The document contains solutions to problems involving logarithmic and exponential expressions. It expresses relations in logarithmic form, converts logarithmic forms to exponential forms, finds values of logarithms and their bases, and solves for values of x in logarithmic equations. The problems cover topics such as properties of logarithms, evaluating logarithmic expressions, and manipulating logarithmic and exponential terms.

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3. 1.
Express the following relations in the logarithmic forms:
Ans.
HOME
Solution:
3 81
log 81 4
3
4


Logarithmic form :
(i)

4. 1.
Express the following relations in the logarithmic forms:
Ans.
HOME
Solution:
5 1
log 1 0
5
0


Logarithmic form :
(ii)

5. 1.
Express the following relations in the logarithmic forms:
Ans.
HOME
Solution:
1
5
32 2
1
5

32 2
log 2
32
5


Logarithmic form
:
(iii)

6. 1.
Express the following relations in the logarithmic forms:
Ans.
HOME
Solution:
3
1
1
125
log
125
5
5
3
 
 
Logarithmic form :
(iv)

7. 1.
Express the following relations in the logarithmic forms:
Logarithmic form :
Ans.
HOME
Solution:
a c
c b
a
b


log
(v)

8. 2.
Express the following logarithmic forms in the exponential forms:
log 64 6
Ans.
HOME
Solution:
6
2
2 
64

exponential form :
(i)

9. 2.
Express the following logarithmic forms in the exponential forms:
exponential form :
Ans.
HOME
Solution:
1
1
81
3
4
81
log
4
3

 

(ii)

10. 2.
Express the following logarithmic forms in the exponential forms:
log 1 0
Ans.
HOME
Solution:
1
0 

a
a
exponential form :
(iii)

11. 2.
Express the following logarithmic forms in the exponential forms:
3
exponential form :
Ans.
HOME
Solution:
1
1

125
25
2
125
log
25
3
2


  




(iv)

12. 2.
Express the following logarithmic forms in the exponential forms:
P R
log 
exponential form :
Ans.
HOME
Solution:
R
Q
Q 
P
(v)

13. 3.
Find the value of x :
log 128
Ans.
HOME
Solution:
log 128
 
2 128
, 2 
2
, 7
7
2
2



or
or x
x
x
x
x

(i)

14. 3.
Find the value of x :
log 81 4
log 81 4
Ans.
HOME
Solution:
81
x
 
4 4
4
or x
, 
3
, 
3


or x
x
x

(ii)

15. 3.
Find the value of x :
Ans.
HOME
Solution:
log  
3
x
log 3
or x
1

125
 5

,
1
5
,
3
3
5
5

 

or x
x
x

(iii)

16. 3.
Find the value of x :
log 243 10
log 243 10
Ans.
HOME
Solution:
10 5
3
x
 
or x
, 
3
or x
, 3
1
2
10

, 3
1
5




or x
x
x

(iv)

17. 3.
Ans.
HOME
Find the value of x :
Solution:
log 49
7

log 49
7
 
 

7 49
1

2 2
, 7 7
2
1
,
or
or x
2

 
,  2  2 
4
or x
x
x
x
x

(v)

18. 3.
Find the value of x :
log 7 5 2
10
 
x
log 7 5 2
2
10
  
10 7 5
or x
, 100  7 
5
or , 100  5 
7
x
or x
Ans.
HOME
Solution:
 
 
, 105 7
105
,
or x
7
or , 15
x


, 
15

 
or x
x
x

(vi)

19. 3.
Find the value of x :
log 0.0001
log 0.0001
10
log 10
4


x
x
(vii) x
or x
, 4log 10
Ans.
HOME
Solution:
 
 
 
, 4  log 10 1
10
10
10
10
  
 



or x

20. 3.
5

 
Ans.
HOME
Find the value of x :
Solution:
log 0.25 4
log 0.25 4
 

1
2


5


5
10
x
or x
or x
,
10
,
10
25
100
,
0.25
2
1
2
2
4
2
4


 

















 



or x
x
x

(viii)

21. 4.
Find the value of logarithms of :
(i) to the base of
625 5
log 625
5
log 5
, 4log 5
Ans.
HOME
Solution:
, 4  log 5 1
5
5
4
5




or
or

22. 4.
Find the value of logarithms of :
343 7
log 343
7
log 7
, log 7
, log 7
6
, 6log 7 log 7 1
Ans.
HOME
Solution:
 
 
   
, 6
7 7
7
2 3
7
3
7
or
or
or
or
to the base of





(ii)

23. 4.
Find the value of logarithms of :
to the base of
 
Ans.
HOME
Solution:
0.1 9 3
 
log 0.1
1
let x let
9 3
 
1


2 2
4
5
1
1
2

, 3 3 3
1

2 2
2

, 3 3
5

  


, 3 3
2
5
5
,
 
or
or
or
or
or x
, 2
2
2
9
, 9 3
9
log
1
9
0.1
9
log
2 2
.
9 3
.
9 3
.
    
 

   
 

  

 











or x
x
x
x
x


(iii)

24. 4.
Find the value of logarithms of :
to the base of
1728 2 3
log 1728
2 3
 
log 2 3
6 3
let log 2  3

x
6 3
x
, 2 3 2 3
 
 
x
, 2 3 2 3
x
, 2 3  2 
3
Ans. HOME
Solution:
 
 
 
   
  6
6
 
,    6
2 3 2 3

, 6
3
2
6
2 3
6 3
2 3



 
or
or
or
or
or x
x

(iv)

25. 4.
Find the value of logarithms of :
to the base of
2401 7
log 2401
7
3
log 7
let log 7
x
x
3 4
, 7 7
1

3 4
   

, 7 7
Ans.
HOME
Solution:
 
4
1
,
or
or
or x
3
, 12
4
7
4
7
3
3
3


 




or x
x

(v)

26. 4.
Find the value of logarithms of :
to the base of
2 4
Ans.
HOME
Solution:
log 2
let x
4
8
4


, 4 
2
 
4
8
8


2 8
, 2 
2
, 2 
2

or
or
or
or x
, 2  
8
8
2
,
log 2
2 8
-8

   
or x
x
x
x

(vi)

27. 4.
Find the value of logarithms of :
Ans. HOME
Solution:
to the base of
81 9
log 81
9
let log 81
x
9
 3

x
, 9 81
1

3 2
  


, 9 9
1
3 2
, 9 9
2
1
,
or
or
or
or x
3


, 2 3 6
3
3
3
  
 



or x
x
x

(vii)

28. 4.
Find the value of logarithms of :
to the base of
5 0.008
log 5
0.008
let x
x
, 0.008 5
1

 
1
1
1
Ans.
HOME
Solution:
   
8
1
1

 



 
, 5 5
, 5 5
1
1
1
1
6
or
or
or
or
or
or
or x
or x
,
6
1


2 3
, -
2
, -3
5
5
,
5
125
,
5
1000
,
log 5
2
-3
2
-3
2
3
2
2
3
0.008
 






 


 

 





or x
x
x
x
x
x

(viii)

29. 5.
Find the base when:
is the logarithamof
3 343
let base x
 
log 343 3
3
or x
, 
343
3 3
Ans.
HOME
Solution:
or x
, 
7
, 
7

or x
x
(i)

30. 5.
Find the base when:
is the logarithamof
4 144
let base 
x
 
log 144 4
4
or x
, 
144
4 4 2
or x
, 2 3
 
 
or x
, 2 3
or x
, 2 3
Ans.
HOME
Solution:
 
 
 
 
or x
, 
2 3
, 2 3
4
4
4
4 4
2
2
4 4


 
or x
x
(ii)

31. 5.
Find the base when:
is the logarithamof
let base 
x
  
1
1
Ans.
HOME
Solution:
 
1
3
1
1


 
1
or x
or ,
x
or x
, 
3
, 27
3
,
3
3
log
1
3
3

3
3
3









or x
x
(iii)

32. 5.
Find the base when:
let base 
x
  
Ans.
HOME
Solution:


1
or ,
x a
or x a
a
or x
a
a
is the logarithamof
x



 
,
1
,
1
1
log
1
1
1 1
(iv)

33. 6.
Find the simplest value of :
log 5 log 27
log 27
10
log 3
3log 3
3 25
log 5
10
log 5
10
log 5


Ans.
HOME
Solution:
3
2
or,
10
2log 5
10
log 3
or,
log 5
log 3
or,
log 25
log 3
or,
10
10
2
10
3
10
10
10
10

(i) 

34. 6.
Find the simplest value of :
log 27 if log 3
8 2
log 27
10
log 8
10
log 3
log 2
3log 3
10
3log 2
or,
or,
or,
or, log 3
Ans.
HOME
Solution:
a
a  
a

or, log 3
2
2
10
3
10
3
10

(ii)

35. 6.
log
log
log
1
2
1
x
x
log
Ans.
HOME
Find the simplest value of :
(iii)
x x a
log if log
Solution:
2 2 2

1
log
x
log
x

x
2 1
2
3
2
2
1
1
2
1
log
log
log
log 2
or,
log 2
or,
log 2
or,
log 2 2
or,
log 2 2
or,
x
x


1
1
1
3
or,
3
or,
log
3
or,
log 2
3
or,
3log 2
or,
3log 2
or,
log 2
or,
2
2
3
a
a
x
x
x





36. 7.
Prove that log(1 23)  log1 log2  log3
proved
HOME
Solution:
L H S
. . .
log(1  2 
3)
log(6)
log(1  2 
3)
log1  log 2 
log 3
or,
or,
or,

37. 8.
Express M in term of N
M N
log 3log 1
2
1
 
M N
 
M N
or, log  log 
1


or, log log 3 log 1
solved HOME
Solution:
given :
 
3 3
1
1
1
2 3
M N
or,  
3
2
2
3
3


3 2 6
1
2
3
2 3
3
3
3
2
3
3 3
3 9
or,
3
or,
3
or,
log 3log 1
2
1




N N
M
N
M
N
M
M N a
a
 
   

 




(i)

38. 8.
Express M in term of N
(ii)  
N M
log 3 2log
10 10
N  
M
N M
or, log  log 
3
N M
or, log (  ) 
3
N M
or, log (  )  3 
log 10
or, log ( ) log 10
2 3
or,  
10
or,  
1000
1000
solved
HOME
Solution:
given :
N M
N M
N
M
N
M
N M
1000
or,
l og 3 log
2
2
3
10
2
10
10
2
10
2
10
2
10 10
2
10 10
 

 

39. 9.
Prove that :
log
log
x
a 
x
let y a
x
y a
 
log log
. ..(i)
log
x
x a
a a
log log
a a
x a

  
log 1 [ log 1]
a a
y x
log log
a a
y x
x
a x
proved
HOME
Solution:
or,
log
or, a
f rom (i)
a
a
a





(i)

40. 9.
Prove that :
a
2log 2
2log
x 
a
let y x
a
y x
 
log log
. ..(i)
2log
a
a x
x x
2log log
x x
a x

  
2log 1 [ log 1]
x x
y a
log 2log
x x
y a
or, log log
x x
2
y a
a
2log 2
x a
proved
HOME
Solution:
or,
2
or, x
f rom (i)
x
x
x






(ii)

41. 9.
y x
log log
log
y
x y
(iii) 
z x
 
log log
. ..(i)
log
y
a a
a
y x
log log
a a
log log


z y
log log
a a
a a
log
x
a
z y
y x
log log
proved
HOME
Prove that :
Solution:
or,
log
x
or, a a
f rom (i)
a
a
a a
x y
x y
let z x





42. 9.
m n n m
(iv)   
log log log log
a b a b
n
b
m
b
L H S
m 
n
a b
m
a
n
a


n m R H S
proved
HOME
Prove that :
Solution:
log
log
log
log
. . .
log log
log
log
log
log
or,
or,
or, log  log 
. . .
a b

43. 9.
log 3  log 2 
1
1
2 3
. . .
log 3 log 2
proved
HOME
Prove that :
Solution:
2 3
or, 1 . . .
]
1
log
[ log
log 3
or, log 3
2
2
R H S
a
b
L H S
b
a

 


(v)

44. 9.
b c a
log  log  log 
1
a b c
. . .
b c a
log  log 
log
log
c
proved
HOME
Prove that :
Solution:
log
b
or, 1 . . .
]
log
log
[ log
log
log
log
log
or,
R H S
b
a
b
a
c
b
a
L H S
a
a b c

  

(vi)

45. 9.
Prove that :
   
2 2
log log log( ) log
. . .
x y
log log
x


x y x y a b a b a b
or, log  log log  log [   (  )( 
)]
x


proved
HOME
Solution:
   
  
or, log( )log . . .
2 2
2 2
R H S
y
xy
L H S
y
x y xy
  

 


 

 

 

(vii)

46. 10.
Find the value of :
(i) 
log 729 9 (27)
4 3
log 729 9 (27)
4
4

or, log 3 3 (3 )
4 6 3  2 
4
3
or, log 3 3 3
or, log 3 3
4 6 2
3
or, log 3
or, log 3
or, log 3
4
6
4
solved HOME
Solution:
given :
   
 
 
or, 1
3
4
3
4
3
6
3
4 3
3
6 2 1 3
3
3
1
3
4 3
3
1
3


 








47. 10.
Find the value of :
log log log 81
3 2 3
log log log 81
3 2 3
or, log log log 3 [ 
3 
81]
or, log log 8log 3 [ log 1]
3 2 3
or, log log 2
or, log 3
solved
HOME
Solution:
given :
   
 
or, 1
3
3
3 2
8 8
3 2 3

a a 
(ii)

48. 10.
Find the value of :
log 16 
log 9
2 3
log 16 log 9
2 3
or, log 2  log 3 [ 
2  16 3 
9]
or, 8log 2 4log 3 [ log 1]
solved
HOME
Solution:
given :
       
   
8
4
or,
or, 2
2 3
4 8 4
3
8
2
 

a
and
a 
(iii)

49. 10.
Find the value of :
b c d a
(iv)   
log log log log
a b c d
b c d a
log  log  log 
log
a b c d
log
b
solved
HOME
Solution:
given :
or, 1
log
log
log
d
log
log
log
log
or,
a
d
c
c
b
a
  

50. 10.
Find the value of :
(v)  
log 27 log 8 log 1000
log1.2
log 27  log 8 
log 1000
3 3 3
log 3  log 2 
log 10
12
3




log 3 log 2 log10
3
3
log10
2
 

log 3 3log 2
2
 
 
log 3  2log 2 
log10
 
log 3  2log 2 
log10
solved
HOME
Solution:
given :
 
 
3
3
 
 
3
2
or,
log 3 2log 2 log10
2
or,
log 3 2log 2 log10
2
or,
log 3 2 log10
3
or,
log12 log10
or,
10
log
or,
log1.2
2
2
2 3
 

51. 10.
Find the value of :
(vi)    
log 4 log 5 log 6 log 7 log 3
3 4 5 6 7
log 4  log 5  log 6  log 7 
log 3
log 4
solved
HOME
Solution:
given :
or, 1
log 3
log 7
log 7
log 6
log 6
log 5
log 5
log 4
log 3
or,
3 4 5 6 7
   

52. 10.
Find the value of :
log 6 6 6...
x
let  6 6 6... 
. ..(i)
2
x
or, 6 6 6 6...
 
x x
or, 
6 from (i)
or, 6
  
log 6 6 6... log
6 6
solved
HOME
Solution:
given :
or, log 6 1
6
2
6



x
x
(vii)

53. 11.
25
   
25
  
16 12 7

5
25
5
16
16

16
2
2
25
16
. . .

24


 
64










or, log 2 5 3
81
81


3
3

81
81

28 7







proved HOME
Prove that:
Solution:
 
 
or, log 2 5 1
or, log10
or, 1 . . .
2 5
2 3
3 5
or, log 2
2 5
2 3
3 5
or, log 2
80
24
15
or, log 2
80
log
24
log
15
or, log 2 log
80
7 log
24
12log
15
log 2 16log
1
80
7 log
24
12log
15
log 2 16log
1 64 36 28 24 16 7 28 16 12
28
36 12
16 16
7
4
4
12
3
2
16
4
16 12 7
R H S
L H S

 
 
 

  








 

 


  

 


   
 



 

 


 
 




 
 

 


      
(i)

54. 11.
  
 
7 2 3
25

25


 

 
7 2
5
5
4

6 2
2 3
10

10

 
5 2

7 7
5 2
7 7
5 2
25
25
10
. . .
10



12
12

7 4 3 7 6 12 2 12 14







or, log 5 2 3
81
81
81
3
3
 
proved
HOME
Prove that:
Solution:
 
 0 1 0

 
or, log 5 2 3
or, log 1  2 
1
or, log2 . . .
81

2 5
5
3
or, log
2 5
2 3
3
or, log
3

2 5
2 3
3
or, log
80
24
9
or, log
80
log
24
log
9
or, log
80
3log
24
2log
9
7 log
log 2
80
3log
24
2log
9
7 log
1
12 3
4
14
12 3
6 2
14
3
4
4
2
3
2
7 2 3
R H S
L H S

 


 

  





 

 












 

 


  

 


 

  
 









 

 

     
(ii)

55. 11.
  
 
7 5 3
25

25


 
10
5
16

16
2
2
5
81
81
81
25
25
16
. . .
16




12

 
28





28 15 12 10 7 3 12 7 5






or, log 2 5 3
proved
HOME
Prove that:
Solution:
3
81
 
 
 
or, log 2  5 
3
or, log 2  1 
1
or, log2 . . .
3
2 5
2 3
3 5
or, log
2 5
2 3
3 5
or, log
80
24
15
or, log
80
log
24
log
15
or, log
80
3log
24
5log
15
7log
log 2
80
3log
24
5log
15
7log
1 0 0
12 3
15 5
7 7
3
4
4
5
3
2
7
4
7 5 3
R H S
L H S

 

 

 












 

 


  

 


   
 


 

 







 

 

 

     
(iii)

56. 11.
32
  
32

5
2
2
2

 

  

5
3
75
5
5

 
3 5
3 5
3 5
5
75
. . .
75

 
5
5
2
4

2 5
2


1 4 5 2 2 5 4




or, log 3 5 2
2
32



proved HOME
Prove that:
Solution:
 
 
 
or, log 3  5 
2
or, log 1  1 
2
or, log2 . . .
3
5
2
or, log
3
3
2
or, log
3
3
2
or, log
243
log
9
log
16
or, log
243
log
9
2log
16
log
log 2
243
log
9
2log
16
log
0 0 1
5
2
4
2
5
4
4
2
5
4 2
R H S
L H S

 
 

  
 

 

 

 

 

 

 

 




 



 

 

   
(iv)

57. 11.
(v) x y z
y z z x x y
log  log log  log log 
log
  
y z z x x y
log  log log  log log 
log
let k  x  y 
z
k x y z
or, log log
1
y z z x x y
log  log log  log log 
log
  
y z z x x y
log  log log  log log 
log
k x y z
or, log  log  log 
log
k y z x z x y x y z
or, log  log  log log  log  log log  log 
log log
k y x z x z y x y x z y z
or, log  log log  log log  log log  log log  log log 
log log
proved
HOME
Prove that:
Solution:
 
     
k
or, log 
0
k
or, log log1
k
or, 1
y z z x x y
log log log log log log
or,   
1


  
x y z

58. 11.
Prove that:
1
  
xy yz zx
1
1
. . .
1
xyz xyz xyz
xy yz zx
  
xy yz zx
or, log ( )
xyz
xyz
or, log ( )
1
log ( )
xyz
or, 2log log 1
proved
HOME
Solution:
 
or, 2
log
1
log
or, log log log
1
log ( )
log ( )
2
log ( )
log ( )
log ( )
2

 






 
xyz a
a
b
xy yz zx
L H S
xyz xyz xyz
xyz a
b
a
xyz xyz xyz


(vi)

59. 11.
a b c
  
3 3 3
b c a
. . .
a b c
log log log
3 3 3
c
b
  
1
log
log
b
proved
HOME
Prove that:
Solution:
log
a
log
1
1
27
or,
3
a
1
3
3
or,
log
log
c
3log
3log
3log
or,
log
log
log
log
log
log
or,
1
27
log log log
3 3 3
 
 






 
a
c
b
a
b
a
a
c
b
L H S
b
b c a

(vii)

60. 11.
. . .
2 3
a a a a
log  log  log  ... 
log
n
a a a n a
or, log  2log  3log  ... 
log
or, 1  2  3  ... 
log
proved
HOME
Prove that:
Solution:
 
 



 



( 
1)

    
( 1)
2
log
2
or,
log
2
1
log log log ... log
2 3
n n
a sum of natural number
n n
n a
L H S
a
n n
a a a a
n

(viii)

61. 11.
x y z x y z
(ix)     
log log log log log log
a b c b c a
proved
HOME
Prove that:
Solution:
z
c
z
a
. . .
x y z
log  log 
log
a b c
y
 
b
y
 
c
x
a
x
b
log
log
log
log
log
log
x y z
L H S
log
log
log
or, log log log
b c a
or,
log
log
log
or,
 

62. 11.
Prove that:
x y z
log  log  log  
1
1 1 1
y z x
. . .
x y z
log  log 
log
1 1 1
log
y
 
log
y
log
1
x
x
1
log
log
log
proved
HOME
Solution:
1
or, 1
1
1
1
or,
z
log
z
log
1log
log
1log
1log
or,
log
log
log
or,
1
log
1
log
1
log
or,
1 1 1











 
  
x
z
z
y
y
x
x
z
y
x
z
y
L H S
y z x
(x)

63. 11.
Prove that:
x y z
  
2 2 2
x y z
. . .
x y z
log log log
2 2 2
x y z
  
1
1
1
1
 
proved
HOME
Solution:

 
1
1
8
or,
log 1
2
2
2
or,
1
1
1
2log
1
1
2log
2log
or,
1
log
log
log
log
log
or,
8
log log log
2 2 2
  





 
a
x y z
b
a
x y z
L H S
a
x y z
a
b
x y z


(xi)

64. 12 (a)
log 25
8
log 25
log 8
log 5
2
log 2
3
2log 5
3log 2
10
2
2log

3log 2

2 log10 log 2
3log 2
2 1 
0.3010
2 0.6990
solved
HOME
log 2 0.3010 , log 25. 8 If  find the valueof
 
 
 
0.9030
1.3980
0.9030
1.548
log 2 0.3010 given
3 0.3010
Solution: 
















65. 12 (b)
log 3 , log 5 , log 8. 30 30 30 If  a and  b find the valueof

30 
log 8
log 2
3log 2

30
30
 
  
  
 


 
3 log 30 log 15
30 30
  
3 log 30 log 3 5
30 30
  
31 log 3 log 5
30 30
  
31 log 3 log 5
   a  b



3 1
15
3log
30 30
30
3
30
solved
HOME
 log 3 and log 5 given 30 30   a  b
Solution:

66. 13. (i)
1
1


If a 2  b 
2  7 ab , show that, log a 
b  log a  log b
a  b 
ab
2 2
a b ab ab ab
or,   2  7 
2
2
 
 
a b ab
or,  
9
 
 
 
ab
ab

a b
1
1
1


 
 
 
a b ab
1

a b  a b
a b
ab
a b
ab
a b
log log
2
1
1
( )
3
3
or,
or, log
log
2
( )
3
or, log
log
3
or, log
3
or,
9
or,
7
2
2
2
2
2 2
  




 





 



2
proved HOME
3



Solution:
given :
[ Adding 2ab on both sides ]
[ a2+b2+2ab=(a+b)2 ]
[ taking log on both sides]

67. 13. (ii)
1
y
x
log log  , 23
2

log    
 
 
 
 
1
 
2 2
25

2
or,
2 2
x y xy xy
or,   2 
25
2 2
x y xy xy
or,   25 
2
2 2
x y xy
or,  
23
y
x
or,  
23
y
x
or, 23
log
5
or, log
log log
5
or, 2log
log log
2
5
log
2 2
2
 

 
 

 


x
y
xy
xy
xy
x y xy
xy
x y
x y
x y
x y
x y
proved HOME
5
x
y
x y showthat
x y
If
Solution:
given :

68. 13. (iii)


If a b a b showthat x x x x x , log log 3 5 5 3  
x x x x
3  5 5 
3
  
x
x
5
3
5
3


x x x x
5  3 5   (3 
)

x x x x
a
a b a b
a
b
b a
5 3 5 3
   
b a
x x
2 2 2
b a
or,
or,
or,
or,
or,
or,
or,
2

2
2
2
2


 




or, log log
b
a
b
x
a
b
a
a
x
b
b
a
a
a
a
a
a
b
x
x
x
x
x
x


or, log log
 

 


 





a
b
a



proved HOME
    
Solution:
given :

69. 13. (iv)
4 4 2 2
a  b 
14
a b
4 4 2 2 2 2 2 2
a b a b a b a b
or,   2  14 
2
 
   
2 2 2 2 2
a b a b
or,  
16
2 2 2 2
a b ab
or,  
4
2 2
a b ab
or,  
4
 a 2 b 2
  ab

 
 
 
or, log  
log 4
e e
2 2
a b a b
or, log   log 4  log 
log
e e e e
2 2 2
a b a b
or, log   log 2  log 
log
e e e e
2 2
a b a b
or, log   2log 2  log 
log
e e e e
or, log  a 2  b 2
  log a  log b

2log 2
e e e e
proved
HOME
14 , log   log log 2log 2 4 4 2 2 2 2
e e e e If a  b  a b showthat a  b  a  b 
Solution:
given :
[ Adding 2a2b2 on both sides ]
[ a4+b4+2a2b2 =(a2+b2)2 ]
[ taking log on both sides]

70. 14. (a)
y
y
x
x
log log log
 
 
 
x k y z
  
z
z
log . ... .(i)
y k z x
log  
. ... .(ii)
z k x y
log . ... .(iii)
     
 
from (i), (ii) and (iii) :
x y z k y z k z x k x y
l og log log
       
 
or, log
x  y  z  ky  kz  kz  kx  kx 
ky
xyz
or, log( ) 
0
or, log( ) log1
or, 
1







xyz
xyz
k
x y
z x
y z
let
HOME
, 1
log log log






prove that xyz
x y
z x
y z
If
Solution:
proved
[ log1=0]

71. 14. (b) (i)
x
y
x
log log log
y
z
z
   
   
   
x k b c a x ak b c
log log . ... .(i)
     
y k c a b y bk c a
log    log  
. ... .(ii)
z k a b c z ck a b
log    log  
. ... .(iii)
     
from (i), (ii) and (iii) :
a log x  b log y  c log
z  ak b  c  bk c  a  ck a 
b
a b c
x  y  z  akb  akc  bkc  bka  cka 
ckb
or, log log log
a b c
x y z
or, log(   ) 
0
a b c
x y z
or, log( ) log1
or, 
1







a b c
x y z
k
a b
c a
b c
let
HOME
, 1
log log log






a b c provethat x y z
a b
c a
b c
If
Solution:
proved
[ log1=0]

72. 14. (b) (ii)
y
x
y
x
log log log

z
z
         2 2

         
         
x k b c b c x b c k b c k b c
         
log log . ... .(i)
2 2
y k c a c a y c a k c a k c a
log     log     
. ... .(ii)
2 2
z k a b a b z a b k a b k a b
log     log     
. ... .(iii)
from (i), (ii) and (iii) :
 b  c  log x   c  a  log y   a  b  log
z  k  b  c   k  c  a   k  a 
b

 b  c   c  a   a 
b

       
       
x y z kb kc kc ka ka kb
or, log log log
       
b  c c  a a 
b
x y z
or, log   
0
b  c c  a a 
b
x y z
or, log log1
      or, 1
2 2 2 2 2 2
2 2 2 2 2 2







b  c c  a a 
b
x y z
k
a b
c a
b c
let
HOME
, 1
log log log






bc ca ab provethat x y z
a b
c a
b c
If
Solution:
[ log1 = 0]
proved

73. 14. (b) (iii)
y
x
y
x
log log log
z
z
   
   
   
x k ry qz p x pk ry qz
     
log log . ... .(i)
y k pz rx q y qk pz rx
log    log  
. ... .(ii)
z k qx py r z rk qx py
log    log  
. ... .(iii)
     
from (i), (ii) and (iii) :
p log x  q log y  r log
z  pk ry  qz  qk pz  rx  rk qx 
py
p q r
x y z kpry kpqz kqpz kqrx krqx krpy
or, log log log
       
 x p y q z
r

 
or, log   
0
p q r
x y z
or, log log1
or, 
1







p q r
x y z
k
qx py
pz rx
ry qz
let
HOME
, 1
log log log






p q r prove that x y z
qx py
pz rx
ry qz
If
Solution:
[ log1 = 0]
proved

74. 15.
log 45
2
 
log(3 5)
2
 
log 3 log 5
 
2log 3 log 5
10


  
2log 3 log10 log 2
   
2 0.4771 1 0.3010
 
0.9542 0.6990
1.6532
2
2log 3 log





 
HOME
If log 2  0.3010, log3  0.4771, log 7  0.8451, f ind the value of :
Solution:
(i)
solved
[ log10 = 1]

75. 15.
log108
2 3
 
log(2 3 )
2 3
 
log 2 log 3
 
2log 2 3log 3
   
2 0.3010 3 0.4771
 
0.6020 1.4313
2.0333

HOME
If log 2  0.3010, log3  0.4771, log 7  0.8451, f ind the value of :
Solution:
(ii)
solved

76. 15.
log 84
2
  
log(2 3 7)
2
  
log 2 log 3 log 7
  
2log 2 log 3 log 7
   
2 0.3010 0.4771 0.8451
   
0.6020 0.4771 0.8451
1.9242

HOME
If log 2  0.3010, log3  0.4771, log 7  0.8451, f ind the value of :
Solution:
(iii)
solved

77. 15.
log 294
  
log(2 3 7 )
  
log 2 log 3 log 7
  
log 2 log 3 2log 7
   
0.3010 0.4771 2 0.8451
   
0.3010 0.4771 1.6902
2.4683
2
2

HOME
If log 2  0.3010, log3  0.4771, log 7  0.8451, f ind the value of :
Solution:
(iv)
solved

78. 15.
log 21.6
216
10
log


 
log 216 log10
 3 3

  
log 2 3 log10
3 3
  
log 2 log 3 log10
  
3log 2 3log 3 log10
    
3 0.3010 3 0.4771 1
   
0.9030 1.4313 1
1.3343






HOME
If log 2  0.3010, log3  0.4771, log 7  0.8451, f ind the value of :
Solution:
(v)
solved
[ log10 = 1]

79. 16 (i)
given
If three positive real numbers a b and c are inG P
, . .,
solved
showthat a b and c are in A P
a, b and c are in G.P.
c
b
 
b
c


log log






 



b a c b
log  log  log 
log
a b c A P.
a
b
b
a
hence, log , log and log are in .
HOME
log , log log . .
Solution:

80. 16 (ii)
log
log
x k
log
 
log
log
log
  
log . ... .(i)
y k
log 
2 . ... .(ii)
z k
log 
3 . ... .(iii)
from (i), (ii) and (iii) :
y x k k k
log  log  2  
. ... .(iv)
z y k k k
log log 3 2 . ... .(v)
from (iv) and (v) :
y x z y
   
log log log log
y
z
z



Hence, x, y and z are in G.P. y
y
x
y
x
k
x y z
let


 

 

 


   
log log
3
2
1
HOME
, , , . .
3
2
1
provethat x y z are inG P
x y z
If  
Solution:

81. 17.
The f irst and the last terms of aG P are a and k respectively
If the number of termbe n provethat
k a
log log
given First term = a
Last term = k
No. of term = n
Common ration = r
. . .

Last terminG P ar
1 k ar
log 
log
k a r
log  log 
log
k a n r
log  log  
1 log
k a n r
log log 1 log
proved HOME
.
log
1
, ,
where r is the commonratio
r
n

 
Solution:
 
 
 
  
n
k a
log log
k a
log log
r
n
r
k ar
n
n
n
n
 

 







1
log
1
log
. .
1
1
1

82. 18.
If the p q and r terms of aG P are a b and c respectively showthat th th th
given
, . . , ,
q  r a  r  p b  p  q c 
pth term = a
qth term = b
rth term = c
let the first term = b
and common ratio = R
pth term = a = b Rp-1
qth term = b = b Rq-1
rth term = c = b Rr-1
Press
( ) log ( ) log ( ) log 0
Solution:

83. since a, b and c are in G.P.
- - - (ii)
- - - (iii)
- - - (i)
b
q p



 
R
1
1 1
R

1
q
1
1
1 1
1
1
R
b
c
b
b


and
b
c
b
R R
R
c
b
R
a
R R
R
a
b
a
r
r q
r q r q
q
q p q p
p

 



  

  


   

   

b
b

Press

84. from (i), (ii) and (iii) :
1 1
b q p c
r q
  
c

 

 
q 
p
q p
b
r q
c
b

b




r  q q  p r  q q 
p
b  b  a 
c
r  q  q  p r  q q 
p
b  a 
c
r  p r  q q 
p
b  a 
c
r p r q q p
r q
r q q p
b a c
a
b
a
b
a
  

 
 










log log log
Press

85. r p b r q a q p c
(  ) log  (  ) log  ( 
) log
r p b q r a p q c
(  ) log   (  ) log  ( 
) log
q r a p q c r p b
(  ) log  (  ) log  (  ) log 
0
q r a r p b p q c
(  ) log  (  ) log  (  ) log 
0 solved
HOME

86. 19. (i)
If x  bc y  ca and z 
ab showthat a b c
1
x y 1
z
1
x y z
1
1
bc ca ab
a b c
bc a ca b ab c
a a b b c c
1
1
1
1
1
. .
  
abc abc abc
a b c
a b c
  
log log log
log
1
1
1
1
log ( )
log ( )
log ( )
1
log ( ) log
log ( ) log
log ( ) log
log ( ) 1
log ( ) 1
log ( ) 1
1
1
1
1
1
1



















abc abc abc
abc
L H S
abc
proved HOME
1
1
( )
log ( ), log ( ) log ,






i
Solution:

87. 19. (ii)
given
If x  log ( bc ), y  log ( ca ) and z 
log ab ,
showthat a b c
( ) 2
x bc
log ( )
a

y ca
log ( )
b

z ab
log ( )
c

hence
1
1
1
bc ca ab
a b c
x y z
x y z
y x
  
1 1
x y
 
 
1 1
( 1) ( 1)
x y z z xy x y
(   2)(  1)  (   
1)
xz  x  yz  y  z   xyz  xz  yz 
z
x  y  2 z  2
 xyz 
z
x  y  2 z  z  2

xyz
x y z xyz
z
z
xy x y
z
z
x y
   


  


 

 















2
2 2
2
( 1)
( 1)( 1)
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
log 1
log 1
log 1
proved
HOME
ii x  y  z  
xyz

88. 20.
given
a a a
. . .
1
   
log log 2 log 3 1
a a a
2 3 4
a
   
a
L H S
log
log
log 2
 
a a
log log 4
a
log 4
a a
log( 4 )
log 4
a
log(2 )
a
a
a
2log 2
log 4
log 2
a


   
a
a
log 3
log 2
log 3
log 4
3
  
2 log 2 log 3
proved
x a
a
2

y log 2
a
a
3

z log 3
a
a
log
4

yz
a
a a
a
a
a
a
a
a
a
a
a
a
xyz
a a
a
2
log 4
2
log 3
2
log 4
1
log 4
1
log 4
log 3
log 2
3 4
3
2






HOME
If x a y a and z a showthat xyz yz a a a log , log 2 log 3 , 1 2 2 3 4     

89. 21.
log  , log  , prove log 
p q p 
given
p
q
1
p q
. . .
1
x
log
log log
x x
1
ab
proved
x 
a
x b
p
q

log
log
L H S
p
1
1
1 1
a b
b a
ab
ab
b a
x x
x
p q
q











1
log
log
1
log
HOME
b a
If x a and x b that x
q

90. 

If log x y a, and log , log log 2 3   
22.
Solution:
given:
x
log( x 2 y 3
) a
- - - (i) b
x


log - - - (ii)
y
  

 


solved
(i)
from
2 3
x y a
log log
 
x y a
2log  3log 
- - - (iii)
x y b
log log
 
log x  log y 
b
- - - (iv)
from (iii) and (iv)
 
y b y a
2 log 3log
  
y b y a
2log 2 3log
  
y a b
5log 2
2
5
log
(ii)
 
a b
y
from

 
and from
a 2
b
b
2 5
3
a b
2
3
5
log
5

log

5
5
5
log
(iv)
a b
y
a b
x
a b b
x



 
 




HOME
  b prove that x and y in terms of a and b
y

 



91. 23.
e
y
2
 2

solved
y y
e 
e
1
If x  y 
y e
y y
e e


y y
y y
e e


y y
e e
e e
y  y y 
y

e  e  e 
e
e e
y
y y
e
y y
y y
e e


y y
e e
y y y y

likewise
 
e  e  e 
e
y y
1
- - - (ii)
e
2
1 1
- - - (i)
2
1 1
e e
y
y y
e e
x
e e
x
x













  





  
y
y y
y
 


1
1
e e
x
x
from and
x
y
y
e
e e
e
x
x
y e y
x
x
x
e
x
e e
e
e e
x
e
e
y
e e
y y y
y
y y


1
1
 



 



  






1
1
1
1
1
log
2
log
2
2 log 2
1
log
log
1
log
2
2
1
(i) (ii)
2
2
HOME
x
x
show that y
e e





1
log
2
log ,
Solution:
given:

92. 24.
log 3 10 Showthat the valueof lies between and
Solution:
HOME
2
.
5
1
2
9 10
log 9 log 10
10 10
- - - (i)
log 3 
1
2log 3 1
1
2
log 3
10
10
2
10



 
and
243 100
log 243 log 100
10 10
log 3 
log 10
5log 3 
2log 10
10 10
- - - (ii)
5log 3 
2
2
5
10
log 3
10
2
10
5
10


 
from i and ii
2
5
log 3
1
2
( ) ( )
10  
proved

93. 25.
Solution:
HOME
log 10 log (32 ) 5, . 6 If b and b find the value of a a a  
a 
log 10 (given)
10
a b
log (32 ) 5 (given)
5
6
a b
(6 ) 
32 - - - (ii)
from (i) and (ii)
5
10

b
a
(6 ) 32

6
2
5
6
2


3
1
1
2
10
5 5
10 5
a
6
32
6
- - - (i)
5
5
5
5
5 5
a
 






 


a
a
a
a
b
a
b
and
b
a
found

94. 26.
Solution:
HOME
found

95. 27. (i) Solve
Solution:
HOME
4
solved
x
x x
2
10
log  log 
10 10 log
2
x

x


10 1
x
x
x
x
x
x

x x
x
x x
2
2
2
10
1
2
10
10


1
2
1
10
2 10
10
10
2
10
10 10
log
log
log
log
log
log
log
log
log
log log




 






 





 

 

 




 


2
2 2
1
x x
log  log 
4
10 10
x
(log ) 4
x
log  4  
2
1

or , 10
x
2
hence
x or
10 10
100
100
log
log
log
log
log
2
2 2
10
2
10
10 10
10
10
10

x or
x x
x
x
x
 






 

96. 27. (ii) Solve
Solution:
log log log 1 2 2 2 x 
 x
log log log 1
2 2 2
 
log log log log 2
2 2 2 2
log log 2
2 2

x
log log  2 
log 2
2 2 2
x
log log 
log 2
log log log 4
2 2 2
HOME solved
log 4
 2

2
4
, 16
2
2 2 2




or x
x
x
x
again
x
x

97. 27. (iii) Solve
Solution:
HOME
log log log 11 8 4 2 x  x  x 
log  log  log 
11
8 4 2
1
   
x x x
1
  
x x x
  
1
1
1

  
2 3 6
6
1
1

solved
1
1
1
1
x
1
x
1
x
1
 

x
log 6
, 64
2
6
log 2
11
11
log 2
11
6
log 2
1 11
2
3
log 2
11
1
log 2
2log 2
3log 2
11
log 2
log 2
log 2
11
log 2
log 4
log 8
6
2
3 2



  

 

 
or x
x
x x x
x
x x x


98. 28.
If a b and c are three consecutive positive integers showthat
log(1 ) 2log
Solution:
HOME
, ,
 
a b and c are three consecutive positiveintegers
proved
ac b
b ac
  1

,
2
2
b ac
log  log(1 
)
b ac
2log  log(1 
)
