SolutionExercise 1.4

8/19/2019 SolutionExercise 1.4

1/22

1

Foundations of Mathematics

Exercise 1.4

1.

Simplify the following:

(a)2

1

2

aa (b)642

aaa (c) 4323 534 abba (d) 5323 642 cbcaba

(e) 726 33 pqq p (f) 22235

2

2

1

4

3

3

2qaaq paap

Solution:

(a) 21

2

aa =

2

12

a

=

2

3

a

(b) 642 aaa = 642 a = 12a

(c) 4323 534 abba = 3243534 ba = 5760 ba

(d) 5323 642 cbcaba = 531213642 cba = 83448 cba

(e) 726 33 pqq p = 1313726 q p = 4484 q p

(f) 2223

5

2

2

1

4

3

3

2qaaq paap = 12132121

5

2

2

1

4

3

3

2 q pa = 34610

1q pa

2.

Simplify the following:

(a) 53)( x (b) 3)3( x (c) 432 )( ba

(d) 202 )4( ba

(e) 332 )( y x (f) 22 )(

1 x

(g) 610 321 p p (h) 372 515 abba (i) abccba 714 23

(j) 21

aa (k)10293 )(4)(20 x x (l) 432 3)(15 abba

(m) 2432 )(218 abcba (n) 342375 )(629 z y x yz x xy (o) 21

2

2

2

1

)(1

a

aa

(p)6 5

3 274

x

x x (q) 3 y y y (r)

ab

ba4 84


2/22

2

Solution:

(a) 53 )( x = 5)3( x = 15 x =

15

1

x

(b) 3)3( x = 33 )()3( x = 327 x

(c) 432 )( ba = 4342 )()( ba = 4342 ba = 128

ba

(d) 202 )4( ba = 22 )4( a = 222 )()4( a = 416 a =4

16

a

(e) 332 )( y x = 33323 )()()1( y x = 96 y x

(f)22

)(

1

x

=

4

1 x

= 4 x

(g) 610 321 p p = 6103

21 p

= 47 p

(h) 372 515 abba = 37125

15 ba

=

43ab

(i) abccba 714 23 = 1112137

14 cba

= ba22

(j) 21

aa = 21

2

1

aa =

2

1

2

1

a

= a

(k) 10293 )(4)(20 x x = 2027 420 x x = 20274

20 x = 75 x

(l) 432 3)(15 abba = 436 315 abba = 43163

51 ba

= 155 ba =b

a55

(m) 2432 )(218 abcba = 22432 218 bacba = 42322218 cba = 45436 cba

(n) 342375 )(629 z y x yz x xy = 3126

375

6

29

z y x

yz x xy

=

331215671

6

29 z y x

=

623 y x =6

23

y

x


3/22

3

(o) 21

2

2

2

1

)(1

a

aa =

21

2

2

2

11

aa

a

=

1)2(2

1

a = 21

3

a =

2

13

1

a

(p)6 5

3 274

x

x x = 6

5

3

1

2

1

32

x x x = 65

3

1

2

1

6

x = 06 x = 6

(q) 3 y y y = 31

2

1

y y y = 31

2

11

y = 611

y

(r)ab

ba4 84 =

21

4

184

ab

ba =

21

21

4

8

4

4

ba

ba = 2

1

4

8

2

1

4

4

ba = 23

2

1

ba

3. Evaluate:

(a) 210 (b) 3)7( (c)2

6

5

(d)

5 3243

(e) 3

2

278

(f)

2

1

416

(g) 32

8

(h)2

1

2516

Solution:

(a) 210 = 210

1

=

100

1

(b) 3)7( = 3)7(

1

=

337)1(

1

=

343

1

(c)

2

6

5

=

2-

-2

6

5=

2

2

5

6=

25

36

=

25

111

(d)5 3243 = 5

3

243 = 53

5 )3(

=

5

35

3 = 33 = 27

(e) 32

278

=

3

2

278

=

3

2

3

3

32

= 3

23

32

=

2

32

= 2

2

32

= 94


4/22

4

(f)2

1

4

16

=

2

1

4

25

=

2

1

2

2

2

5

=

2

12

2

5

=

2

5

(g) 3

2

8

= 3

2

3 )2(

=

3

23

2 = 22 = 2

2

1 = 4

1

(h)2

1

25

16

=

2

1

2

2

5

4

=

21

2

5

4

=

1

5

4

=4

5

4.

Simplify the following and express with positive indices.

(a)

2

21

2

3

2

1

ba

ba (b)

1

42

23

ba

ba (c)

ba

ba

22 (d)

1

11

)(

ba

ba

Solution:

(a)

2

21

2

3

2

1

ba

ba =

22

2

3)1(

2

1

ba =

2

2

1

2

1

ba =

22

12

2

1

ba = 1ab =

b

a

(b)

1

42

23

ba

ba

=

142)2(3][ ba = 165 )( ba =

)1()6()1(5 ba = 65ba = 5

6

a

b

(c)ba

ba

22=

ba

ba

22

11

=ba

ba

ab

22

22

=))(( 22

22

baba

ab

=

))((

))((22 baba

abab

=

))((

))((22 baba

baba

=

22

)(

ba

ba

[ ∵ ))((22 bababa ]

(d)1

11

)(

ba

ba

= ))(( 11 baba =

baba

11)(

=

ab

abba )(

=

ab

abba ))((

=

ab

baba ))(( =

ab

ba 2)(


5/22

5

5. Simplify the following and express with positive indices.

(a)1

10

2

)22(

n (b)

3

12

2

22

n

nn

(c) nnn 3)33( 11 (d) nnn 273)3( 121

(e)11

11

323323

nn

nn (f)

n

nn

5275355 12

(g)11

3

)1(

1

)2(

4

2

2

nn

n

nn

n

(h) nn

n

a

a

1

12

Solution:

(a)1

10

2

)22(

n

=

1)12(2 n = 1)1(2 n = 2

(b)3

12

2

22

n

nn

=

3

1

2

)12(2

n

n

=

3

1

2

2

n

n

=

)3()1(2 nn

=

312 nn

=

22

=

22

1

=

4

1

(c) nnn 3)33( 11 = nn 3)13(3 21 = nn 338 1 = nn 138 = 138 = 3

8

(d) nnn 273)3( 121 = nnn )3(33 3122 = nnn 3122 33 = nn 313 33 = nn 3133 = 13 = 3

1

(e)11

11

323

323

nn

nn

=

)23(3

)23(321

21

n

n

=

)11(3

)7(31

1

n

n

=

11

7

(f)n

nn

527

5355 12

=

n

nn

527

5575 12

=

n

nn

527

575 2

=

n

n

527

)75(5 2

=

27

18

=

3

2

(g)11

3

)1(

1

)2(

4

2

2

nn

n

nn

n

=

)1)(1(

321

2

)2(

2

22

nn

n

nn

n

=

1

621

22

2

2

2

2

n

n

nn

n

=

62

11

2

2

2

22

2

n

n

nn

n

=

)62(1)(1 222 nnnnn

=

6211 222 nnnnn = 62

= 64

(h) nn

n

a

a

1

12

=

n nna )1()12(

=

n nna 112

=

n na 3

=

nna

1

3 )(

=

nna

13

=

3a


6/22

6

6. Solve the following equations for x.

(a) 31

8 x = 32

4 (b) )4(2

2 x

=

28 x (c) 129 x = )81(27 213 x

(d) 11 55 x x = 120 (e) 11 222 x x x = 40 (f) 9)3(1032 x x = 0

(g) 2)2(34 x x = 0 (h) 1)6(76 12 x x = 0 (i) x x 33 = 2

Solution:

(a) 31

8 x = 32

4

3

1

x = 8

432

3

1

x = 3

32

2

2

)2(

3

1

x = 3

3

4

2

2

3

1

x = 3

3

4

2

3

1

x = 35

2

3

3

1

x =

3

3

5

2

x =

52

x =

32

1

(b) )4(2 2 x = 28 x

])2[(2 22 x = 23 )2( x

)2(2 42 x = 632 x

)42(12 x

=

632 x 322 x

=

632 x

∴ 2 x – 3 = 3 x – 6

x = 3


7/22

7

(c) 129 x = )81(27 213 x

122 )3( x = ])3[()3( 21433 x

243 x = )3(3 849 x

243 x = x849

3 2x4

3 = x813

3

∴ 4 x – 2 = 13 – 8 x

12 x = 15

x = 12

15

x = 4

5

(d) 11 55 x x = 120

)15(5 21 x = 120

)24(5 1 x = 120

15 x = 515 x = 15

∴ x – 1 = 1

x = 2

(e) 11 222 x x x

= 40

)122(2 21 x = 40

)5(2 1 x = 40

12 x

= 812

x

=

32

x – 1 = 3

x = 4

(f) 9)3(1032 x x = 0

9)3(10)3( 2 x x = 0

)13)(93( x x = 0

93 x = 0 or 13 x = 0 x3 = 9 or x3 = 1 x3 = 23 or

x3 = 03

∴ x = 2

or

x = 0


8/22

8

(g) 2)2(34 x x = 0

2)2(3)2( 2 x x = 0

2)2(322 x x = 0

2)2(3)2( 2 x x = 0

)22)(12( x x = 0

)12( x = 0 or )22( x = 0

x2

= 1 or x2

= 2 x2

=

02 or x2

=

12

∴ x = 0 or x = 1

(h) 1)6(76 12 x x = 0

1)6(7)6(6 2 x x = 0

1)6(7)6(6 2 x x = 0

]1)6(6)[16( x x = 0

16 x = 0 or 1)6(6 x = 0

x6 = 1 or x6 = 6

1

x6 = 06 or 6 x = 16

∴ x = 0 or x = – 1

(i) x x 33 = 2

)13(3 2 x x = 2

132 x = )3(2 x

1)3(232 x x = 0

2)13( x = 0 ( ∵ 222 2)( bababa )

13 x = 0

x3 = 1 x

3 = 0

3

∴ x = 0


9/22

9

7. Rewrite the following expressions in logarithmic forms

(a) 24 = 16 (b) 34 = 81 (c) 4-3 =64

1

(d) 5-2 x =25

1 (e) 7 x = 1 (f) x2

1

9

Solution:

(a) 24 = 16 4 = log 2 16

(b) 34 = 81 4 = log 3 81

(c) 4-3 =64

1 – 3 = log 4

64

1

(d) 5-2 x =25

1 – 2 x = log 5

25

1

(e) 7 x = 1 x = log 7 1

(f) x2

1

9 2

1 = log 9 x

8. Rewrite the following expressions in exponential forms

(a) log3 243 = 5 (b) log 100 = 2 (c) log 0.2 125 = -3

(d) y = log2 128 (e) y = log3 243 (f) log 2 y2

1

3

4

Solution:

(a)

log3 243 = 5 24335

(b) log 100 = 2 100102

(c) log 0.2 125 = -3 1252.0 3

(d) y = log2 128 1282 y

(e) y = log3 243 2433 y

(f)

log 2 y2

1

3

4

3

42 2

1

y


10/22

10

9. Find the values of the following correct to 3 significant figures

(a) log 4.79 (b) (log 8) (log 9) – log (8 9) (c)

3

7log

3log

7log

(d)

30log

60log

2

4

(e)

(log3 4) (log4 5) (log5 6) (f)

log 2 38 –

log 3 20

1

(g)

10 2 log 7 (h)

log (27 + 43) – (log 27 + log 43)

Solution:

(a) log 4.79 = 0.680

(b) (log 8) (log 9) – log (8 9) = – 0.996

(c)

3

7log

3log

7log = 1.40

(d)

30log

60log

2

4 =

2log30log

4log60log

=30log

2log

4log

60log = 0.602

(e)

(log3 4) (log4 5) (log5 6) = 1.63

(f) log 2 38 – log 320

1 = 7.97

(g) 10 2 log 7 = 49

(h) log (27 + 43) – (log 27 + log 43) = – 1.22


11/22

11

10. Evaluate (a) x = 400log20 (b) x2log = 3

(c) )4(log50 x = 1 (d) 04.0log x = – 2

Solution:

(a) x = 400log20

400 =

x20 ( ∵ y x alog is equivalent to xa y )

220 = x20

∴ x = 2

(b) x2log = 3

x =

32

∴ x = 8

(c) )4(log50 x = 1

x – 4 = 150

x = 54

(d) 04.0log x = – 2

0.04 =

2 x

0.04 =

2

1

x

2 x =

04.0

1

2 x = 25

x = 5

or

–

5

(Rejected, as x is the base and so must be greater than 0)∴ x = 5


12/22

12

11. Simplify the following.

(a) 2log3log7log 555 (b) )4log(log)3log( x x x (c) 75log500log 22

(d) x x log)5log( 3 (e) 6log3log20log 777 (f) log 2 + log(3 x) – log(2 x)

(g) )10log()4log()5log( 22 x x x (h) )3log()3log( 2 x x x (i) )1log()1log( 2 x x

(j) 533 loglog6 x x (k) 25 log7log4 x x (l) log x – 3 log(2 x) + 2 log(4 x)

(m)

)16log(2

1)3log(2 2 x x (n)

x x

1log)4log(

2

3 2 (o)

x x

2log3

2log2

2

Solution:

(a) 2log3log7log 555 = )237(log5 = 42log5

(b) )4log(log)3log( x x x = )]4)()(3log[( x x x = )12log( 3 x

(c) 75log500log 22 =

75

500log 2 =

3

20log 2

(d) x x log)5log( 3 =

x

x35log

= )5log( 2 x

(e) 6log3log20log 777 = )6

320(log7

= 10log7

(f) log 2 + log(3 x) – log(2 x)

=

]2

)3)(2(log[

x

x

=

log 3

(g) )10log()4log()5log( 22 x x x = ]10

)4)(5(log[

2

2

x

x x

= log(2 x)

(h) )3log()3log( 2 x x x =

3

3log

2

x

x x

=

3

)3(log

x

x x

= log x

(i)

)1log()1log( 2

x x =

1

1log

2

x

x =

1

)1)(1(log x

x x = log( x

– 1)


13/22

13

(j) 533 loglog6 x x = 5

3

6

3 loglog x x =

5

6

3log x

x

= x3log

(k)25

log7log4 x x = 7245

)log()log( x x = 1420

loglog x x =

14

20

log x

x =

6

log x

(l) log x – 3 log(2 x) + 2 log(4 x) =

23 )4log()2log(log x x x

= )4log()2log(log

2233 x x x

= )16log()8log(log

23 x x x

=

]8

)16(log[3

2

x x x

= log 2

(m) )16log()3log(2 2

2

1 x x = 2

1

22)16log()3log( x x

= )16log()3log( 21

22

1

22 x x

= )4log()9log( 2 x x

=

x

x

4

9log

2

=

4

9log

x

(n)

x x

1log)4log(

2

3 2

= 123

2 log)4log( x x

= x x log)4log( 23

22

3

=

x x log])2log[( 32

3

2

= x x log)2log( 33

=

)])(8log[(

3

x x

= )8log( 4 x


14/22

14

(o)

x x

2log3

2log2

2

=

32

2

2log

2log

x x

=

3

3

22

2 2

log

2

log x x

=

34

8log

4log

x x

=

34

84log

x x

=

8

4log

3

4

x

x

=

x2

1log

12. Simplify the following.

(a) x

x x nn

log

loglog 1 (b) xy

y x y x

log

)log()log( 223

(c) x

xy y x

log

)log()log( 12 (d)

x

x

16

4log

25

2

Solution:

(a)

x

x x nn

log

loglog 1

=

)log(

)])(log[(

2

1

1

x

x x nn

=

x

x nn

log2

1

)log( 1

=

x

x

log2

1

log

=

2

1

l

= 2

(b) xy

y x y x

log

)log()log( 223

=

])log[(

log

2

1

2

23

xy

y x

y x

=

)log(2

1

)log(

xy

xy

=

2

1

l

= 2


15/22

15

(c) x

xy y x

log

)log()log( 12

=

)log(

)])(log[(

2

1

12

x

xy y x

=

x

y x

log

2

1

]log[ )1(112

=

x

y x

log2

1

)log( 03

=

x

x

log2

1

log 3

=

x

x

log

2

1

log3

=

21

3

= (3)(2)

= 6

(d) x

x

164log

25

2

= 2

125

2164log

x

x

=

x

x

16

4log

2

1 25

2

=

x

x

)2(

)2(log

2

14

252

2

=

x

x

4

410

22

2log2

1

=

)2(log2

1 44102

x x

=

)2(log2

1 462

x

= 2log)46(2

12 x

=

)1)(46(2

1 x

= 3 x + 2


16/22

16

13. Without using calculator, evaluate

(a) 81log3 (b)8log

16log

(c)

9

1log7log15 3

57 (d)

5log6log

125log8log27log

Solution:

(a) 81log3 = 4

3 3log = 3log4 3 = 4

(b)8log

16log

=

3

4

2log

2log

=

2log3

2log4

=

3

4

(c)

9

1log7log15 3

57 =

235

1

73

1log7log15

=

2

37 3log7log5

115

= 3log)2(7log3 37

= )1(2)1(3

= 5

(d)5log6log

125log8log27log

=

5log6log

)125log()8log()27log( 21

2

1

2

1

=

5log6log

)5log()2log()3log( 21

32

1

32

1

3

=

5log6log

)5log(2

3)2log(

2

3)3log(

2

3

=

5log6log

)]5log()2log()3[log(2

3

=

5

6log

5

23log

2

3

=

5

6log

5

6log

2

3

=

2

3


17/22

17

14.

If log 2 = a and log 3 = b, express the following in terms of a and b.

(a) log 12 (b) log 1.2 (c) log 15 (d) log 18

Solution:

(a) log 12 = )322log( = log 2 + log 2 + log 3 = 2 log 2 + log 3 = 2a + b

(b) log 1.2 = )1012log( = log 12 – log 10 = log 12 – 1 = 2a + b – 1

(c) log 15 = )2310log( = log 10 + log 3 – log 2 = 1 + b – a

(d)

18log = )18log( 2

1

= 18log2

1= )32log( 2

2

1

= )3log2(log 2

2

1

= 3log2log 22

1

2

1

= ba 2

1

15. Solve the following equations for x.

(a)

log( x + 2) + log(3 x – 7) = 1 (b) 201

loglog633

x x (c)

2)23(log)5(log 77 x x

(d)

3)1(log2)23(log 22 x x (e) 14)23(log 7

5 x (f) 20)72(log 4

3 x

(g) 2log x = 2)(log x (h) 9)log( log x x (i) 01log2)(log 2 x x

Solution:

(a) log( x + 2) + log(3 x – 7) = 1

log[( x + 2)(3 x – 7)] = 1

)14673log( 2 x x x = 1

)143log( 2 x x = 1

143 2 x x = 10

243 2 x x = 0( x – 3)(3 x + 8) = 0

x – 3 = 0 or 3 x + 8 = 0

x = 3 or x = 3

8

( Rejected as

( x + 2)

and

(3 x –

7)

will be negative and

so log( x + 2) and log(3 x – 7) are undefined )


18/22

18

(b)

633

1loglog

x x = 20

63

1log

x x

= 20

53

1log

x

= 20

5

3log x = 20

x3log5 = 20

x3log = – 4

x = 43 ( ∵ x y alog is equivalent to y

a x )

x =

43

1

x =

81

1

(c) 2)23(log)5(log 77 x x

0)172)(2(

034132

4915132

715132

2)15132(log

2)231015(log

2)]23)(5[(log

2

2

22

2

7

2

7

7

x x

x x

x x

x x

x x

x x x

x x

x + 2 = 0 or 2 x – 17 = 0

x = – 2 or x = 2

17

( Rejected as (5 – x) and (3 – 2 x) will be negative and

so )5(log7 x and )23(log7 x are undefined )


19/22

19

(d) 3)1(log2)23(log 22 x x

0)38)(2(

06198

816823

)12(823

)1(823

8)1(

23

2)1(

23

3)1(

23log

3)1(log)23(log

2

2

2

2

2

3

2

22

2

22

x x

x x

x x x

x x x

x x

x

x

x

x

x

x

x x

x – 2 = 0 or 8 x – 3 = 0

x = 2 or x =

8

3 ( Rejected as ( x – 1) will be negative and so

)1(log2 x is undefined )

(e)

7

5 )23(log x = 14

)23(log7 5 x = 14 ( Remark: As (3 x – 2) must be positive, we can use xn x an

a loglog )

)23(log5 x = 2

3 x – 2 =

25

3 x – 2 = 25

3 x = 27

x = 9

(f) 20)72(log 43 x 204 3)72( x ( Remark: As (2 x + 7) can be negative, we cannot use

72 x 4 203 xn x an

a loglog )

72 x 41

20 )3(

72 x 53 72 x 243

2 x + 7 = 243 or 2 x + 7 = – 243

2 x = 236 or 2 x = – 250

x = 118 or x = – 125


20/22

20

(g) 2log x = 2)(log x

2 log x =

2)(log x

x x log2)(log 2 = 0

(log x)(log x – 2) = 0

log x = 0 or

log x – 2 = 0

x =

010 or log x = 2

x = 1 or x =

210

x = 1 or x = 100

(h) )log( log x x = 9

))(log(log x x = 9 ( ∵ xn x an

a loglog )

2)(log x = 9

∴ log x = 3 or log x = –

3

x = 310 or x = 310

x = 1000 or x =

1000

1

(i)

01log2)(log 2 x x

0)1(log 2 x ( ∵ 222 2)( bababa )

log x + 1 = 0

log x = – 1

x =

110

x =

10

1


21/22

21

16. Solve the following equations for x and give your answer to 2 d.p. if necessary.

(a) 353)10(8

12 x (b) 3115 4 xe (c) 012222 x x

(d) 379ln4 x (e) 1

1ln1

3ln

x

x (f) log(4 x) + 5 log x = 2

Solution:

(a) 353)10(8 12 x

410

32)10(8

12

12

x

x

4log)10log( 12 x (Taking the common log on both sides of the given equation)

(2 x + 1)log 10 = log 4 ( ∵

n

a xlog = xn alog )

2 x + 1 = log 4 ( ∵

log 10 = 1 )

2 x = log 4 – 1

x =

2

14log = –

0.20 (correct to 2 d.p.)

(b)

3115 4

xe

6

305

4

4

x

x

e

e

6ln)ln( 4 xe (Taking the natural log on both sides of the given equation)

(4 – x)ln e = ln 6 ( ∵

n

a xlog = xn alog )

4 – x = ln 6 ( ∵

ln e = 1 )

x = 4 –

ln 6 = 2.21 (correct to 2 d.p.)

(c) 012222 x x

0)42)(32( x x

032 x or 042 x

32 x or 42 x (Rejected as 02 x )

3log)2log( x

x log 2 = log 3

x = 2log

3log = 1.58 (correct to 2 d.p.)


22/22

(d) 379ln4 x

28ln4 x

ln x = 7

x =

7e = 1096.63 (correct to 2 d.p.)

(e) 11

ln13

ln

x

x

11

13

ln

x

x

11

3

1ln

x

11

3

1 e x

3

11 1 e x

3

111

e x

e

e

x 3

31

ee x

33 = 28.95 (correct to 2 d.p.)

(f) log(4 x) + 5 log x = 2

251004

104

2)4log(

2)])(4log[(

2)log()4log(

6

6

26

6

5

5

x x

x

x

x x

x x

x = 1.71 or x = –

1.71

( Rejected as 4 x and x will be negative and

so log(4 x)

and

log x

are undefined )

Documents

SolutionExercise 1.4