1
IOQM 2.0 2025
Basic Mathematics
1.
1
log bc abc
+
1
log ca abc
+
1
log ab abc
has the value
6.
equal to
2.


If log3 ( log2a ) + log 1  log 1 b  = 1 , then the value


3
2 
m
log10   , where m and n are relatively prime
n
positive integers. Find m + n − 100 .
4.
If xy2 = 4 and log3(log2x) + log1/3 (log1/2y) = 1,
then x equals
8.
Find the sum of all integral values of x satisfying
5
( log5 x )2 + log5x   = 1 .
x
9.
Let
If P = log5 (log5 3) and 3C+5 = 405 then C is
The value of


56 + 56 + 56 + 56 + ..... 

is equal to
log 4


3
64
64
64......



5.
7.
–P
equal to
1
. Suppose b = a2 + a3 + a4 + a5
log n 2002
and c = a10 + a11 + a12 + a13 + a14 . Then b − c
equals
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x = 4log2 9
k −1
+7
and
1
y=
and
5 k −1
32log2 3 +1
xy = 4 , then the sum of the cubes of the real
value(s) of k is
Suppose n be an integer greater than 1, let
an =
There is a positive real number x not equal to
1
1
either
or such that
2
20
log20x ( 22x ) = log2x ( 202x ).
The value log20x ( 22x ) can be written as
of ab3 is
3.
DPP-07
10.
The expression
2
2
2

 
 

 log a  +  log b p  +  log c p 
 b   c   a  , wherever
2


 log a p + log b p + log c p 
c
a 
 b
defined, simplifies to
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2
Answer Key
DPP-07
1.
x2 − x − 56 = 0
x2 − 8x + 7x − 56 = 0
x ( x − 8) + 7 ( x − 8) = 0
(01)
We have,
1
1
1
+
+
log bc abc log ca abc log ab abc
( x − 8)( x + 7 ) = 0
= logabc bc + logabc ca + logabc ab
x = 8; −7
= logabcabc = 1
Let y = 64y
y2 = 64y
2.
(09)
log3 ( log3a ) = 1
y2 − 64y = 0
y ( y − 64 ) = 0
y = 0; y = 64
So the given question can be written as,
 56 + x 
log 4 

y 
3
Put x = −7; y = 64
 log3a = 3
2
 a = 33
 a = 27


and log 1  log 1 b  = 1


3
3 
 56 − 7 
We get log 4 
64 
3
 log 1 b = (1/ 3)1
3
 49 
= log 4  
64 
3
1/3
1
b= 
 3
Now,
  1  1/3 
ab = 27     
  3   
 72 
= log 4  2 
8 
3
2
7
= log 4
2 38
Put x = 8; y = 64
3
3
 1  56 + 8 
= log 4  
3 64 
3 
 64 
= log 4  
64 
3
= log3
= log 4 (1) = 0
1
= 27  = 9
3
3.
(04)
P = log5 (log53 )
−p
3C+5 = 405
− log5 (log 53 )
3C+5
C+
3
5.
1
log53
C+log53
3
= 405
= 405
log35
3C.3
3
= 405
(01)
log10 n
1
=
logn 2002 log10 2002
log2 + log3 + log4 + log5
b=
log2002
an =
a log a = b
b
= 405
5.3 = 405
C
c=
3C = 81 = 3 4
C=4
log 10 + log11 + log12 + log13 + log14
log2002
b−c =
4.
(00)
Let x = 56 + x
log(x  8  4  5 /10 1112 13 14)
log 2002
= log 2002 (2002)−1 = −1
x2 = 56 + x
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3
6.
(12)
Define a to be log20x ( 22x ) = log 2x ( 202x ) ,
what we are looking for. Then, by the definition
of the logarithm,
(20x)a = 22x

a
(2x) = 202x.
Dividing the first equation by the second equation
11
gives us 10a =
, so by the definition of logs,
101
11
. This is what the problem asked for,
a = log10
101
11
so the fraction
gives us m + n = 112 .
101
7.
(64)
or log3 ( log 2 x ) − log3 ( log1/2 y ) = 1
( (
or log3 log2 4 / y2
k −1

=4
5 k −1
32log2 3 +1
Now as we know 4 = 22 and 32 = 2 25 so we can
further write the equation as
1
+7) 2
=4
1
2
Now by using the logarithm power rule
(loga x p = ploga x) we can further write the
1/2
above equation as
2
2
3
2
+7
5log2 (3k−1 +1) 5
( )
or log ( 4 / y ) = −3( log y)
or log ( 4 / y ) + ( log y ) = 0
2
4log2 9
k −1
or log2 4 / y2 = 3(log1/2 y )
2
k −1
x = 4log2 9 +7
1
y=
log2 5 3k −1 +1
32
xy = 4 – –(i)
We can write (i) as by substituting x and y
xy = 4
22log2 (9
)) − log (log y ) = 1
3
and the inverse property we will further solve the
equation and then find the value of k.
Complete step by step answer.
Given functions,

2
or 4y = 1 or y = 1/ 4  x = 64
1
2x log2 (9k−1 +7)
2 2
1
5 log2 (3k−1 +1)
2 5
=4
−1
8.

=4
k −1
2log2 (3 +1)
Now by applying the inverse property of
logarithm (blogb x = x) in the above obtained
(06)
( log5 x )2 + log5x   = 1
5
x
5 

 log5 x 
( log5 x ) + 
 =1
 log5 ( 5x ) 


equation we can further write
2

 1 − log5 x 
 =1
 1 + log5 x 
( log5x )3 + ( log5 x )2 + 1 − log5 x = 1 + log5x
( log5x )3 + ( log5 x )2 − 2log5x = 0
let log5 x = t
)
(
)
log2 3k−1 +1
=4
9k −1 + 7
=4
3k −1 + 1
Hence by further solving (cross multiplying) this
we get

(
)
(3) ( ) + 7 = 4 (3 + 1)
 3  3 + 7 = 4 (3  3 + 1)
(3 )  3 + 7 = 4 (3  3 + 1)
 9k−1 + 7 = 4 3k −1 + 1
t3 + t 2 − 2t = 0
t ( t − 1)( t + 2) = 0  t = 0,1, −2
9.
(
log2 9k−1 +7
2
2
( log5 x )2 + 
 x = 1, 5,
2log2 (9k +7)
2 k −1
1
25
(09)
Hint: in this question the value of x and y are
given and the relation between the x and y are
also given so by using the logarithm power rule
k −1
2k
−2
k 2
−2
k
k
−1
−1
Now let 3k = p , so we can further write the
equation as
(
)
p2  3−2 + 7 = 4 p  3−1 + 1
Hence by further solving, we get
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4

p2
p 
+ 7 = 4  + 1
9
3 

p2
4
+7= p+4
9
3
10.
(01)
2
p2 4
− p+3=0
9 3
p2 − 12p + 27 = 0
Now we solve the obtained quadratic equation to
find the value of P,
 p2 − 12p + 27 = 0



1
T1 =  log a P  =
2

 
a
 b 
 logp b 


1
=
,log = x,logp b
2
logpa − ( logp b )
(
)
= y and log p c = z
 T1 =
1
, T2 =
(x − y)2
 p2 − 3p − 9p + 27 = 0
 p(p − 3) − 9(p − 3) = 0
 (p − 3)(p − 9) = 0
and T3 =
Now, since 3k = p , hence we get the value of k
Hence E =
as When p = 3
So, 3k = 31
Therefore k = 1
when p = 9
So, 3k = 32
Therefore k = 2
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1
1
(y − z)2
1
(z − y)2
1
=1
 1
1
1
+
+

2
2
2
 (x − y) (y − z) (z − y)
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