Chapter -24 Equations InEquations, Log & Polynomials
EXPONENTIAL AND LOGRITHMIC FUNCTION
Exponential Function : A function f  x  defined by f  x   a x , a  0, x  R is called an exponential
function. The positive number a is called the base of the function. The negative base exponential
function is not a real function.
Results on a x when a  1.
(i)
a x  0 for all x.
(ii)
a x is monotonically increasing.
(iii)
From the graph it follows that :
x
ax
 ,0 
 0,1
 0,1
1,a 
1, 
 a,  
Results on a x : When 0  a  1
(i)
a x  0 for all x
(ii)
a x is monotonically decreasing for all x.
(iii)
From the graph, we have
x
ax
 ,0 
1, 
 0, 
 0,1
Logrithmic Function : If y  a x then we say x  log a y which is the inverse function. On
interchanging x and y the inverse function is given by y  log a x . log a x is defined if
x  0, a  0, a  1and if log a x  y then x  a y .
Results :
(i)
log a  xy   log a x  log a y if x, y  0, a  0, a  1 . Domain is curtailed.
(ii)
x
log a    log a x  log a y if x, y  0, a  0, a  1 . Domain is curtailed.
 y
(iii)
loga x y  y loga x if x  0, a  0, a  1
(iv)
log a 1  0 if a  0, a  1
(v)
log a a  1 if a  0, a  1
491
(vi)
log y x 
1
, x, y  0, x  1, y  1
log x y
(vii)
log y x 
log z x
, x, y, z  0; y  1, z  1
log z y
(viii)
a loga x  x, x  0, a  0, a  1
2.
Analytical Results on log a x when a  1
(i)
log a x is monotonically increasing i.e., if x1  x2  0 then log a x1  log a x2 .
(ii)
If log a x  y , then x  a y and if log a x  y then 0  x  a y .
(iii)
From the graph we have,
x
log a x
 0,1
 ,0 
1,a 
 0,1
 a,  
1, 
(base change formula)
(inversion formula)
Analytical results on log a x when 0  a  1
(i)
log a x is monotonically decreasing i.e., if x1  x2  0, then log a x1  log a x2 .
(ii)
loga x  y  0  x  a y ;loga x  y  x  a y
(iii)
From the graph, we have
x
log a x
 0,1
 0, 
1, 
 ,0 
Monotonicity of the function log x a :
The function log x a is defined if a  0, x  0, x  1. If a  1 then the function is monotonically
decreasing in  0,1
1,  . If
0  a  1, then it is monotonically increasing in  0,1
1,  .
Thus log 5.1 3  log5.01 3; log 0.11 3  log 0.01 3
In above illustrations a  3, which is greater than one. Note that monotonically decreasing character
of log x 3 is there if either both numbers are taken in  0,1 or both taken if 1,  . Since the function
is discontinuous at 1.
492
1
1
1
1
1
Again taking a  , log5.1  log5.01 ; log0.11  log0.01 showing that log x a is monotonically
3
3
3
3
3
increasing in  0,1
1,  .
SOLVED EXAMPLES
log 2 24 log 2 192

3
log96 2
log 2 2
Example 1 :
Prove that
Solution :
Changing to base 10 LHS

 


 
3
5
6
2
log 24 log 96 log192 log12 log 2  3 log 2  3  log 2  3 log 2  3



2
log 2 log 2
log 2 log 2
 log 2

 3log 2  log35log 2  log3   6log 2  log3 2log 2  log3 
2
 log 2
15  log 2 2  8  log 2  log3   log32  12  log 2 2  8  log 2  log3   log32  



2
2
log
2
log
2





3  log 2 
 log 2
2
2
3
Example 3 :
If log10 15  a, log 20 50  b, show that log9 40 
Solution :
First equality gives log3  log5  a
5b
2  a  2b  ab  1
(i)
The second equality can be written as (on base 10)
log 50
log 5  log10
b
b
log 20
log 2  log10
log 5  1
log 5  1
log 5  1
b
b
b
10
log10  log 5  1
2  log 5
log  1
5
On solving for log5 , we get log5 
From (i), we have log3  a 
Now log 9 40 

2b  1
1 b
2b  1 a  ab  2b  1

1 b
1 b
log 40 log 4  log10

log 9
log 32
2  log10  log5  1
2log3

(ii)

(iii)
2log 2  1 2  log10 / 5  1

2log3
2log3
5b
3  2 log 5

using (ii) and (iii)
2
a

2
b  ab  1
2 log 3

493

Example 5 :
Solve the equation 6.4 x  13.6 x  6.9 x  0,
Solution :
4
3
On dividing by 6 , we get 6    13  6.    0
6
2
x
x
x
x
6
2
Putting    t , we get  6t  13   0
t
3
x
3
2
3 2
2
    , or  x  1 or  1
 6t  13t  6  0  t  ,
2
3
2 3
3
2
Example 8 :
Solve log 2  3  x   log 2 1  x   3
Solution :
The equation is defined if 3  x  0 and 1  x  0, if these conditions are satisfied
(i)
then the given equation can be written as
log 2  3  x 1  x   3
(ii)
  3  x 1  x   23  x2  4 x  5  0
  x  5 x  1  0  x  1,5
By direct verification we note that only x  1 satisfies the original equation.
5 does not satisfy the inequalities 3  x  0, 1  x  0
NOTE :
(a) It is advisable to check all the roots in case of equations inequations involving
logs. For any root the expression whose logs have been taken in the original
equation must be positive. Further for any root the base should be a non unity
positive number. If a root does not satisfy these conditions it should be considered
as an extraneous root.
(b) By writing log x  log y  log xy, in above examples we extended the
domain of the equation, since log xy is not only defined for x, y  0 , but is also
defined if x  0, y  0 on the contrary log x  log y is defined if and only if
x  0, y  0 . This explains the appearance of an extraneous root 5.
You must note that 5 is a root of (ii) but it is not a solution of the original
equation (i).
Example 12 :
Solution :
Solve the inequation 2x  2 x1  3  0
2
On putting 2 x  t , we get t   3  0
t
On multiplying by t , we get (note that t  0 )
  t 2  3t  2   0 or  t  1 t  2   0
494
 t  1,2   1  2x  2
On taking logs to base 2, we get log1  x  log 2  0  x  1
Note : From the inequality a  c  b we can conclude that log a  log c  log b
provided (1) a, b, c  0. (2) log is taken at a base which is greater than unity.
Example 18 : Solve 2  log 2  x 2  3x   0
Solution :
The inequation makes sense if x 2  3x  0
 x   , 3
 0,  
(i)
For these values the given inequation is equivalent to
log 2  x 2  3x   2  x 2  3x   2   4
2
 x2  3x  4  0   x  4  x  1  0
 x   4,1
(ii)
Taking the intersection with (i) we get the required solution as  4, 3
Example 19 : Solve log 2 x 
Solution :
 0,1.
2
log 2 x  1
On putting t  log 2 x, the given inequation becomes
t
 t  2 t  1  0
t2  t  2
2
2
0
t 
0 
t 1
t 1
t 1
t 1
 t   , 1
 x   2
1
1,2
 log 2 x  1 or 1  log 2 x  2
or 21  x  22  x 
1
1

or 2  x  4  x   , 
2
2

 2,4
But, as the original inequation makes sense if x  0 the required solution set is
 1
 0, 
 2
 2,4.
Example 21 : Solve log x3  x  1  2
Solution :
The inequation makes sense if x 1  0, x  3  0, x  3  1
 x  1, x  3, x  4  x  3, x  4  x   3, 4 
 4,  
The base of given log x  3 in the given inequation lies between 0 and 1 if
x   3, 4  and is greater than unity if x   4,   . Therefore we solve the given
inequation over the two intervals separately. Over  3, 4  the inquation is equivalent
to
x 1   x  3  x2  7 x  10  0   x  2 x  5  0
2
 x   2,5  Solution set S1 on  3, 4  is S1   3,4  itself.
Over  4,  the inequation is equivalent to  x  1   x  3
495
2
  x  2  x  5  0  x   ,2 
 5,  
S2   5,  
The required solution of the given inequation is S1
  3,4 
S2
 5,  
EXERCISE
1.
Prove the following :
(i)
log3 .5log 25 27 
(iii)
log3 135 log3 5

3
log15 3 log 405 3
2.
If log5  a,log3  b, show that log 30 8 
3.
log 30 3  a,log 30 5  b, show that log30 8  31  a  b 
4.
If log196  a,log56  b, show that log .175  5a  6b  4
5.
If log12 18  a,log 24 54  b, show that ab  5  a  b   7
6.
If log 3 7  a,log 7 5  b,log 5 4  c, show that log3 12  abc  1.
7.
If a 2  b2  7ab and a, b  0, show that log
3
2
(ii)
log 3 5.log 4 9.log 5 2  1
(iv)
log  tan1  log  tan 2   ......  log  tan89   0
3 1  a 
1 b
4
ab 1
  log a  log b 
3
2
OBJECTIVE EXERCISE
125 625
1.
The value of log 5
2.
(a)
725
(b)
6
The value of log 2 10  log8 125 is
(a)
3.
(b)
is
1
(c)
3125
(d)
5
(c)
0
(d) 1  2log 2 5
(c)
2a
3 a
(d) 5 
If log12 27  a , then log 6 16 equals
(a)
4.
1  log 2 5
25
1 a
a
(b)
 3 a 
4

 3 a 
If  log5 k  log3 5 logk x   k , then the value of x equals
496
 2a 

 2a
5.
6.
(a)
k3
(b)
5k
k5
(c)
Given that log p x  α and log q x  β , the value of log p / q x equals
(a)
αβ
β α
(b)
β α
αβ
(d)
3k
(d)
αβ
αβ
αβ
αβ
(c)
If x is a positive number different from 1 such that log a x,logb x and log c x are in A.P.,
then
c 2   a.c 
(a)
7.
8.
log a b
b
(b)
If x is a real number and y 

1
2
e
x
ac
2
 e  x  , then

(c) b  a.c

(a)
x can be either log y  y 2  1 or log y  y 2  1
(b)
x can only be log y  y 2  1
(d)
x can be either log y  y 2  1 or log y  y 2  1





x can only be log y  y 2  1
(c)

(d) None of these


The remainder obtained when the polynomial x  x3  x9  x 27  x81  x 2
243
by x  1 is
(a)
6x  1
is divided
2
9.
π
12
4x
(c)
sin 2 x
  81
(d)
cos2 x
6x
 30 is
π
π
(c)
(d) None of these
6
8
It is given that the expression ax 2  bx  c takes negative values for all x  7 . Then
(a)
the equation ax 2  bx  c  0 has equal roots
(b)
a is negative
(c)
11.
5x  1
The smallest positive solution of the equation  81
(a)
10.
(b)
(b)
a and b are both negative
a
The minimum value of the quantity
(d)
2
None of these
 3a  1 b2  3b  1 c 2  3c  1
abc
, where
a , b and c are positive real numbers, is
12.
113
(a)
(b)
125
(c)
25
23
If log 30 3  a and log30 5  b , then log30 8 is equal to
(a)
13.
a  b (b)
3 1  a  b 
(c)
8
1  a  b 
3
(d)
(d)
27
1
1  a  b 
2
The remainder R  x  obtained by dividing the polynomial x100 by the polynomial
x 2  3 x  2 is
(a)
2100  1
(b)
(c)
2100 x  3.2100
(d)
497
2
2
100
 1 x  2  299  1
100
 1 x  1 299  1

14.
lying
15.
The number of values of k for which the equation x2 – 3x + k = 0 has two distinct roots
in the interval (0, 1) is
(a)
three (b)
two
(c)
infinite
(d)
no value of k will satisfy
2
If ,  are the roots of the equation ax + bx + c = 0, then the value of 3 + 3 is

16.
17.
18.
n
19.
20.
21.
22.
23.
24.
25.
(a)
cx 2  ax  b  0
(b)
bx 2  ax  c  0
(c)
cx2  bx  a  0
(d)
ax 2  cx  b  0
If the equation (  1) x2  (  1) x  (  1)  0 has real roots , then  can have any value
in the interval
(a)
(b)
(c)
(d)
(3,3)
(0, 3)
(1/ 3, 2)
(1/ 3,3)
Let |x| and [x] denote the fractional integral part of a real number x respectively . Then
solutions of 4|x| = x+[x] are
2
4
5
(a)
(b)
(c)
(d)
2, 0
 ,0
 ,0
0,
3
3
3
The value of ‘a’ for which the equations x3+ax+1=0 and x4 + ax + 1 = 0 have a common
roots is
(a)
2
(b)
-2
(c)
0
(d) none of these

log x2 4 x 5

The real roots of the equation 7 7
 x 1
(a)
1 and 2
(b)
2 and 3
(c)
3 and 4
(d)
4 and 5
The minimum value of (x – p)2 + (x – q)2 + (x – r)2 will be at x =
pqr
3 pqr
(a)
(b)
(c)
pqr
(d)
p2 + q2 + r2
3
If two equations x2 + a2 = 1 – 2ax and x2 + b2 = 1 – 2bx have only one common root, then
(a)
a–b=1
(b)
a – b = -1
(c)
a–b=2
(d)
a - b=
1
26.
27.
28.

 3abc  b3
3abc  b3
3abc  b3
a 3  b3
(b)
(c)
(d)
3abc
a3
a3
a3
If the ratio of the roots of the equation x2 + bx + c = 0 is as that of x2 + qx + r = 0, then
(a)
r2b = qc2
(b)
r2c = qb2
(c)
c2r = q2b
(d)
b2r = q2c
2
If the sum of the roots of the quadratic equation ax + bx + c = 0 is equal to the sum of the
square of their reciprocals, then
(a)
ab2, ca2, bc2 are in A.P
(b)
a2b, c2a, b2c are in A.P
2
2
2
(c)
ab , ca , bc are in G.P
(d)
none of these
If f(x) = 2x3 + mx2 – 13x + n and 2, 3 are roots of the equation f(x) = 0, the values of m and
are
(a)
5, -30
(b)
-5, 30
(c)
5, 30 (d)
none of these
1 1
, are the roots of
If
,
are the roots of ax 2  bx  c  0, and
 
(a)
1
1
1
equals

 ...... 
log 2 x log3 x
log1998 x
(a)
-1
(b)
0
(c)
1
(d)
198
The number of values of k for which (log x)2 – log x – log k = 0 is/are:
(a)
1
(b)
2
(c)
3
(d)
4
If x  1998! , then value of the expression
7
4
The value of ' c ' for which α 2  β 2  , where  and  are the roots of
2 x 2  7 x  c  0, is
498
(a)
29.
0
(b)
2
(c)
4
(d)
6
(c)
a 5
(d)
2a5
If x  2ax  10  3a  0 for all x  R, then
2
(a)
5 a  2
(b)
a5
( a  b  c)
always belongs to :
(ab  bc  ca)
(c)
(d)
[2,3]
[3, 4]
2
30.
If a, b, c are sides of a triangle, then
(a)
31.
[1, 2]
(b)
[4,5]
If  ,  are the roots of the equation  ( x 2  x)  x  5  0 and if 1 and 2 are two values
of  obtained from
(a)
4192
  4
 
  , then 12  22 equals:
2 1
  5
(b)
4144
(c)
4096
(d)
4048
6.
12.
18.
24.
30.
(a)
(a)
(b)
(a)
(c)
ANSWERS
1.
7.
13.
19.
25.
31.
(d)
(a)
(b)
(c)
(c)
(d)
2.
8.
14.
20.
26.
(b)
(b)
(d)
(d)
(c)
3.
9.
15.
21.
27.
(b)
(d)
(c)
(c)
(b)
4.
10.
16.
22.
28.
499
(d)
(c)
(d)
(b)
(d)
5.
11.
17.
23.
29.
(a)
(c)
(a)
(b)
(a)