486-10g92 =log9 4~6 =log93

5=510g93 =Slog9. J 9 =510g9(9)2 = 5 ( 1 10g9(9)) = i log9(9) = i

h. 210g 3 5 - 310g3 2 = log352 - log323 = log 3 e ni. tlogs 2 -tlogs 27 = logs 2! -logs (27)t = 10gs.J2 -log5 3 = logs ( f)

b. 10g10. f0 .g. Given that loglo 3 = 0.477 and ]oglo 4 = 0..6021. Evaluate 3. log10 144

SOLUTION

3. loglO 144 = loglo (9 x 16) = 10glO9 + 10glO16 = loglO 32+ 10glO 4

2= 210g1 O 3 + 210g1 O 4 = 2(0.4 77) + 2(0.602 J) = 2,] 6

c. Joglo 2



Ih. loglO.Ji2 = loglO1i2 "'tloglO12 =t1oglO(3x4) "'t(JogI03+ logJO4) =t(0.477+0.6021) =0.54

Ic. JogIO 2 = loglO J4 = loglO 42 =~loglO 4 = t(0.602]) = 0.301

INDEX LAW (Use index law to eliminate the logarithmic expression).

If an expression has the form Iloga b = c I, it can be expressed as laC = b I i.e log, b = c is equivalent to aC = b

Eg. Use Index law to find x for each of the following:

a. log2 (2x - 1 ) = = 3

d. Jog3(x+l])=2Iog32+310g35

SOLUTION a. log2 (2x - 1 ) = 3 Refer to Index law: a=2, b= 2x-1 , c= 3

aC =b => 23= 2x -1 => 8=2x -I => 9 = 2x => x = 4.5

h. log3(4x+7)=4 Refer to Index law: a=3, b=4x+7, c=4

aC

= b => 34

= 4-x+ 7 => 81 = 4x + 7 => x = 18.5

c. !og2(2x+I)=log2 4+1og2 3 => ]og2(2x+l)=log212

logl(2x+l)=1og112 1'-v-- '-v-' => (2.x + 1 ) = 1 2 => x", 5 .5

Equate 'liked' terms

d. log3(x+ll)=21og32+31og35 => log3(X+l)=Jog322

+log}53

=> log3(x+l)=log3(4x125)

=> log 3(x + 1) = log 3(500) . Equate "liked' terms: x + 1=500 => x = 499

e. log2 (x+ 70) = 21og2 3+ 4 => log2 (x + 70) = Jog2 32+4 => log2 (x +70)-log2 9 = 4 => log2 ( X ~ 7 0 ) = 4

Index Law: => 24 = x + / O => x = 74

Eg. If 31og2 A = (~) + Jog2(H + 1 ) . Make H the subject.

SOLUTION log2 A3

:= : ( ~) + log2 (H + 1 ) => log2 A3

- Jog2 (H + 1 ) = = - & - = " >

Index-Law: => i~ )= H < l = > H + 1= /;) => H = /;) -1

(A3 ) _ E

log2 H +1 -G

=> H=A3.2(-!)-1

Eg. If 210ge (x + 7)2 = II-loge.:!:. Make y the subject.y

SOLUTION 21oge(X+7)2=11-1oge(;) =>2Jog e(X+7)2+)oge( ; )=l l => logeex+7)4 +IOge(-;)=II

Eg. Find the value of x for which: (I0g2 x i - 2(log2 x) - 3 = 0

SOLUTION The above expression is in the form of a quadratic equation.

Let a==log2x => a2-2a-3=0 => (a-3)(a+l)=0 => a=3 and a=-1

Sub a=og-» back! • log2x = -1 => 2-1 = x x =0.5

Eg. Find the value of x for which: (loglO X2) 2 -(JogIO x9) + 2 = 0

SOLUTION (loglO x2i-loglO x

9) + 2 = 0 => (21oglO x)2 - (logJO x

9) + 2 = 0



=> (2)2 (loglO X )2 - (log)_QX9

) + 2 = 0 => 4(loglO xi - 9(log10 x) +2= 0

Let a=)oglox => 4a2-9a+2=0 => (4a-1)(a-2)=0 => a=t and 0=2

ISub a = JoglOx backl • log., x = t => 104 = x => x = 1.78 • loglO X = 2 => 10

2=X X = 100

LOGARlTHl\1IC EXPRESSION INVOLVING THE CHANGE OF BASE

Note: log1o = JlogJ and loge = jlogl in the calculator!

a. log4(7)

SOLUTION

h. Jogl1(108)

Eg. Use a calculator, and by converting to loge or JogIO , evaluate:

a .- loglO(7)- 4

b . I (l08) - loglO(J08)-1 95 c . I (205) = logJO205) = 1 16Jog4(7) - log]o(4) -1. ogll - loglO1 1 ) __; . .og98 loglo(98)'

d . [age (5) =] .61 loge(i)=-J.38 f.loge8

e. log38 = loge3 = 1.8927

Galeulator has both-log., or log. => any of them can beused=> answer would be the same. Notice that for

(a), (b) and (c), the base of 10g]O was used whereas (f), the base of log. was used.

LOGARITHMIC GRAPHS

• Logarithmic graphs have vertical asymptotes but they have no horizontal asymptotes.

(an asymptote is a line in which the graph will approach but will never touch).

• A simplest l og graph is shown: • The graph can be reflected about X or Y axis to

Y j . p roduce the fo llow ing:

~ y = IOge(x) y = loge( -x).. X

( It has a vertical ----~&+-~:=--------+ xasym pto te at x = = 0 Y =- IOge( -x) y = - IOge(X)

• General equation: !Y = aloge(nx - b) + cJ with vertical asymptote by letting (nx - b) = 0 => I x = * 1

Eg. Sketch the following graph 3. y : : = loge (2x - 3) c. y=31og e(4-3x)+1

SOLUT I ON a. x=~

II

y

c.x=1

3

I

I

--------~~-~;;:

I

II

y

Vertical asymptote:

Let zx - 2= 0

;:> X ",12

Vertical asymptote:

Let (7x+11)= 0

= > = = - ¥ -

Y intercept (let x = 0 )

y = - (!)Jog. (11)

Vertical asymptote:

Let (4 - 3x) = 0

4;:> X= 3 '

Y intercept (let x= 0)

y = = 3Ioge(4)+1

x intercept (let y=0)

log,(2;;: - 3) = 0

(Index law)

,,°=2;;:-3 =) 1=2;;:-3 =>;;:=2

x intercept (let y =0)

- ( ~ )lo g Q (7x+ 11) = 0

= = > log,(7x+11)= 0

(Index law) eO = 7x + 11 => x=-~

K intercept (let y = 0)

0= 3Iog,(4- 3x)+ 1

- ~ = log,(4- 3x)

_ 1 . ( _ 1 )e 3=4-3x =>x=~ 4-e:1



TRANSFORMATION OF A LOGARITHMIC GRAPH

General equation ofa logarithmic graph revisited: I y = aloge(nx -b)+cl

• Effect of the constant "a": Stretching the graph by "a" times away from the X axis (dilate vertically)

• Effect of the constant "n": Stretching the graph by "~ " times envoy from the Y axis (dilate horizontally)

• Effect of the constants "band e": translating (moving) the function vertically and horizontally. "b" translates the

Junction in the horizontal direction and "c" translates the function in the vertical direction.

• Effect on reflection:- A -sign in front of the function (i . e. infront oj "a 'j will reflect the whole/unction about the X axis (upside down).

- A -sign in front of the term containing x i. e (nx) will reflect the function about the Yaxis (flip horizontally).

TRANSFORMA nON ORDER REVISTED

• When transforming, perform DILATION or REFLECTION first then finally TRANSLATION.

Eg. Describe the transformation for each oftbe following from the basic graph of y = loge (x)

a. y=210ge{x-2)+3 b. y=-(t)log.(2x-5) c. y=loge(2-7x)-5

SOLUTION

a. y =210ge(x -2)-+-3-

• Dilate by ( x 2 ) away from x axis

• Translate by 2 units to the right and 3 units up

c. y=loge(2-7x)-5

• Dilate by ( x ' ; ) = ( x ~ ) away from y axis

• Reflect about the Y axis

(due to the -ve sign in front of7x)

• Shift to the right by t units

(let 2 - 7x =0=> x = t) and 5 units down.

b. y = -(t)log~(2x-5)

• Dilate by ~ away from x axis

• Dilate by ( x - ; ) = ( x ~ ) away from y axis

• Reflect about the x axis then shift to the right by ~ units (let 2x - 5 = 0 => x = f )

INVERSE OF LOGARITHl\1IC GRAPHS

Eg..Find the equation of the inverse and state its domain and range for each:

a. !:x>-±--?Nwhere f(x) ",,]oge(2x +1) b. f:x<2--?R where f(x)=-t1oge(2-x)

SOLUTION a. Sketch the graph, its range would be y ER .,To find inverse, swap x and y :

X = loge(2)(1 + 1 ) => x = 10ge(2y-1 + 1) => e X = 2y-l + 1 => I y - I =ylwhere y-l = inverse function.

X Y

Swap domain and range: x> - t becomes y -I > -t for inverse, y ER becomes x ER for inverse

Thus inverse function is y-I = e X ; 1with x ERand y-l > - t ob. Sketch the graph, its range would be y E R . To find inverse, swap x and y:

X=-t1oge(2-X) =>x=-t1oge(2-y-l) =>-2x=loge(2-y-J) => e-2x=2_y-l => ly-J=2_e-2X Ix y-l

Swap domain and range: x < 2 becomes y -1 < 2 for inverse, y ER becomes x E R for inverse

Thus inverse function is y -1 = 2 - e -2.:< with x ERand y -] < 2,



Exercises Year 12 Mathematical Methods

UNITS 3 AND,4 MATHS METHODS (CAS) - SET 5

1. Evaluate: a. log1 O

1000 c. log5 ( l i s )

2. Evaluate: a, log, (t) + log, 12

.3 . Evaluate:

avlog, 81 -log9 27

b l og227

. Jog23

d. 510gl0 i f lO

b. log31 08 + log)l 00 - 210g3 20 C. 310g5 2 + log, 10 - 210g5 20

4 ..Given that loglo 2 = O J ; Joglo 5 = 0 ,7 ; find:

3. Joglo 25 b. loglO 20 C. loglo 4 0

5. Convert the following expression to JogIO:

a. log311 b. log920

6, Find the value of x for each:

d. log225

a. Jog, ( 7 ) = x

7. Solve for x:

a. log, x = log; 6 + log, 2

b. log2(2x-l)=3 c.. log e (2 x) co 0.5 d. log e (3 x -1) = 0,5

b. log, ( 5x + 2) -log3 ( X - 2) = 2 C. 210gl0 X = 210g l0 2 + 10glO(X+ 1 )

8. If loglo A= Tx + log] 0 Z , make Z the subject.

9. Find the values ofx for which:

2 3a. (1og10 x) -log10 X + 2:::: 0

10. State all transformation required for each equation from the basic graph of y = loge (x) .

a. y =t loge (x ) b. y = loge (2x) c. y = -210ge (x -1) d. y = loge (3- x)

11. Without the calculator, sketch the graph, find x and y intercepts (if any), and state domain and range for each of

the following functions:

a. fCx)=loge(x-3) b. f(x)=-loge(2x-l) c. f(x)=-ioge(3-4x)+1

12. Find the equation of the inverse and state its domain and range for each:

a. j:x>-2----';N where j(x)=tloge(2+x)-6 b.. j:x<1----';R where fex)=-loge(3-2x)

Answers: La. 3- Ib. 2 " c.-3 2.a. 3 b. 3 c.] d · t 3.a. t b.3 c.-I 4.a. 1.4 b.1.3 c.1.6

5

. lo glO I I

.a. JaglD 3 6.3. X =-3 b. x=1 c. x = l .eO.

52

eO S + 1

d. x=-37.a. 12 h. 5 c.2+2Ji 8. Z =A x 1 O ~Tx 9.3. X = 10, x =100

Ib. x = 3, X=3~4

10.a. Dilate by ~ away from X axis b . Dilate by ~ away from Y axis

c. Dilate by 2 away from X axis; then a reflection about X axis; followed by a translation of 1 unit to the right.

d ..A reflection about the Y axis followed by a translation of 3 unit to the right.

11 . a. Y ):=)I b. Y x intercept: x = 1

domain: x > trange: y ER

C 4 Y 4 . : . = ~" ' , 1

I

domain: x <. ~

range y E R

X Inter = e'- 3-4

Y Inter = -109a(3) + 1

x intercept: x = 4

domain: x>. 3

range: 1 E R

12.2. Y-I= e 3{x+6) - 2; x E R, y-I >-2 Y-l <1 xER2 '

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