CAT Logarithm Questions PDF [Most Important]

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CAT logarithms questions
CAT logarithms questions

CAT Logarithm Questions PDF [Most Important]

The logarithm is one of the most important topics in the CAT Quantitative Ability Section. You can check out these Logarithm questions in CAT Previous year papers. If you want to learn the basics, you can watch these videos on Logarithm basics. This article will look into some important Logarithm Questions for CAT. These are good sources for practice; If you want to practice these questions, you can download this CAT Logarithm Most Important Questions PDF below, which is completely Free.

Download Logarithm Questions for CAT

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Question 1: For a real number a, if log15a+log32a(log15a)(log32a)=4 then a must lie in the range

a) 2<a<3

b) 3<a<4

c) 4<a<5

d) a>5

1) Answer (C)

View Video Solution

Solution:

We have :log15a+log32a(log15a)(log32a)=4
We get (logalog 15+logalog32)logalog 15× logalog32  =4
we get loga(log32 +log 15)=4(log a)2
we get (log32 +log 15)=4loga
=log480=loga4
=a4 =480
so we can say a is between 4 and 5 .

Question 2: If log2[3+log3{4+log4(x1)}]2=0 then 4x equals

2) Answer: 5

View Video Solution

Solution:

We have :
log2{3+log3{4+log4(x1)}}=2
we get 3+log3{4+log4(x1)}=4
we get log3(4+log4(x1) = 1)
we get 4+log4(x1) = 3
log4(x1) = 1
x-1 = 4^-1
x = 14+1=54
4x = 5

Question 3: If 5log101+x+4log101x=log1011x2, then 100x equals

3) Answer: 99

View Video Solution

Solution:

5log101+x+4log101x=log1011x2

We can re-write the equation as: 5log101+x+4log101x=log10(1+x× 1x)1

5log101+x+4log101x=(1)log10(1+x)+(1)log10(1x)

5=log101+x+log101+xlog101x4log101x

5=5log101x

1x=110

Squaring both sides: (1x)2=1100

  x=11100=99100

Hence, 100 x =100× 99100=99

Question 4: The value of loga(ab)+logb(ba), for 1<ab cannot be equal to

a) 0

b) -1

c) 1

d) -0.5

4) Answer (C)

View Video Solution

Solution:

On expanding the expression we get 1logab+1logba

or 2(logab+1logba)

Now applying the property of AM>=GM, we get that  (logab+1logba)21 or (logab+1logba)2 Hence from here we can conclude that the expression will always be equal to 0 or less than 0. Hence any positive value is not possible. So 1 is not possible.

Question 5: If loga30=A,loga(53)=B and log2a=13, then log3a equals

a) 2A+B3

b) 2A+B3

c) A+B23

d) A+B32

5) Answer (A)

View Video Solution

Solution:

loga30=A or loga5+loga2+loga3=A………..(1)

loga(53)=B or loga3loga5=B………….(2)

and finally loga2=3

Substituting this in (1) we get loga5+loga3=A3

Now we have two equations in two variables (1) and (2) . On solving we get

loga3=(A+B3)2 or log3a=2A+B3

Checkout: CAT Free Practice Questions and Videos

Question 6: If log45=(log4y)(log65), then y equals

6) Answer: 36

View Video Solution

Solution:

log 52log2 =log y2log2log 52log6

log 36 = log y;  y =36

Question 7: If Y is a negative number such that 2Y2(log35)=5log23, then Y equals to:

a) log2(15)

b) log2(13)

c) log2(15)

d) log2(13)

7) Answer (B)

View Video Solution

Solution:

2Y2(log35)=5Y2(log32)

Given, 5Y2(log32)=5(log23)

=> Y2(log32)=(log23)=>Y2=(log23)2

=>Y=(log23) or (log23)

since Y is a negative number, Y=(log23)=(log213)

Question 8: Let x and y be positive real numbers such that
log5(x+y)+log5(xy)=3, and log2ylog2x=1log23. Then xy equals

a) 150

b) 25

c) 100

d) 250

8) Answer (A)

View Video Solution

Solution:

We have, log5(x+y)+log5(xy)=3

=> x2y2=125……(1)

log2ylog2x=1log23

=>  yx  23

=> 2x=3y   => x=  3y2

On substituting the value of x in 1, we get

  5x24=125

=>y=10, x=15

Hence xy=150

Question 9: If p3 = q4 = r5 = s6, then the value of logs(pqr) is equal to

a) 4710

b) 245

c) 165

d) 1

9) Answer (A)

View Video Solution

Solution:

Given that, p3 = q4 = r5 = s6

p3=s6

p = s63 = s2   …(1)

Similarly, q = s64 = s32   …(2)

Similarly, r = s65   …(3)

logs(pqr)

By substituting value of p, q, and r from equation (1), (2) and (3)

logs(s2s32s65)

logs(s4710)

4710

Hence, option A is the correct answer.

Question 10: 1log21001log4100+1log51001log10100+1log201001log25100+1log50100=?

a) 12

b) 10

c) 0

d) −4

10) Answer (A)

View Video Solution

Solution:

We know that 1logab = logxalogxb

Therefore, we can say that 1log2100 = log102log10100

1log21001log4100+1log51001log10100+1log201001log25100+1log50100

log102log10100-log104log10100+log105log10100-log1010log10100+log1020log10100-log1025log10100+log1050log10100

We know that log10100=2

12[log102log104+log105log1010+log1020log1025+log1050]

12[log1025205041025]

12[log1010]

12

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