LOGARITHMS AND INDICES - TestlabzMath Class IX 2 Question Bank Q.1. (C) By converting to exponential...

31
Math Class IX 1 Question Bank LOGARITHMS AND INDICES Q.1. (A) Convert each of the following to logarithmic form. (i) 25 5 2 = (ii) 27 1 3 3 = - (iii) 4 (64) 3 1 = (iv) 1 6 0 = (v) 0.01 10 2 = - (vi) 4 1 4 1 = - Ans. We know that log b a a x b x = = (i) 25 5 2 = 2 25 log 5 = (ii) 27 1 3 3 = - 3 27 1 log 3 - = (iii) 4 ) 64 ( 3 1 = 3 1 4 log 64 = (iv) 1 6 0 = 0 1 log 6 = (v) 01 . 0 10 2 = - 2 ) 01 . 0 ( log 10 - = (vi) 4 1 4 1 = - 1 4 1 log 4 - = Q.1. (B) Convert each of the following to exponential form. (i) 4 81 log 3 = (ii) 3 2 4 log 8 = (iii) 3 8 1 log 2 - = (iv) 2 (0.01) log 10 - = (v) 1 5 1 log 5 - = (vi) log a 1 = 0 Ans. (i) 4 81 log 3 = 4 3 81 = (ii) 3 2 4 log 8 = 2 3 4 8 = (iii) 3 8 1 log 2 - = 3 1 2 8 - = (iv) 2 ) 01 . 0 ( log 10 - = 2 10 0.01 - = (v) 1 5 1 log 5 - = 1 1 5 5 - = (vi) 0 1 log = a 0 1 a =

Transcript of LOGARITHMS AND INDICES - TestlabzMath Class IX 2 Question Bank Q.1. (C) By converting to exponential...

Page 1: LOGARITHMS AND INDICES - TestlabzMath Class IX 2 Question Bank Q.1. (C) By converting to exponential form find the value of each of the following. (i) log 264 (ii) log 832 (iii) 9

Math Class IX 1 Question Bank

LOGARITHMS AND

INDICES Q.1. (A) Convert each of the following to logarithmic form.

(i) 2552

= (ii) 27

13 3

=− (iii) 4(64)3

1

= (iv) 160

=

(v) 0.01102

=− (vi)

4

14

1=

Ans. We know that logb

aa x b x= ⇒ =

(i) 2552= ∴ 225log5 = (ii)

27

13 3

=− ∴ 3

27

1log3 −=

(iii) 4)64( 3

1

= ∴ 3

14log64 = (iv) 16

0= ∴ 01log6 =

(v) 01.0102

=− ∴ 2)01.0(log10 −= (vi)

4

14 1

=−

∴ 14

1log4 −=

Q.1. (B) Convert each of the following to exponential form.

(i) 481log3 = (ii) 3

24log8 = (iii) 3

8

1log2 −= (iv) 2(0.01)log10 −=

(v) 15

1log5 −=

(vi) loga 1 = 0

Ans.

(i) 481log3 = 43 81∴ = (ii) 3

24log8 =

2

3 48∴ =

(iii) 38

1log2 −= 3 1

28

−∴ = (iv) 2)01.0(log10 −= 210 0.01−

∴ =

(v) 15

1log5 −=

1 1

55

−∴ = (vi) 01log =a 0 1a∴ =

Page 2: LOGARITHMS AND INDICES - TestlabzMath Class IX 2 Question Bank Q.1. (C) By converting to exponential form find the value of each of the following. (i) log 264 (ii) log 832 (iii) 9

Math Class IX 2 Question Bank

Q.1. (C) By converting to exponential form find the value of each of the

following.

(i) 64log2 (ii) 32log8 (iii) 9

1log3 (iv) (16)log0.5

(v) (0.125)log2 (vi) 7log7

Ans. (i) Suppose x=64log2 , then 222222642 ×××××==x ⇒

622 =

x ∴

6=x Hence, 664log2 =

(ii) Suppose ,32log8 x= then 328 =x ⇒ 53

22 =x ∴ 53 =x ⇒

3

5=x

Hence, 3

532log8 =

(iii) Suppose ,9

1log3 x= then 2

23

3

1

9

13

−===

x ∴ 2−=x . Hence,

29

1log3 −=

(iv) Suppose ,)16(log 5.0 x= then 16)5.0( =x ⇒ 2222

2

1×××=

x

⇒ 4

22 =−x

∴ 44 −=⇒=− xx . Hence, 4)16(log 5.0 −=

(v) Suppose ,)125.0(log2 x= then 3

2

1

8

1

1000

125.0125.02 ====

x

⇒ 3

22−

=x

3−=∴ x . Hence, 3)125.0(log2 −=

(vi) Suppose ,7log7 x= then 1777 ==

x 1=∴ x

Hence, 17log7 =

Q.2. Find the value of x, when :

(i) x = −2log 2 (ii) x =log 9 1 (iii) x=9log 243

(iv) x =3log 0 (v) x= −4log 32 4 (vi) x − =3

log ( 1) 2

(vii) x − =2

7log (2 1) 2 (viii) x =

3log 64

2

Page 3: LOGARITHMS AND INDICES - TestlabzMath Class IX 2 Question Bank Q.1. (C) By converting to exponential form find the value of each of the following. (i) log 264 (ii) log 832 (iii) 9

Math Class IX 3 Question Bank

Ans. (i) 2log2 −=x ∴ x=−2

2 ⇒ 4

1

22

1

2

12

==x . Hence, .4

1=x

(ii) 19log =x ∴ 91

=x 9=⇒ x . Hence, 9=x

(iii) x=243log9 ∴ 2439 =x ⇒ 2(3 ) 3 3 3 3 3x

= × × × × ⇒ 5233 =

x

⇒ 52 =x ⇒ 5.22

5==x . Hence, 2.5x =

(iv) 0log3 =x

⇒ 130

=⇒= xx Hence, 1=x )1( 0=x∵

(v) 432log4 −= x

∴ 324 4=

−x ⇒ 542)22( =×

−x ⇒ 5422)2( =

−x ⇒ 58222 =

−x

On comparing both sides 582 =−x

⇒ 2

13132 =⇒= xx . Hence,

13

2x =

(vi) 2)1(log3

=−x

∴ 1)3( 2−= x ⇒

122(3 ) 1 3 1x x= − ⇒ = − ∴ 413 =+=x . Hence, 4x =

(vii) 2)12(log2

7 =−x

∴ 12722

−= x ⇒ 12492

−= x ⇒ 14922

+=x ⇒ 5022

=x

⇒ 252

502==x ⇒ 25 5x = = ±

Hence, 5x = ±

(viii) 2

364log =x

∴ 642

3

=x ⇒ 3/233/2)4()64( ==x 1644

23

23

===

×

Hence, .16=x

Page 4: LOGARITHMS AND INDICES - TestlabzMath Class IX 2 Question Bank Q.1. (C) By converting to exponential form find the value of each of the following. (i) log 264 (ii) log 832 (iii) 9

Math Class IX 4 Question Bank

Q.3. If x p=10log and y q=10log , show that p qxy

+= (10) .

Ans. ∵ px =10log

∴ xp

=10 ...(i)

Similarly, qy =10log

yq

=10 ...(ii)

Multiplying equation (i) by (ii), we get

qpqpxy

+=×= )10(1010 Hence, qp

xy+

= )10(

Q.4. Given x a y b= =10 10log , log ,

(i) Write down a+110 in terms of x. (ii) Write down b2

10 in terms of y.

(iii) If P a b= −10log 2 , Express P in terms of x and y.

Ans. 10 10log , logx a y b= =

∴ xb

=10 and yb

=10

(i) 11101010 ×=

+ aa xx 1010 =×= )10Put ( xa

=

(ii) bbb101010

2×= )10( y

b=

2yyy =×=

(iii) baP −= 2log10 ∴ Pba

=−2

10

⇒ 210 10

a bP = ÷

ba 10)10( 2÷=

yx ÷=2 (Put x

a=10 and )10 y

b=

2x

Py

=

Q.5. Evaluate the following without using log tables :

(i) log42

1log82log5 −+ (ii) log182log3log25log8 −++

(iii) 243

32log

9

52log

16

75log +− (iv) −

3 15log2 + log25 + log49 log28

2 2

(v) 30

1loglog36

2

1log52log2 −−+ (vi) −log (1.2) + 2log(0.75) log(6.75)

Page 5: LOGARITHMS AND INDICES - TestlabzMath Class IX 2 Question Bank Q.1. (C) By converting to exponential form find the value of each of the following. (i) log 264 (ii) log 832 (iii) 9

Math Class IX 5 Question Bank

Ans. (i) 4log2

18log5log2 −+ 2

1

2)4log(8log)5log( −+= ( log log )

nm n m=∵

2

825log2log8log25log

×=−+=

[∵ mnnm logloglog =+ and log log log ]m

m nn

− =

2log100 log10 2log10 2= = = =

(ii) 18log3log225log8log −++2log8 log 25 log(3) log18= + + −

18log9log25log8log −++=

××=

18

9258log

2log100 log10 2log10 2= = = =

(iii) 243

32log

9

5log2

16

75log +−

243

32log

9

5log

16

75log

2

+

−= (∵ )loglog

naa mmn =

243

32log

9

5

9

5log

16

75log +×−=

243

32log

81

25log

16

75log +−=

75

3216log log25 243

81

= + (∵ )logloglog nmn

maaa −=

243

32log

25

81

16

75log +×=

243

32log

25

81

16

253log +×

×=

243

32log

1

81

16

13log +×

×=

243

32log

16

243log +=

×=

243

32

16

243log 2log)21log( =×=

Page 6: LOGARITHMS AND INDICES - TestlabzMath Class IX 2 Question Bank Q.1. (C) By converting to exponential form find the value of each of the following. (i) log 264 (ii) log 832 (iii) 9

Math Class IX 6 Question Bank

(iv) 28log49log2

125log

2

32log5 −++

28log)49log()25log(2log 2

1

2

3

5−++= 28log7log)5log(32log

3−++=

28log7log125log32log −++=32 125 7

log log100028

× ×= =

3log10 3log10 3= = =

(v) 30

1log36log

2

15log2log2 −−+

30

1log)36log(5log2log 2

1

2−−+= ]30log1[log6log5log4log −−−+=

30log1log6log5log4log +−−+=4 5 30

log log1006 1

× ×= =

×

2log10 2log10 2= = =

(vi) )75.6log()75.0log(2)2.1log( −+

)75.6log()75.0log()2.1log( 2−+= log(1.2) log(0.5625) log(6.75)= + −

1.2 0.5625

log6.75

×=

12 5625 100log

10 10000 675

× ×=

× ×11.0log

10

1log −===

Q.6. Express each of the following as a single logarithm :

(i) 23log(1.5)log36log82log 10101010 −−+ (ii) 3log22log51 +−

(iii) 12log32log52log 1010102 +−+ (iv) 52log9log2

12 1010 −+

(v) 12log62log8log4

19log

2

110101010 −++

(vi)

+

91

55log

77

130log

13

112log 101010

Ans. (i) 2log3)5.1(log36log8log2 10101010 −−+

3

1010102

10 2log)5.1(log36log)8(log −−+=

8log5.1log36log64log 10101010 −−+=

×

××=

×

×=

815

103664log

85.1

3664log 1010 192log10=

Page 7: LOGARITHMS AND INDICES - TestlabzMath Class IX 2 Question Bank Q.1. (C) By converting to exponential form find the value of each of the following. (i) log 264 (ii) log 832 (iii) 9

Math Class IX 7 Question Bank

(ii) 2log35log21 +−32

2log5log10log +−= [∵ ]110log =a

8log25log10log +−=

8log)25log10(log +−=

8log25

10log −= 8

25

10log ×=

16log

5=

(iii) 12log3log25log2 101010 +−+

10log2log)3(log)5(log 10102

102

10 +−+= ]110log[ 10 =∵

10log2log9log25log 10101010 +−+= 1125log2

10925log 1010 =

××=

(iv) 5log29log2

12 1010 −+

210

2

1

1010 )5(log)9(log100log −+= [∵ ]2100log10 =

25log3log100log 101010 −+=

12log25

3100log 1010 =

×=

(v) 12log6log281log4

19log

2

110101010 −++

12log)6(log)81(log)9(log 102

104

1

102

1

10 −++=

12log36log3log3log 10101010 −++=

27log12

3633log 1010 =

××=

(vi)

+

91

55log

77

130log

13

11log2 101010

+

=

91

55log

77

130log

13

11log 1010

2

10

+=

91

55log

77

130log

169

121log 101010

×=

91

55log

77

130

169

121log 1010

Page 8: LOGARITHMS AND INDICES - TestlabzMath Class IX 2 Question Bank Q.1. (C) By converting to exponential form find the value of each of the following. (i) log 264 (ii) log 832 (iii) 9

Math Class IX 8 Question Bank

91

5577

130

169

121

log10

×

=55

91

77

130

169

121log10 ××=

2log10=

Q.7. Evaluate the following without using log tables :

(i) log27

log81 (ii)

log32

log128

(iii) 3log

log27 (iv)

log27

log3log9 −

Ans. (i) 3

4

3log3

3log4

3log

3log

27log

81log3

4

===

(ii) 5

7

2log5

2log7

2log

2log

32log

128log5

7

===

(iii) 61

23

2

1

3

3log2

1

3log3

3log

3log

3log

27log

2

1

3

====

(iv) 3

1

3log3

3log

3log3

3log3log2

3log

3log3log

27log

3log9log3

2

==−

=−

=−

Q.8. Solve for x :

(i) x − =10log ( 10) 1 (ii) x − =2

log ( 21) 2

(iii) x x− + + =log ( 2) log ( 2) log5

(iv) x x+ + − = +log ( 5) log ( 5) 4log2 2log3

(v) x x+ − − =log ( 4) log ( 4) log2

(vi) x x+ − − =log ( 3) log ( 3) 1

(vii) x=log81

log27 (viii) x=

log128

log32

Ans. (i) 1)10(log10 =−x

⇒ 10log)10(log 1010 =−x )110log( =∵

⇒ 2010101010 =⇒+=⇒=− xxx

(ii) 2)21(log 2=−x

Page 9: LOGARITHMS AND INDICES - TestlabzMath Class IX 2 Question Bank Q.1. (C) By converting to exponential form find the value of each of the following. (i) log 264 (ii) log 832 (iii) 9

Math Class IX 9 Question Bank

⇒ 100log)21(log 102

10 =−x )2log( 100 =∵

⇒ 100212

=−x

⇒ 1212110022

=⇒+= xx ⇒ 11=x

(iii) 5log)2(log)2(log =++− xx

⇒ 5log)2()2(log =+×− xx [∵ ]logloglog mnnm aaa =+

⇒ 2log ( 4) log5x − =

⇒ 94554222

=⇒+=⇒=− xxx ⇒ 3)3()( 22=⇒= xx

(iv) 3log22log4)5(log)5(log +=−++ xx

⇒ 3log22log4)5()5(log +=−×+ xx [∵ ]logloglog mnnm aaa =+

⇒ 24223log2log])5()[(log +=−x

⇒ 2log ( 25) log 16 log9x − = +

⇒ )916(log)25(log2

×=−x [∵ ]logloglog mnnm aaa =+

⇒ 144log)25(log 2=−x ⇒ 14425

2=−x ⇒ 25144

2+=x

⇒ 13)13()(169222

=⇒=⇒= xxx

(v) 2log)4(log)4(log =−−+ xx ⇒ 2log)4(

)4(log =

+

x

x

⇒ )4(2424

4−=+⇒=

+xx

x

x

⇒ 482824 −−=−⇒−=+ xxxx ⇒ 1212 =⇒−=− xx

(vi) 1)3(log)3(log =−−+ xx

⇒ 1)3(log)3(log 1010 =−−+ xx

⇒ 1)3(

)3(log10 =

+

x

x [∵ ]logloglog

n

mnm aa =−

⇒ 10)3(

)3(10log

)3(

)3(log 1010 =

+⇒=

+

x

x

x

x

⇒ )3(10)3( −=+ xx ⇒ 3 10 ( 3)x x+ = −

⇒ 3 10 30 10 30 3x x x x+ = − ⇒ − = − −

⇒ 9

33339339 =⇒=⇒−=− xxx

⇒ 11 2

33 3

x x= ⇒ =

Page 10: LOGARITHMS AND INDICES - TestlabzMath Class IX 2 Question Bank Q.1. (C) By converting to exponential form find the value of each of the following. (i) log 264 (ii) log 832 (iii) 9

Math Class IX 10 Question Bank

(vii) x=27log

81log⇒

)333(log

)3333(log

27log

81log

××

×××=⇒= xx

⇒ 3

4

3log3

3log4

3log

3log3

4

==⇒= xx

(viii) x=32log

128log

⇒ )22222(log

)2222222(log

32log

128log

××××

××××××=⇒= xx ⇒

2log5

2log7

2log

2log5

7

=⇒= xx

⇒ 5

21

5

7=⇒= xx

Q.9. Given log x m n= + and log y m n= − , express the value of x

y2

10log in

terms of m and n.

Ans. nmx +=log (Given) ...(i)

nmy −=log (Given) ...(ii)

2

2

10log log10 log

xx y

y= − (∵ )logloglog nm

n

maaa −=

2loglog10log yx −+=

yx log2log10log −+=

yx log2log1 −+= ]110log[ =∵

Putting the value of nmx +=log and nmy −=log , we get

)(2)(1 nmnm −−++= nmnmnm 31221 +−=+−++=

Page 11: LOGARITHMS AND INDICES - TestlabzMath Class IX 2 Question Bank Q.1. (C) By converting to exponential form find the value of each of the following. (i) log 264 (ii) log 832 (iii) 9

Math Class IX 11 Question Bank

Q.10. If a=10log 2 and b=10log 3 ; express each of the following in terms of ‘a’

and ‘b’:

(i) log12 (ii) log2.25 (iii) log5.4 (iv) log60

(v) 8

1log3 (vi) 18log (vii)

4

9log

Ans. a=2log10 ...(i)

b=3log10 ...(ii)

(i) 12log )32(log)322(log 2×=××=

3log2log2

+= 3log2log2 += ]loglog[ mnm an

a =∵

3log2log2 1010 +=

ba +×= 2 [Putting the value of a=2log10 and ]3log10 b=

ba += 2

(ii) 25.2log425

925log

100

225log

×

×==

2

2

3log

4

9log

==

]2log3[log22

3log2 −== )(2]2log3[log2 1010 ab −=−=

[Putting ]2log,3log 1010 ab ==

ab 22 −= 2( )b a⇒ −

(iii) 10log54log10

54log4.5log −== 10log)3332(log 10−×××=

10log)32(log 103

10 −×= 13log2log3

1010 −+= ]110[log10 =

13log32log 1010 −+= 1313 −+=−×+= baba

(iv) )3210(log60log 10 ××=

3log2log10log 101010 ++= ]logloglog[log 10101010 nmllmn ++=

ba ++= 1 [Putting a=2log10 and ]3log10 b=

1++= ba

Page 12: LOGARITHMS AND INDICES - TestlabzMath Class IX 2 Question Bank Q.1. (C) By converting to exponential form find the value of each of the following. (i) log 264 (ii) log 832 (iii) 9

Math Class IX 12 Question Bank

(v) 4

4

8

25log

8

25log

8

13log 10 ×==

22222

100log

32

100log 1010

××××==

51010510 2log100log

2

100log −== [∵ ]logloglog nm

n

maaa −=

2log52 10−= a×−= 52 [Putting ]2log10 a=

a52 −=

(vi) 2

1

)18(log18log = 2

1

)233(log ××=

)23(log2

1)23(log

22

1

2×=×= ]2log3[log

2

1 2+=

]2log3log2[2

1+= 2log

2

13log2

2

1+×=

2log2

13log2

2

1+×= 2log

2

13log +=

+=

2

ab

(vii)

2

2

3log

22

33log

4

9log

=

×

×=

32log 2 [log3 log 2] 2( )

2b a= = − = −

Q.11. If 0.3010,log2 = find the value of .243

32log

9

52log

16

75log

+−

Ans. 243

32log

9

5log2

16

75log +−

243

32log

9

5log

16

75log

2

+

−=

75 25 32

log log log16 81 243

= − +81

25log

243

32log

16

75log −+=

75 32 25

log log16 243 81

= × −

××=

25

81

243

32

16

75log

3010.02log ==

Q.12. If 0.9030,log8 = find the value of :

(i) log4 (ii) 32log (iii) (0.125)log

Ans. 9030.08log =∵ ⇒ 9030.0)2(log 3= ⇒ 9030.02log3 =

∴ 3010.03

9030.02log ==

(i) 2log22log4log 2== 6020.0)3010.0(2 ==

Page 13: LOGARITHMS AND INDICES - TestlabzMath Class IX 2 Question Bank Q.1. (C) By converting to exponential form find the value of each of the following. (i) log 264 (ii) log 832 (iii) 9

Math Class IX 13 Question Bank

(ii) 52

1

2log2

1)32(log32log ==

)3010.0(2

52log

2

5== 7525.01505.05 =×=

(iii)

==

8

1log

1000

125log)125.0(log

9030.0)3010.0(32log3)2(log2

1log

33

−=−=−==

=

Q.13. If 1.4313log27 = , find the value of :

(i) log9 (ii) log30

Ans. 4313.127log = ⇒ 4313.13log 3= ⇒ 4313.13log3 =

⇒ 477.03

4313.13log ==

(i) 9542.0)4771.0(23log23log9log 2====

(ii) 4771.114771.010log3log)103(log30log =+=+=×=

Q.14. Show that log3log2log13)2(1log ++=++

Ans. 3log2log1log)321(log6log)321(log ++=××==++ Hence, proved.

Q.15. (i) If log logm n m n+ = +log ( ) , show that n

mn

=− 1

(ii) If log loga b

a b+

= +

1log ( ),

2 2 show that a b ab+ =

1( ) .

2

Ans. (i) nmnm loglog)(log +=+

⇒ )(log)(log nmnm ×=+

∴ mnnm =+

⇒ nmnm −=−

⇒ (1 )m n n− = −

⇒ 1)1(1 −

=−−

−=

−=

n

n

n

n

n

nm . Hence,

1−=

n

nm

Page 14: LOGARITHMS AND INDICES - TestlabzMath Class IX 2 Question Bank Q.1. (C) By converting to exponential form find the value of each of the following. (i) log 264 (ii) log 832 (iii) 9

Math Class IX 14 Question Bank

(ii) )log(log2

1

2log ba

ba+=

+⇒ 2

1

)(log)(log2

1

2log abab

ba==

+

⇒ log log2

a bab

+ =

Comparing both sides, we get

abba

=+

2 or abba =+ )(

2

1. Hence, .)(

2

1abba =+

Q.16. If log loga b a b+ = +log ( ) , find a in terms of b.

Ans. baba loglog)(log +=+

⇒ abba log)(log =+ [∵ ]logloglog nmmn aaa +=

⇒ abba =+ ⇒ baba −=−

⇒ baab −=+− ⇒ bba −=−− )1( ⇒1

)1(−

=⇒=−b

babba

Q.17. If a b

mbc ca

= =

2 2

l log , log and c

nab

=

2

log,, find the value of l m n+ + .

Ans. ab

cn

ca

bm

bc

al

222

log,log,log ===

ab

c

ca

b

bc

anml

222

logloglog ++=++

××=

ab

c

ca

b

bc

a 222

log

1loglog222

222

==

cba

cba

10

10log

10

101log =×=

10log10log −= 011 =−=

Q.18. Prove that :

log loga

a b abb

− = ⋅

2 2( ) ( ) log log ( )

Ans. L.H.S. 22 )(log)(log ba −=

]log[log]log[log baba −+=

⋅=

b

aab log)(log

[∵ nmmn aaa logloglog += and ]logloglog nmn

maaa −=

Page 15: LOGARITHMS AND INDICES - TestlabzMath Class IX 2 Question Bank Q.1. (C) By converting to exponential form find the value of each of the following. (i) log 264 (ii) log 832 (iii) 9

Math Class IX 15 Question Bank

)(loglog abb

a⋅

= ...(i)

RHS )(loglog abb

a⋅

= ...(ii)

From (i) and (ii),

LHS = RHS

Q.19. (i) If a a+ = − −log ( 1) log (4 3) log3; find a

(ii) If log logy x− − =2 3 0, express x in terms of y.

(iii) Prove that: 2).log(13125log 1010 −=

Ans. (i) 3log)34(log)1(log −−=+ aa

⇒ 3

)34(log)1(log

−=+

aa

⇒ 3

)34()1(

−=+

aa ⇒ 34)1(3 −=+ aa ⇒ 3433 −=+ aa

⇒ 3343 −−=− aa ⇒ 6−=− a ⇒ 6=a

(ii) 03loglog2 =−− xy

⇒ 3loglog2 =− xy ⇒ 3loglog 2=− xy ⇒ 3log

2

=x

y

⇒ 1000loglog2

=x

y⇒ 1000

2

=x

y⇒ xy 10002

=

⇒ 21000 yx = ⇒ 1000

2yx =

(iii) )2log1(3125log 1010 −=

LHS 5log35log125log 103

1010 ===

RHS 5log3)5(log32

10log3)2log1(3 10101010 ==

=−=

∴ LHS = RHS

Page 16: LOGARITHMS AND INDICES - TestlabzMath Class IX 2 Question Bank Q.1. (C) By converting to exponential form find the value of each of the following. (i) log 264 (ii) log 832 (iii) 9

Math Class IX 16 Question Bank

Q.20. If log loga x b y= =2 3

, and loga b z− =2 3

3 2 6 , express y in terms of x and z.

Ans. 2 3log , loga x b y= = ⇒ zba log623

32=−

⇒ zyx log6log2log3 =− ⇒ 623 logloglog zyx =−

⇒ 6

2

3

loglog zy

x= ⇒ 6

2

3

zy

x= ⇒ 362

xzy = ⇒6

32

z

xy =

⇒ 2

1

6

3

=

y

xy ⇒

33

232

3

xy x z

z= = ÷ Hence,

3

32y x z= ÷

Q.21. If log loga b

a b−

= +1

log ( ),2 2

show that a b ab+ =2 2

6 .

Ans. log )log(log2

1

2ba

ba+=

−)(log

2

1ab= 2

1

)(log ab= ∴ 2

1

)(2

abba

=−

Squaring both sides, we get

212

2( )2

a bab

− =

⇒ ababba

=−+

4

222

⇒ ababba 4222

=−+

⇒ ababba 2422

+=+ ⇒ abba 622

=+

Q.22. If 23ab,ba22

=+ show that: logb)(loga2

1

5

balog +=

+

Ans. abba 2322

=+

⇒ abababba 223222

+=++ (Adding ab2 both sides, we get)

⇒ ababba 25222

=++ ⇒ abba 25)( 2=+

⇒ abba

=+

25

)( 2

⇒ abba

=

+2

5

Taking log both sides, we get

)(log5

log

2

abba

=

+ [∵ ]loglog mnm a

na =

⇒ )(log5

log2 abba

=+

⇒ )(log2

1

5log ab

ba=

+⇒ )log(log

2

1

5log ba

ba+=

+

Page 17: LOGARITHMS AND INDICES - TestlabzMath Class IX 2 Question Bank Q.1. (C) By converting to exponential form find the value of each of the following. (i) log 264 (ii) log 832 (iii) 9

Math Class IX 17 Question Bank

Q.23. Solve for x and y if 0x > and :0y > 2.2log2y

xloglogxy =+=

Ans. log log 2log 2 2x

xyy

= + =

(i) 10log2122log =×==xy 100log10log 2==

∴ 100=xy ...(i)

(ii) 22log2log =+y

x

100log2loglog2

=+y

x )2100log( =∵

100log2log2

=×y

x⇒ 100log

4log =

y

x⇒ 100

4=

y

x

⇒ yxyx 251004 =⇒= ...(ii)

From eqn. (i), we get

10025 =⋅ yy ⇒ 222 )2(410025 ±==⇒= yy

⇒ 2=y )0( >y∵

∴ 502

1001002 ==⇒=× xx

Hence, .2,50 == yx

Q.24. If m = log20 and n = log25, find the value of x, if: 2log ( 1) 2 .x m n+ = −

Ans. 25log,20log == nm

nmx −=+ 2)1(log2

⇒ 25log20log2)1(log2

−=+x ⇒ 25log)20(log)1(log22

−=+x

⇒ 25log400log)1(log2

−=+x ⇒ 2 400log ( 1) log

25x + =

∴ 22)4(16)1( ==+x ∴ 41 =+x ⇒ 314 =−=x ∴ 3=x

Page 18: LOGARITHMS AND INDICES - TestlabzMath Class IX 2 Question Bank Q.1. (C) By converting to exponential form find the value of each of the following. (i) log 264 (ii) log 832 (iii) 9

Math Class IX 18 Question Bank

Q.25. If

−−+−

45

7log2log3log16log2log7 log n= +1 , find the value of n.

Ans. nlog145

7log3log216log2log7log +=−−+−

⇒ nlog10log45

7log3log16log2log7log 2

+=−−+− )110log( =∵

⇒ nlog10log45

7log9log16log2log7log +=−−+−

⇒ )10(log)10(log792

45167log nn =×=

××

×× ∴

792

4516710

××

××=n

⇒ 479210

45167=

×××

××=n . Hence, 4=n

Q.26. Evaluate :

(i) 2

3

9 (ii) 3

2

8 (iii) 3

4

8−

(iv) 5

3

(243)

(v)

3

2

(27)

1

(vi) 5

2

32)(− (vii) 3

1

125

8−

(viii) 3

1

(0.001)−

Ans. (i)

3 3 3 3 32 1

22 2 2 2 19 (3 3) (3 ) 3 3× ×

= × = = = [ ( ) ]m n mna a=∵

2733333

=××==

(ii) 3

23

3

2

33

2

3

2

2)2()222(8×

==××= [ ( ) ]m n mna a=∵

422

==

(iii) 3

4

3

4

)222(8−−

××= 3

43

3

4

32)2(

−×−

== [ ( ) ]m n mn

a a=∵

16

1

2222

1

2

12

4

4=

×××===

− 1n

na

a

− =

(iv) 5

3

5

3

)33333()243(−−

××××= 5

35

5

3

53)3(

−×−

==

3

3

3

1)3( ==−

27

1

333

1=

××=

1n

na

a

− =

Page 19: LOGARITHMS AND INDICES - TestlabzMath Class IX 2 Question Bank Q.1. (C) By converting to exponential form find the value of each of the following. (i) log 264 (ii) log 832 (iii) 9

Math Class IX 19 Question Bank

(v) 3

2

3

2)27(

)27(

1=

1 n

na

a−

=

3

2

33

2

)3()333( =××= 9333323

6

=×=== [ ( ) ]m n mn

a a=∵

(vi) 5

2

5

2

)]2()2()2()2()2[()32( −×−×−×−×−=−

5

25

5

2

5)2(])2[(

×

−=−= 4)2()2()2(2

=−×−=−= [ ( ) ]m n mn

a a=∵

(vii) 3

1

3

1

555

222

125

8−−

××

××=

2

5

5

2

5

2

5

213

3

13

13

=

=

=

=

−×−−

(viii) 3

1

3

1

)1.01.01.0()001.0(

−−

××=

101

101

10

1

1

1.0

1)1.0()1.0(

13

13

=====−

−×

Q.27. Evaluate the following :

(i) 2

1

03

22

16

9583

4

1−−

+××−

(ii) 03

2

2

1

3(27)(0.01)4

1×−+

(iii)

÷

×

−− 3

2

3

4

3

2

5

9

25

16

81 (iv)

12 2

23(64) × 2 ÷ 70

(v) 3

2

2

3

4

3

216

343

9

49

81

16

÷

×

(vi)

2 03

1 3

4

64 1 25÷ +

125 64256

625

Page 20: LOGARITHMS AND INDICES - TestlabzMath Class IX 2 Question Bank Q.1. (C) By converting to exponential form find the value of each of the following. (i) log 264 (ii) log 832 (iii) 9

Math Class IX 20 Question Bank

(vii)

3

1

2

3

1

2

1

5

2

(64)(2)

(8)(4)(32)

−−

÷

×× (viii) 3

1

3

1

)532()532( +−

Ans. (i)

122203

1 93 8 5

4 16

− −

− × × +

2

1

03

22

4

3

4

35)222(3

2

1

2

1−−

×+××××−

×=

−×−×

+××−

=

2

12

03

2

3)2(2

4

35)2(3

2

1 ])([ mnnm aa =∵

12

4

4

3123

2

1−−

+××−

= ]1[

0=a∵

3

4143)2( 4

+××−=3

412)2222( +−×××=

=

m

m

aa

1∵

3

44

3

41216 +=+−=

3

15

3

16

3

412==

+=

(ii) 03

2

2

1

3)27()01.0(4

1×−+

03

2

2

12

12

3)333()1.01.0(2

1×××−×+

=

03

2

32

12

2

12

3)3()1.0(2

1×−+

=

−××

0211

3)3()1.0(2

1×−+

=

− ])([ mnnm aa =∵

191.0

1

2

1×−+=

==

−1and

1 0a

aa

m

m∵

2

119

2

1109

1

10

2

1=−=−+=

Page 21: LOGARITHMS AND INDICES - TestlabzMath Class IX 2 Question Bank Q.1. (C) By converting to exponential form find the value of each of the following. (i) log 264 (ii) log 832 (iii) 9

Math Class IX 21 Question Bank

(iii)

÷

×

−−3

2

3

4

3

2

5

9

25

16

81

÷

×

××

×××

×××=

−−−

32

3

4

3

2

5

33

55

2222

3333

÷

×

=

−×

−× 3

2

32

4

34

5

2

3

5

2

3

3 3 33 5 2

2 3 5

− − = × ÷

÷

×

=

333

5

2

5

3

3

2

××÷

×××

××

××=

5

2

5

2

5

2

5

3

5

3

5

3

333

222

18

27

27

8

8

125

125

27

27

8=×=

××=

(iv) 2

1

023

2

72)64(

÷×

2

1

23

2

12

1)444(

÷×××=

2

1

3

2

31

22

1)4(

÷

××=

2

1

3

23

14

14

×

÷×=2

1

2

1

21

4

1161

4

14

−−

××=

÷×=

2

1

2

1

)22(

1

4

1)4(

2

12

2

1

2

12

1

==

×

===

×

Page 22: LOGARITHMS AND INDICES - TestlabzMath Class IX 2 Question Bank Q.1. (C) By converting to exponential form find the value of each of the following. (i) log 264 (ii) log 832 (iii) 9

Math Class IX 22 Question Bank

(v) 3

2

2

3

4

3

216

343

9

49

81

16

÷

×

3

2

2

3

4

3

666

777

33

77

3333

2222

××

××÷

×

××

×××

×××=

3 3 24 2 34 2 32 7 7

3 3 6

= × ÷

3

23

2

32

4

34

6

7

3

7

3

2××

−×

÷

×

=

233

6

7

3

7

3

2

÷

×

=

66

77

333

777

2

33

×

×÷

××

×××

=

36

49

27

343

222

333÷×

××

××=

49

36

27

343

8

27××=

2

131

2

63==

(vi)

0

3

4

1

3

2

64

25

625

256

1

125

64

+

÷

0

3 3

2

4

1

3

2

4

5

5555

4444

1

555

444

+

×××

×××

÷

××

××=

−0

4

14

3

23

4

5

5

4

1

5

4

+

÷

=

1

5

4

1

5

4

4

14

3

23

+

÷

=

×

−×

]1[ 0=a∵

Page 23: LOGARITHMS AND INDICES - TestlabzMath Class IX 2 Question Bank Q.1. (C) By converting to exponential form find the value of each of the following. (i) log 264 (ii) log 832 (iii) 9

Math Class IX 23 Question Bank

14

5

4

51

5

4

1

5

42

1

2

=+

÷

=

=

m

m

aa

1∵

4

12

4

91

4

51

5

4

16

25==+=+×=

(vii)

3

1

2

3

1

2

1

5

2

)64()2(

)8()4()32(−

÷

××

3

1

2

3

1

2

1

5

2

)444()2(

)222()22()22222(−

××÷

×××××××××=

3

1

32

3

1

32

1

25

2

5

)4()2(

)2()2()2(−

÷

××=

−×

×

−××

÷

××=

3

13

2

3

13

2

12

5

25

)4()2(

)2()2()2(

122

112

12

112

)2(2

2

42

222−−

+−

−−

÷

=

÷

××= 4

1

22

22

222

2

=

÷

=−−

(viii) 3

1

3 )532()532( +−

1

])([ mmm abba =∵

3

1

)]532)(532[( +−= 3

1

22])5()32[( −=

3

1

3

1

)27()532( =−= 3)3()333( 3

1

33

1

==××=

Q.28. Simplify :

(i) n n

n n

+

− −

×

×

1

1 1

3 9

3 9 (ii)

n n

n n n n

+ +

− + −÷

1 1

( 1) 1 1

3 9

3 (3 )

(iii) a b a b− − −

+ ⋅ +1 1 1

( ) ( ) (iv) n n

n n

+ +− ×

× − ×

3 1

2

5 6 5

9 5 5 2

Ans. (i) 121

12

11

1

)3(3

)3(3

93

93−−

+

−−

+

×

×=

×

×

nn

nn

nn

nn 2 2

1 2 2

3 3

3 3

n n

n n

+

− −

×=

×

33

23

221

22

3

3

3

3−

+

−+−

++

==n

n

nn

nn

5332333 ==

+−+ nn

Page 24: LOGARITHMS AND INDICES - TestlabzMath Class IX 2 Question Bank Q.1. (C) By converting to exponential form find the value of each of the following. (i) log 264 (ii) log 832 (iii) 9

Math Class IX 24 Question Bank

(ii) 11

1

)1(

1

)3(

9

3

3−+

+

+

÷nn

n

nn

n

1

)1)(1(1

1

11

)1(

1

)33(

3

3

3

9

)3(

3

32 +

−+

+

+

−+

+

×

×=×=n

nn

nn

n

n

nn

nn

n

)1(2

11

12

)1()(1

3

3

3

3

)3(

3

3

32

2

22

2 +

+

+

+

×=×=n

n

nn

n

n

n

nn

n

)22()(11

22

1122

2

23

3

3

3

3 +−−−−++

+

+

=×=nnnnn

n

n

nn

n

9

1

33

1

3

133

2

22211 22

====−−−+−−++ nnnnn

(iii) )()( 111 −−−+⋅+ baba

+⋅

+=

baba

11

)(

1

ab

ab

baab

ab

ba

)(

)(

111

)(

1 +⋅

+=

×+×⋅

+=

abab

ba

ba

1)(

)(

1=

+⋅

+=

(iv) 2

13

2559

565

×−×

×−++

nn

nn

)49(5

)65(5

4559

5655 21121

−=

×−×

×−⋅=

+++

n

n

nn

nn

19)5(5

)19(55

)5(5

)625(551

=⋅

=−⋅

=n

n

n

n

Q.29. If a b c= ⋅ ⋅1960 2 5 7 , calculate the value of a b c− −

⋅ ⋅2 7 5 .

Ans. cba7521960 ⋅⋅=

⇒ 1960752 =⋅⋅cba

⇒ 775222752 ×××××=⋅⋅cba

⇒ 213752752 ××=⋅⋅

cba ...(i)

Comparing powers of 2, 5 and 7 on both sides of eqn. (i), we get

2,1,3 === cba

Hence, value of:

213572572

−−−−⋅⋅=⋅⋅

cba

232352

7

5

17

2

1

×

=××=

200

7

258

7

55222

7=

×=

××××=

Solve for x : (Q.30 to Q.32)

Page 25: LOGARITHMS AND INDICES - TestlabzMath Class IX 2 Question Bank Q.1. (C) By converting to exponential form find the value of each of the following. (i) log 264 (ii) log 832 (iii) 9

Math Class IX 25 Question Bank

Q.30. Evaluate

xa b

b a

=

1 3

Ans.

x

a

b

b

a31−

= ⇒

13)31(2

1−−−

=

=

xx

b

a

b

a

b

a

Comparing both sides, we get

2

31

2

13

2

113 =+=⇒=− xx

Hence, 3 1 1

2 3 2x = × =

Q.31. Evaluate

x−

=

1

32 27

3 8

Ans.

3

3

31

3

2

3

)2(

)3(

8

27

3

2

===

−x

⇒3

3

1

3

2

3

2−

=

x

Comparing both sides, we get

8199133

1−=+−=⇒−=−⇒−=

−xx

x. Hence, 8−=x

Q.32. Evaluate

x−−

=

1 13 27

5 125

Ans.

11

125

27

5

3−−

=

x

131

3

32

1

5

3

5

3

5

3−−−

=

=

x

32

1

5

3

5

3−

=

x

Comparing both sides, we get

6132

1−=−⇒−=

−x

x⇒ 516 −=+−=x . Hence, 5−=x

Page 26: LOGARITHMS AND INDICES - TestlabzMath Class IX 2 Question Bank Q.1. (C) By converting to exponential form find the value of each of the following. (i) log 264 (ii) log 832 (iii) 9

Math Class IX 26 Question Bank

Q.33. Solve for x:

(i) x x− +− =

2 14 2 0 (ii) x x

=2

3 : 3 9 : 1

(iii) x x x2 +18× 2 + 4× 2 = 1 + 2 (iv) x x

−2 +2 3

2 + 2 4× 2 = 0

(v) x x– 3 +14( 3) = ( 3)

Ans. (i) 02412

=−+− xx ⇒ 12

24+−

=xx

⇒ 12

24+−

=xx ⇒ 12 2)22( +−

=×xx ⇒ 1)2(2

22+−

=xx ⇒ 142

22+−

=xx

⇒ 2 4 1x x− = + ⇒ 2 1 4x x− = + ⇒ 5x =

(ii) 1:93:32

=xx

⇒ 931

9

3

3 22

=⇒=− xx

x

x

=

−nm

n

m

aa

a

⇒ 3332

×=− xx ⇒ 2

332

=− xx ⇒ 2

2=− xx

⇒ 022

=−− xx ⇒ 0222

=−+− xxx ⇒ 0)2(1)2( =−+− xxx

⇒ 0)2)(1( =−+ xx . If 0)2( =−x then 2=x . If 0)1( =+x then 1−=x

(iii) xxx212428

12+=×+×

+ ⇒ xxx 21224)2(8 12+=××+×

⇒ xxx2128)2(8

2+=×+× ⇒ 01228)2(8

2=−−×+×

xxx

⇒ 01)18(2)2(8 2=−−+×

xx ⇒ 0127)2(82

=−×+xx

]2 Putting[ yx

=

⇒ 017)(8 2=−×+× yy ⇒ 0178 2

=−+ yy

⇒ 0188 2=−−+ yyy ⇒ 0)1(1)1(8 =+−+ yyy

⇒ 0)18)(1( =−+ yy ⇒ 0)1( =+y and 018 =−y

⇒ 012 =+x and 0128 =−×

x

⇒ 12 −=x

and 128 =×x ⇒ 12 −=

x and

8

12 =

x

⇒ 12 −=x

and 3

2

12 =

x

⇒ 12 −=x

and 3223

−=⇒=−

xx

Page 27: LOGARITHMS AND INDICES - TestlabzMath Class IX 2 Question Bank Q.1. (C) By converting to exponential form find the value of each of the following. (i) log 264 (ii) log 832 (iii) 9

Math Class IX 27 Question Bank

(iv) 02422 322=×−+

+xx

⇒ 2 2(2 ) 2 2 4 2 2 2 0x x+ ⋅ − × × × = ⇒ 2(2 ) 2 4 32 0·x x

+ − =

⇒ 03224)2( 2=−×+

xx ⇒ 03242=−+ yy [Putting ]2 y

x=

⇒ 032482

=−−+ yyy ⇒ 0)8(4)8( =+−+ yyy

⇒ 0)4)(8( =−+ yy ⇒ 08 =+y and 04 =−y

⇒ 8−=y and 4=y ⇒ 82 −=x and 2

22 =x

⇒ 2=x

(v) 143)3()3(

+−=

xx

1

4

13

2

1

33

+−

=

xx

⇒1

4

13

2

1

33+×−×

=

xx

⇒ )1(4

1)3(

2

1+×=−× xx ⇒

4

)1(

2

)3( +=

− xx

⇒ )1(2)3(4 +=− xx ⇒ 22124 +=− xx

⇒ 12224 +=− xx ⇒ 72

14142 =⇒=⇒= xxx

Q.34. Solve for x if :(i)x+

=

12

53 32( 4) ( 8) (ii) x xp p q q p q

− −× = × ≠

1 2 1 2; .

Ans. (i) 532

12

3 )8()4( =

+x

⇒ 532

12

3 )222()22( ××=×

+x

⇒ 53 32

12

3 2)2()2( =

+x

2 1 32

53 2 32 (2 )x

+

= ⇒

2 12

1 53 22 (2 )x

+

=

⇒ 512

12

3

2

22×

+

=

x

⇒ 52

12

3

2

22 =

+x

⇒ 52

12

3

2=

+x ⇒ 5

3

2

1222

=

×+× x

⇒ 53

14=

+x⇒ 15143514 =+⇒×=+ xx

⇒ 1441154 =⇒−= xx ⇒2

13

2

7

4

14=⇒=⇒= xxx

Page 28: LOGARITHMS AND INDICES - TestlabzMath Class IX 2 Question Bank Q.1. (C) By converting to exponential form find the value of each of the following. (i) log 264 (ii) log 832 (iii) 9

Math Class IX 28 Question Bank

(ii) )(;2121 qpqqpp xx≠×=×

−− ⇒ 2

1

212

1

21qqpp

xx×=×

−−

⇒ 2

121

2

121 +−+−

=

xx

qp ][ nmnmaaa

+=×

But this is true only

If 02

121 =+− x ]1[

00== qp∵

⇒ 2

112 −−=− x ⇒

2

32

2

122

−=−⇒

−−=− xx

⇒ 22

3

2

32

×=⇒= xx ⇒

4

3=x

Q.35. Prove that :

(i)

a b b c c aa b c

b c a

x x x

x x x

+ + +

⋅ ⋅ =

1

(ii)

1 1 1a b cab bc ca

b c a

x x x

x x x

⋅ ⋅ =

1

(iii)

( ) ( ) ( )a b c b c a c a ba b c

b c a

x x x

x x x

+ − + − + −

⋅ ⋅ =

1

Ans. (i) LHS

ac

a

ccb

c

bba

b

a

x

x

x

x

x

x+++

=

))(())(())(( acaccbcbbabaxxx

+−+−+−⋅⋅=

222222 accbba

xxx−−−

⋅⋅=

====−+−+− 10222222

xxaccbba RHS

(ii) LHS ca

a

cbc

c

bab

b

a

x

x

x

x

x

x

111

=

caac

bccb

abba

xxx

1)(

1)(

1)( −−−

⋅⋅=

ca

ac

bc

cb

ab

ba

x

−+

−+

=

abc

acbcbabac

x

)()()( −+−+−

=

Page 29: LOGARITHMS AND INDICES - TestlabzMath Class IX 2 Question Bank Q.1. (C) By converting to exponential form find the value of each of the following. (i) log 264 (ii) log 832 (iii) 9

Math Class IX 29 Question Bank

====

−+−+−

10

xx abc

abbccaabbcca

RHS

(iii) LHS

)()()( bac

a

cacb

c

bcba

b

a

x

x

x

x

x

x−+−+−+

=

))(())(())(( bacacacbcbcbabaxxx

−+−−+−−+−⋅⋅=

abbcaccaabcbbcacbax

+−−++−−++−−=

222222

=== 10

x RHS

Q.36. Simplify :

(i)

2 2 2 2 2 2a ab b b bc c c ca aa b c

b c a

x x x

x x x

+ + + + + +

⋅ ⋅

(ii)

2 2 2 2 2 2a ab b b bc c c ca aa b c

b c a

x x x

x x x

− + − + − +

− − −

× ×

Ans. (i)

222222 acac

a

ccbcb

c

bbaba

b

a

x

x

x

x

x

x++++++

222222

)()()(acacaccbcbcbbababa

xxx++−++−++−

××=

))(())(())(( 222222 acacaccbcbcbbababaxxx

++−++−++−××=

333333 accbba

xxx−−−

××=

10333333

===−+−+−

xxaccbba

(ii)

222222acac

a

ccbcb

c

bbaba

b

a

x

x

x

x

x

x+−

+−

+−

×

×

222222

)()()(acacaccbcbcbbababa

xxx+−++−++−+

××=

))(())(())(( 222222 acacaccbcbcbbababaxxx

+−++−++−+××=

333333 accbba

xxx+++

××=333333 accbba

x+++++

=)(2 333 cba

x++

=

Page 30: LOGARITHMS AND INDICES - TestlabzMath Class IX 2 Question Bank Q.1. (C) By converting to exponential form find the value of each of the following. (i) log 264 (ii) log 832 (iii) 9

Math Class IX 30 Question Bank

Q.37. If ,x ya b b c= = and z

c a= , prove that : xyz = 1.

Ans. We are given that cbbayx

== , and ac z=

bax

=

⇒ yzxyzba = (Raising to the power yz both sides)

⇒ zzyxyzcba == )( )( cb

y=∵

⇒ aaxyz

= )( ac z=∵

⇒ 1aa

xyz=

Comparing both sides, we get

⇒ 1=xyz or zca =

zyb )(=

yzb= )(

ybc =∵

yzxa )(=

xyza= )( xab =∵

Comparing both sides, we get 1=xyz

Q.38. If x y za b c= = and 2 ,b ac= prove that :

2.

xzy

x z=

+

Ans. Let kcbazyx

===

∴ yx kbka

11

, == and zkc

1

=

Now acb =2

⇒ zxy kkk

112

1

×=

⇒ zxy kk

112+

=

Comparing both sides, we get

xzzxyxz

zx

yzxy2)(

2112=+⇒

+=⇒+=

Hence, zx

xyy

+=

2.

Page 31: LOGARITHMS AND INDICES - TestlabzMath Class IX 2 Question Bank Q.1. (C) By converting to exponential form find the value of each of the following. (i) log 264 (ii) log 832 (iii) 9

Math Class IX 31 Question Bank

Q.39. If x y z= =2 3 12 , show that :

z y x= +

1 1 2.

Ans. Let ,1232 kzyx

=== then

xxkk

1

22 =⇒= , yykk

1

33 =⇒=

zzkk

1

1212 =⇒= . Now 3232212 2×=××=

⇒ yxz kkk

1

2

11

)( ×= ⇒ yxz kkk

121

×= ⇒ yxz kk

121+

=

Comparing both sides, we get

yxz

121+=

Q.40. If x y z−= =2 3 6 show that :

x y z+ + =

1 1 10.

Ans. Let ,632 kzyx

===− then zyx kkk

111

6,3,2−

===

Now 632 =×

⇒ zyx kkk

111 −

=× ⇒ zyxkk

111−+

=

Comparing both sides, we get

zyx

111−=+ ⇒ 0

111=++

zyx