Chapter 13: Logarithms
13
1.
2.


3.
Logarithms
2x.3x+4 = 7x
Taking log on both the sides, we get
loge 2x + loge 3x+4 = loge 7x
 x loge 2 + (x + 4) loge 3 = x loge 7
 x loge 2 + x loge 3  x loge 7 = 4 loge 3
 x loge 6  x loge 7 = 4 loge 3
4log e 3
x=
log e 7  log e 6
5.
1/ 2

log 7 log 5 ( x 2  5  x )  0
1/2
0
 log5(x + 5 + x) = 7
 (x2 + 5 + x)1/2 = 51
 (x2 + x + 5) = 25
 x2 + x  20 = 0
 (x  4)(x + 5) = 0  x = 4, 5
log 2 .log 3 ....log100 100
= log 2 log 3 ....log 99 99
1
..2
98.
= log 2 log 3 ....log 98 98
=



log 2 2log 5  2log 2 


=
2 
 25  
 log  10  
=
log y = log 2  y = 2
6.
Since, c2 = a2 + b2  c2  b2 = a2 ….(i)
log a
log a

log c  b a  log c  b a
log(c  b) log(c  b)
=
2log a  log a
2log c  b a  log c  b a
log(c  b)log(c  b)

=
=
log100 100
 

log a log(c 2  b 2 )

2log a log a
log a.log a 2
2log a log a
….[From (i)]
log 99 99
….[ log100 100 = 1]
log 2  2log5  2log 2 


2  log5  log 2 

.21
..
98
99
1
..2
98.
1/ 2
 1/3  

 16 log 2.5 1/ 2  
log 2.5 

 11/3  
=  (0.16)
 




 100 

1
1
 16 
log2.5   . log 
log y =

2
2
 100 
1
 4 
log2.5 2.log  
2
 25 
 log 2
.(log 4  log 25)
=
2log (2.5)
x = loga bc
 1 + x = 1 + logabc
 1 + x = loga a + loga bc = loga abc
(1 + x)1 = logabc a
Similarly, (1 + y)1 = logabc b and
(1 + z)1 = logabc c
(1 + x)1 + (1 + y)1 + (1 + z)1
= logabc a + logabc b + logabc c
= logabc abc = 1
2
4.
1
1 1 1


log 2.5   2  3 ....   2
3 3 3


Let y =  (0.16)


=1
7.
1
= log2 log3 32 = log2 21 log3 3
= log2 2 = 1

Let y = 340
Taking log on both the sides, we get
log y = log 340
 log y = 40 log 3  log y = 19.08
Number of digits in y = 19 + 1 = 20
1
Std. XI : Triumph Maths
1
1
1
1


 ....... 
log 2 n log 3 n log 4 n
log1983 n
8.
= log n 2  log n 3  log n 4  ......  log n 1983
= log n (2.3.4....1983)  log n (1983!)  log n n  1
log0.3 (x  1) < log0.09 (x  1)
log ( x  1)
 log0.3 (x  1) <
log (0.3) 2
9.
1
log0.3 (x  1)
2
1
 log0.3 (x  1)  log0.3 (x  1) < 0
2
 log0.3 (x  1) < 0
 (x  1) > (0.3)0
x>2
 x  (2, )
 log0.3 (x  1) <
10.
log 3 x  log 3 x  log 3 4 x  log 3 8 x  ...  4
1
1
1
 log 3 x  log 3 x  log 3 x  log 3 x  ...  4
2
4
8
 1 1 1

 log 3 x  1     ....   4
 2 4 8

 1 
 log 3 x 
4
 11 / 2 
 log3 x = 2
 x = 32 = 9
2