1
IOQM 2024
Basic Mathematics
1.
1
log bc abc
+
1
log ca abc
+
1
log ab abc
has the value
6.
equal to
2.
The value log20x ( 22x ) can be written as
m
log10 , where m and n are relatively prime
n
positive integers. Find m + n −100 .
–P
If P = log5 (log5 3) and 3C+5 = 405 then C is
equal to
4.
5.
7.
If xy2 = 4 and log3(log2x) + log1/3 (log1/2y) = 1,
then x equals
8.
Find the sum of all integral values of x
2
5
satisfying ( log5 x ) + log5x = 1 .
x
9.
Let x = 4log2 9
The value of
56 + 56 + 56 + 56 + .....
is equal to
log 4
3
64 64 64......
Suppose n be an integer greater than 1, let
1
. Suppose b = a 2 + a 3 + a 4 + a 5
an =
logn 2002
k−1
+7
1
and y =
and
32
xy = 4 , then the sum of the cubes of the real
log2 5 3k−1 +1
value(s) of k is
and c = a10 + a11 + a12 + a13 + a14 . Then b − c
equals
There is a positive real number x not equal to
1
1
either
or such that
2
20
log20x ( 22x ) = log2x ( 202x ).
If log3 ( log2a ) + log 1 log 1 b = 1 , then the value
3
2
of ab3 is
3.
DPP-05
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
2
Answer Key
DPP-05
1.
2.
3.
4.
5.
6.
(01)
(09)
(04)
(00)
(01)
(12)
7.
8.
9.
10.
(64)
(06)
(09)
(01)
Hint Solution
5
1.
3C.3log3 = 405
We have,
1
1
1
+
+
log bc abc log ca abc log ab abc
5.3C = 405
3C = 81 = 3 4
C=4
= logabc bc + logabc ca + logabc ab
= logabc abc = 1
2.
4.
x 2 = 56 + x
log3 ( log3a ) = 1
x 2 − x − 56 = 0
log3a = 32
x 2 − 8x + 7x − 56 = 0
a = 33
a = 27
and
x ( x − 8) + 7 ( x − 8) = 0
( x − 8)( x + 7) = 0
log 1 log 1 b = 1
3
3
Let y = 64y
x = 8; −7
y2 = 64y
log 1 b = (1/ 3)
1
y2 − 64y = 0
3
y ( y − 64) = 0
1/3
1
b=
3
Now,
y = 0; y = 64
So the given question can be written as,
1 1/3
3
ab = 27
3
56 + x
log 4
y
3
Put x = −7; y = 64
3
56 − 7
We get log 4
64
3
1
= 27 = 9
3
Hence the correct option is A
3.
Let x = 56 + x
72
= log 4 2
38
2
7
= log 4
2 38
P = log5 (log53 )
−p
3C+5 = 405
− log5 (log 53 )
3C+5
1
C+
log53
3
= 405
5
Put x = 8; y = 64
= 405
3C+log3 = 405
49
= log 4
64
3
alog ab = b
1 56 + 8
= log 4
3 64
3
3
( log5x )3 + (log5x )2 + 1 − log5x = 1 + log5x
( log5x )3 + (log5x )2 − 2log5x = 0
64
= log 4
64
3
= log3
let log5 x = t
= log 4 (1) = 0
t 3 + t 2 − 2t = 0
3
5.
an =
t ( t −1)( t + 2) = 0 t = 0,1, −2
log10 n
1
=
logn 2002 log10 2002
b=
log2 + log3 + log4 + log5
log2002
c=
log 10 + log11 + log12 + log13 + log14
log2002
b−c =
log(x 8 4 5 /10 1112 13 14)
log 2002
= log2002 (2002)−1 = −1
1
25
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
and the inverse property we will further solve the
equation and then find the value of k.
Complete step by step answer.
Given functions,
x = 1, 5,
9.
k−1
6.
Define a to be log20x ( 22x ) = log2x ( 202x ) ,
what we are looking for. Then, by the definition
of the logarithm,
a
(20x) = 22x
a
(2x) = 202x.
Dividing the first equation by the second equation
11
gives us 10a =
, so by the definition of logs,
101
11
a = log10
. This is what the problem asked for,
101
11
so the fraction
gives us m + n = 112 .
101
7.
or log3 ( log2 x ) − log3 (log1/2 y) = 1
( (
))
or log3 log2 4 / y2 − log3 (log1/2 y) = 1
( ) = 3(log y )
or log ( 4 / y ) = −3( log y )
or log ( 4 / y ) + ( log y ) = 0
or log 2 4 / y
2
1/2
x = 4log2 9 +7
1
y=
5 k −1
32log2 3 +1
xy = 4 – –(i)
We can write (i) as by substituting x and y
xy = 4
k−1
1
2log2 (9k−1 +7) 2
2
3
2
2
or 4y = 1 or y = 1/ 4 x = 64
8.
( log5x )2 + log5x = 1
5
x
5
log5 x
( log5 x ) +
=1
log5 ( 5x )
2
1 − log5 x
=1
1 + log5 x
( log5 x )2 +
=4
1
5log2 (3k−1 +1) 5
2
Now by using the logarithm power rule
(loga xp = ploga x) we can further write the
above equation as
2
2
+7
=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
2
2
4log2 9
1
2x log2 (9k −1 +7)
2
2
1
5 log2 (3k−1 +1)
2 5
=4
−1
2log2 (9k +7)
=4
k −1
2log2 (3 +1)
Now by applying the inverse property of
logarithm (blogb x = x) in the above obtained
equation we can further write
)
(
)
log2 3k−1 +1
2
(
log2 9k−1 +7
2
=4
9k −1 + 7
=4
3k −1 + 1
4
(p − 3)(p − 9) = 0
Hence by further solving (cross multiplying) this
we get
(
(3) ( ) + 7 = 4 ( 3
Now, since 3k = p , hence we get the value of k
)
+ 1)
as
When p = 3
9k −1 + 7 = 4 3k −1 + 1
2 k −1
k −1
So, 3k = 31
Therefore k = 1
when p = 9
So, 3k = 32
Therefore k = 2
(
)
(3 ) 3 + 7 = 4(3 3 +1)
32k 3−2 + 7 = 4 3k 3−1 + 1
k 2
−2
k
−1
Now let 3k = p , so we can further write the
2
equation as
−2
(
−1
10.
)
p 3 + 7 = 4 p 3 +1
2
Hence by further solving, we get
(
p
p
+ 7 = 4 + 1
9
3
2
(
))
= y and logpc = z
p2
4
+7= p+4
9
3
2
p 4
− p+3=0
9 3
2
p −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
log
p b
1
=
,log = x,logp b
2
logp a − logp b
T1 =
1
, T2 =
(x − y)2
and T3 =
1
1
(y − z)2
1
(z − y)2
Hence
E=
1
=1
1
1
1
+
+
2
2
2
(x − y) (y − z) (z − y)
p2 − 3p − 9p + 27 = 0
p(p − 3) − 9(p − 3) = 0
PW Web/App - https://smart.link/7wwosivoicgd4
Library- https://smart.link/sdfez8ejd80if
0
