Exponentials and Logs€¦ · Exponentials and Logs Paper 1 Section A Each correct answer in this...

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Exponentials and Logs

Paper 1 Section A

Each correct answer in this section is worth two marks.1. Simplify log4 8 + log4 2 − 3 log5 5.

A. − 12

B. −1

C. log4

(

165

)

D. log4

(

16125

)

Key Outcome Grade Facility Disc. Calculator Content SourceB 3.3 A/B 0.53 0.33 NC A28 HSN 17


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Questions marked ‘[SQA]’ c© SQAAll others c© Higher Still Notes

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2. Solve logb x − logb 7 = logb 3 for x > 0.

A. x = 21

B. x = 10

C. x =73

D. x =37

Key Outcome Grade Facility Disc. Calculator Content SourceA 3.3 A/B 0.64 0.59 CN A28, A32 HSN 175


3. Solve loga 5 + loga x = loga 20 for x > 0.

A. x =14

B. x = 4

C. x = 15

D. x = 100

Key Outcome Grade Facility Disc. Calculator Content SourceB 3.3 A/B 0.7 0.45 CN A28, A32 HSN 111


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4. The diagram shows the graph of y = 3ekx .PSfrag replacements

O x

y

3

y = 3ekx

(4, 18)

What is the value of k?

A. 32e

B. 14 loge 6

C. 14 loge 15

D. 118 loge

43

Key Outcome Grade Facility Disc. Calculator Content SourceB 3.3 A/B 0.43 0.31 NC A30 HSN 095


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5. Solve 3 loga 2 = 12 for a .

A. a = 64

B. a = 36

C. a = 49

D. a = 116

Key Outcome Grade Facility Disc. Calculator Content SourceA 3.3 A/B 0.56 0.77 CN A31 HSN 112


[END OF PAPER 1 SECTION A]

Paper 1 Section B6.[SQA] Evaluate log5 2 + log5 50 − log5 4. 3

Part Marks Level Calc. Content Answer U3 OC32 C NC A28 2 2000 P1 Q91 A/B NC A28

•1 pd: use loga x + loga y = loga xy•2 pd: use loga x − loga y = loga

xy

•3 pd: use loga a = 1

•1 log5 100 − log5 4•2 log5 25•3 2

7. Evaluate log4 16 + 2 log3(3√

3) . 3

Part Marks Level Calc. Content Answer U3 OC33 A NC A28 5 OB 11-004

•1 ss: use laws of logs•2 ss: use laws of logs•3 pd: complete

•1 log4 16 = 2•2 log3(9 × 3)

•3 log4 16 + 2 log3(3√

3) = 5

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8. Evaluate log3 18 + log3 6 − log3 4 + log5√

5. 4

Part Marks Level Calc. Content Answer U3 OC32 B CN A28 Ex 3-3-22 C CN A28 7

2

•1 pd: use loga x + loga y = loga xy•2 pd: use loga x − loga y = loga

xy

•3 pd: use loga xk = k loga x•4 pd: use loga a = 1

•1 log3(18 × 6)•2 log3(

18×64 )

•3 log3 33 + log5 5 12 = 3 log3 3 + 1

2 log5 5•4 3 + 1

2 = 72

9. The expression 4 loga(2a) − 3 loga 2 can be written in the form 2n + loga n , wheren is a whole number.

Find the value of n . 3

Part Marks Level Calc. Content Answer U3 OC33 B CN A28 n = 2 Ex 3-3-4

•1 pd: use log laws•2 pd: process•3 ic: state n

•1 4(loga 2 + loga a) − 3 loga 2•2 4 + 4 loga 2 − 3 loga 2 = 4 + loga 2•3 n = 2

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10. The graph below shows the curve with equation y = log8 x .PSfrag replacements

O x

y

(8, 1)

1

y = log8 x

Sketch the graph of y = log8

(

1x2

)

. 3

Part Marks Level Calc. Content Answer U3 OC33 A CN A28, A3 sketch OB 11-003

•1 ss: use laws of logs•2 ic: know to reflect and scale•3 ic: annotate sketch

•1 y = −2 log8 x•2 reflect in x-axis and scale•3 show (1, 0) and (8,−2)

PSfrag replacements

O x

y

(8,−2)1

y = log8( 1

x2

)

11. The diagram shows the curve with equation y = log2 x .

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O x

y

1

y = log2 x

Sketch the curve y = log2( 2

x)

. 4

Part Marks Level Calc. Content Answer U3 OC34 A CN A29, A28 sketch AT064

•1 ss: use law of logs•2 ic: interpret reflection•3 ic: interpret translation•4 ic: annotate sketch

•1 y = log2 2 − log2 x•2 reflect curve in y-axis•3 shift 1 unit up•4 decreasing log curve thro’ (1, 1)

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12. Sketch and annotate the curve with equation y = loge(x + 1)2 . 4

Part Marks Level Calc. Content Answer U3 OC34 A CN A29, A28 sketch WCHS U3 Q10

•1 ss: use log law•2 ic: interpret translation•3 ic: interpret scaling•4 ic: sketch with points annotated

•1 y = 2 loge(x + 1)•2 y = loge x translated 1 unit left•3 then made twice as tall•4 sketch, with points (0, 0) and

(e − 1, 2)

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Higher Mathematics


13. The diagram below shows the graph of the function f (x) = ex .

PSfrag replacements

O x

y

P

Q

y = f (x)

The points P and Q have x -coordinates −1 and 1 respectively.

Straight lines are drawn from the origin to P and Q as shown.

(a) Show that OP and OQ are perpendicular. 3

(b) Show that the area of triangle OPQ is 1 + e2

2e . 3

(c) The function g is defined by g(x) = f (x − 2) + 1.(i) Sketch the curve with equation y = g(x) .

(ii) The graphs of y = f (x) and y = g(x) intersect when x = a .

Show that ea =e2

e2 − 1 .

Hence express a in the form A + B loge(e − 1) + C loge(e + 1) , stating thevalues of A , B and C . 9

Part Marks Level Calc. Content Answer U3 OC3(a) 3 B CN G5, G2 proof AT040(b) 3 B CN G1 proof(ci) 3 C CN A3 sketch(cii) 6 A CN A30, A28 A = 2, B = C = −1

•1 ic: obtain coordinates of P and Q•2 pd: find gradients•3 ss: use m × m⊥ = −1

•4 pd: compute side length•5 pd: compute side length•6 ss: use area formula and complete

•7 ss: translate parallel to x-axis•8 ss: translate parallel to y-axis•9 ic: annotations•10 ss: form equation•11 ic: complete•12 ss: know to take logs•13 pd: use laws of logs•14 pd: use laws of logs•15 ic: state A, B, C

•1 P(

−1, 1e)

, Q(1, e)•2 mOP = − 1

e , mOQ = e•3 mOP × mOQ = −1 so OP ⊥ OQ

•4 OP =√

1 + 1/e2

•5 OQ =√

e2 + 1•6 Area = 1

2 × OP × OQ = (1 + e2)/2e

•7 shift 2 units to right•8 shift 1 unit up•9 P′(1, 1/e + 1), Q′(3, e + 1)•10 f (x) = g(x)•11 ex = e2/(e2 − 1)•12 x = loge

(

e2/(e2 − 1))

•13 · · · = loge e2 − loge(e2 − 1)•14 · · · = 2 − loge(e + 1) − loge(e − 1)•15 A = 2, B = C = −1

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14.[SQA] Given x = log5 3 + log5 4, find algebraically the value of x . 4


15.[SQA]



16.[SQA] Find x if 4 logx 6 − 2 logx 4 = 1. 3

Part Marks Level Calc. Content Answer U3 OC33 C NC A32, A28, A31 x = 81 2001 P1 Q8

•1 pd: use log-to-index rule•2 pd: use log-to-division rule•3 ic: interpret base for logx a = 1 and

simplify

•1 logx 64 − logx 42

•2 logx64

42

•3 all processing leading to x = 81

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17. Solve loga 12 + loga x − 2 loga 2 = 6 for x in terms of a , where a > 1. 4

Part Marks Level Calc. Content Answer U3 OC34 B CN A32, A28, A31 a = 1

3 a6 WCHS U3 Q6

•1 ss: use log law•2 ss: use log law•3 ss: know to convert log to

exponential•4 pd: complete

•1 loga 12 + loga x − loga 22

•2 loga(12x

4 )•3 3x = a6

•4 a = 13 a6

18. Solve log8 x − log6 4 = log6 9 for x > 0. 3

Part Marks Level Calc. Content Answer U3 OC33 C CN A32, A28, A31 x = 64 Ex 3-3-6

•1 ss: use laws of logs•2 pd: use laws of logs•3 pd: use laws of logs

•1 log8 x = log6 9 + log6 4 = log6 36•2 log8 x = 2•3 x = 82 = 64

19.[SQA] The graph illustrates the law y = kxn .If the straight line passes throughA(0·5, 0) and B(0, 1) , find the values ofk and n . 4

PSfrag replacements

O

xy

A(0·5, 0)

B(0, 1)

log5 y

log5 x

Part Marks Level Calc. Content Answer U3 OC34 A/B NC A33 y = 5x−2 2002 P1 Q11

•1 ic: interpret graph•2 ss: use log laws•3 ss: use log laws•4 pd: solve log equation

•1 log5 y = −2(log5 x) + 1•2 log5 y = log5 x−2 + . . .•3 . . . + log5 5•4 y = 5x−2

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20. The graph below shows a straight line in the (log4 x, log4 y)-plane.

PSfrag replacements

O

xy

(7, 10)

3log4 x

log4 y

Find the equation of the line in the form y = kxn , where k, n ∈ R . 5

Part Marks Level Calc. Content Answer U3 OC35 A NC A33, G3, A28 y = 64x OB 11-002

•1 ic: interpret graph (gradient)•2 ic: interpret graph (complete eqn)•3 ss: use log laws•4 ss: use log laws•5 ic: complete

•1 gradient = 1•2 log4 y = log4 x + 3•3 log4 y = log4 k + n log4 x•4 log4 k = 3 ⇒ k = 43

•5 y = 64x

[END OF PAPER 1 SECTION B]

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Paper 2

1. Solve log2 x2 − 3 log2 x = 4 − log2 7 for x > 0. 3

Part Marks Level Calc. Content Answer U3 OC33 A CN A28, A31 x = 7

16 OB 11-001

•1 ss: use log law•2 ss: use log law•3 ss: know to convert log to

exponential and complete

•1 log2 x2 = 2 log2 x•2 log2 7 − log2 x = log2(

7x )

•3 7x = 24, so x = 7

16

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2. The graph of the function f (x) = loge x is shown in the diagram below.

PSfrag replacements

O x

y

A(1, 0)

B(e2, b)

y = f (x)

The point B has coordinates (e2, b) .

(a) Write down the value of b . 1

(b) The function g is defined by g(x) = − f (x − 2) . Sketch the graph of y = g(x) . 3

(c) The graphs of y = f (x) and y = g(x) intersect at C. The x -coordinate of C isof the form x = m +

√n .

Determine the values of m and n . 6

Part Marks Level Calc. Content Answer U3 OC3(a) 1 C CN A2 2 AT010(b) 3 C CN A29 sketch(c) 6 A CN A32, A34 m = 1, n = 2

•1 ic: interpret graph

•2 ic: reflection•3 ic: horizontal translation•4 ic: annotate sketch

•5 pd: expression for g(x)•6 ss: equate•7 ss: use log law•8 ss: convert from log•9 pd: solve quadratic equation•10 ic: interpret solution

•1 b = 2

•2 reflect in x-axis•3 shift 2 units to right•4 show A′(3, 0) and B′(e2 + 2,−2)

•5 g(x) = − log2(x − 2)•6 loge x = − loge(x − 2)•7 loge x = loge(x − 2)−1

•8 x = 1/(x − 2)•9 x = 1 ±

√2

•10 m = 1, n = 2

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3.[SQA]PSfrag replacements

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4.[SQA]



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5.[SQA]



6.[SQA] Before a forest fire was brought under control, the spread of the fire was describedby a law of the form A = A0ekt where A0 is the area covered by the fire when itwas first detected and A is the area covered by the fire t hours later.

If it takes one and a half hours for the area of the forest fire to double, find thevalue of the constant k . 3

Part Marks Level Calc. Content Answer U3 OC33 A/B CR A30 k = 0·46 2001 P2 Q9

•1 ic: form exponential equation•2 ss: express exp. equ. as log

equation•3 pd: solve log equation

•1 2A0 = A0ek×1·5

•2 e.g. 1·5k = ln 2•3 k = 0·46

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7. A population of bacteria is growing in such a way that the number of bacteria Npresent after t minutes is given by the formula N(t) = 32e0·01225t .

(a) State N0 , the number of bacteria present when t = 0. 1

(b) The “e -folding” time, l minutes, is the length of time until N(l) = eN0 .Find the e -folding time for this population correct to 3 decimal places. 3

Part Marks Level Calc. Content Answer U3 OC3(a) 1 C CN A6 N0 = 32 Ex 3-3-3(b) 3 B CR A30 l = 81·633 (3 d.p.)

•1 ic: interpret formula

•2 ic: interpret N(l)•3 ss: form equation•4 pd: solve

•1 N0 = N(0) = 32

•2 N(l) = eN0 = 32e•3 32e0·01225l = 32e•4 0·01225l = 1 ⇒ l = 81·633 (3 d.p.)

8.[SQA]



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9.[SQA]




Oxy


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11.[SQA]



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12.[SQA]



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13.[SQA]



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14.[SQA]



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15.[SQA]



16.[SQA] Find the x -coordinate of the point where the graph of the curve with equationy = log3(x − 2) + 1 intersects the x -axis. 3

Part Marks Level Calc. Content Answer U3 OC32 C CN A31 2002 P2 Q71 A/B CN A32 x = 2 1

3

•1 ss: know to isolate log term•2 pd: express log equation as exp. equ.•3 pd: process

•1 log3(x − 2) = −1•2 x − 2 = 3−1

•3 x = 2 13

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17.[SQA]



18.[SQA]



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19. A sequence is defined by the recurrence relation un+1 = 12 un + 3 with u0 = log3 4.

(a) Show that u1 = log3 54. 4

(b) Find an expression for u2 in the form log3 a . 3

(c) Find the value of u2 correct to two decimal places. 4

Part Marks Level Calc. Content Answer U3 OC3(a) 4 A CN A11, A28 proof AT051(b) 3 A CN A11, A28 log3(81

√6)

(c) 4 A CR A31, A28 4·82 to 2 d.p.

•1 ic: find u1•2 ss: use law of logs•3 ss: convert constant to log•4 ss: use law of logs

•5 ic: find u2•6 ss: use law of logs•7 pd: complete

•8 ss: know to change base•9 ss: use law of logs•10 pd: process•11 ic: state value

•1 u1 = 12 log3 4 + 3

•2 · · · = log3(41/2) + 3•3 · · · = log3 2 + log3 33

•4 · · · = log3 54

•5 u2 = 12 log3 54 + 3

•6 · · · = log3(541/2) + log3 33

•7 · · · = log3(27√

54) (= 81√

6)

•8 3u2 = 81√

6.•9 u2 loge 3 = loge 81

√6

•10 u2 = (loge 81√

6)/(loge 3)•11 · · · = 4·82 to 2 d.p.

20.[SQA]



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21.[SQA]



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22.[SQA] The results of an experiment give rise to the graph shown.

(a) Write down the equation of the line interms of P and Q . 2

PSfrag replacements

O

xy

P

Q

1·8

−3

It is given that P = loge p and Q = loge q .

(b) Show that p and q satisfy a relationship of the form p = aqb , stating thevalues of a and b . 4

Part Marks Level Calc. Content Answer U3 OC3(a) 2 A/B CR G3 P = 0·6Q + 1·8 2000 P2 Q11(b) 4 A/B CR A33 a = 6·05, b = 0·6

•1 ic: interpret gradient•2 ic: state equ. of line

•3 ic: interpret straight line•4 ss: know how to deal with x of

x log y•5 ss: know how to express number as

log•6 ic: interpret sum of two logs

•1 m =1·83 = 0·6

•2 P = 0·6Q + 1·8

Method 1

•3 loge p = 0·6 loge q + 1·8•4 loge q0·6

•5 loge 6·05•6 p = 6·05q0·6

Method 2ln p = ln aqb

•3 ln p = ln a + b ln q•4 ln p = 0·6 ln q + 1·8 stated or implied

by •5 or •6

•5 ln a = 1·8•6 a = 6·05, b = 0·6

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23.[SQA]



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24. The results of an experiment were noted as follows.

x 1·70 2·10 2·50 2·90log10 y 2·14 1·96 1·75 1·53

The relationship between these data can be written in the form y = abx where aand b are constants.

Find the values of a and b and hence state a formula relating the data. 6

Part Marks Level Calc. Content Answer U3 OC36 A CR A33, A28 y = 1023·29 × (0·31)x WCHS U3 Q13

•1 ss: know to take logs•2 pd: use laws of logs•3 ic: interpret equation•4 pd: find gradient•5 pd: start to find other constant•6 ic: complete, and state equation

•1 log10 y = log10(ax)•2 log10 y = (log10 b)x + log10 a•3 gradient of line is log10 b•4 m =

1·53 − 2·142·90 − 1·70 = −0·51 (2 d.p.),

so b = 10−0·51 = 0·31 (2 d.p.)•5 (1·70, 2·14) : 2·14 = −0·51× 1·70 + log10 a•6 a = 1023·29 (2 d.p.)

so y = 1023·29 × (0·31)x

25.[SQA]



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26.[SQA]



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Oxy


[END OF PAPER 2]

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Exponentials and Logs€¦ · Exponentials and Logs Paper 1 Section A Each correct answer in this...

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