*Sunday, July 5th, 2020*

Questions on Logarithm have been asked in exams like CAT and XAT almost every year. More often than not, they are on the easier side but students get scared because they do not understand the concepts properly and hence are unable to attempt them under pressure. In this post, we will try to help CAT aspirants overcome that fear.

Let us start with some basics about logarithms.

When a^{x } = N , then we say that x = logarithm of N to the base a and write it as x = log_{a}N . In simple words, it represents the power to which a number must be raised. Let me expand on that by giving a simpler example.

If we are asked, what would be the result if ‘a’ is multiplied with itself ‘b’ times; then your answer would be x = a*a*a*…. b (times). This can also be written as a^b. This is also known as ‘a raised to the power of b’

If we are asked, which number multiplied with itself ‘b’ times, will result in a; then you are asked for the value of x such that x*x*x*… b(times) = a

=> x^b = a

=> x = a^(1/b)

This is also known as ‘bth root of a’

If swe are asked, how many times should you multiply ‘a’ with itself to get ‘b’, that is where the concept of logarithm comes into the picture. You are asked for the value of ‘x’ such that

a*a*a…. x (times) = b

=> x = Log _{a} b

This is also known as ‘**Log b to the base a’**

Examples to illustrate this :

How many times should you multiply 2 with itself to get 8? Answer is Log _{2} 8 = 3

How many times should you multiple 5 with itself to get 625? Answer is Log _{5} 625 = 4

Another way to understand this would be:

**If a ^{x } = N, then x = log_{a}N**

Where N is a POSITIVE number , “a” is a positive number OTHER THAN 1.

Since the log of a number is a value, it has 2 parts:

- Integral part known as
**Characteristic** - Decimal part known as
**Mantissa**

For example, Log 27 = 3 Log 3 = 3*0.4771 = 1.4313

In this case, the characterstic is 1 and the mantissa is 0.4313

There are 2 types of logarithms that are commonly used on the basis of bases:

**Natural logarithm**: base of the number is “e” .**Common logarithm**: Base of the number is 10 . When the base is not mentioned , it can be taken as 10.

The following can be derived from the above properties.

**Example:**

- 3
^{4}= 81 , log_{3}81 = 4. - log
_{3}10 = ? ( given values of log_{3}2 and log_{3}5)

log_{3} 10 = log_{3} ( 2* 5) = log_{3} 2 + log_{3} 5.

**SOME POINTS TO REMEMBER :**

- Characteristic of a number greater than unity for a common base is positive and is
**1****less**than the number of digits in the integral part.For example : Characteristic of log 1000 = 3 which is 1 less than the number of digits in 1000. - For a number between 0 and 1 , the characteristic is negative and its magnitude is
**1****more**than the number of zeros after the decimal point. For example : Characteristic of log 0.001 = -3. - log( x – y ) ≠ logx – logy
- log( x + y ) ≠ logx + logy

The questions on logarithms are generally very direct , but can be increased in difficulty level by introducing the concept of number of digits.

Firstly, have a look at the log values of some numbers ( base 10 ). The values are always mentioned in the question , but it is still advisable to memorise the values of numbers till 10.

Number | Value |

1 | 0 |

2 | 0.301 |

3 | 0.4771 |

4 | 0.602 |

5 | 0.698 |

6 | 0.778 |

7 | 0.845 |

8 | 0.903 |

9 | 0.954 |

Let us find the number of digits in 3 ^{100}.

x = 3 ^{100}

Log x = 100 Log 3

= 47.71

Number of digits here will be 47 + 1 = 48.

**Problem 1** : If **a ^{x } = b^{y}**

Then :

a.) Log (a/b)= x/y

b.) Log(a) / log (b) = x/y

c.) Log(a) / log (b) = y/x

d.) None of these

(**IIFT**)

**Solution** : option C is the correct answer

Take log on both sides to obtain the answer

**Problem 2** : The value of the expression ∑ 1 / log_{i} 100 ! ( where i= 2 to 100 ) is

a.) 01

b.) 1

c.) 1

d.) 10

e.) 100

**( XAT 2014 )**

**Solution**: Option C is the correct answer.

The given expression is : ∑ 1 / log_{i} 100 !

Expand the above expression.

= 1 / log_{1} 100 ! + 1 / log_{2} 100 ! + 1 / log_{3} 100 ! + 1 / log_{4} 100 ! + ……. + 1 / log_{100} 100 !

Using logarithm property, the above expression becomes :

= log_{100!} 2 + log_{100!} 3 + …… + log_{100!} 100

= log_{100!} 2 * 3* 4 * …… 100

= log_{100!} 100!

= 1

**Problem 3 : **If log_{13}log_{21 }{ √x+21 + √x } = 0, then the value of x is

a.) 21

b.) 13

c.) 81

d.) None of the above

**( IIFT 2013 )**

** **

**Solution :** log_{13}log_{21 }{ √x+21 + √x } = 0

This implies, log_{21 }{ √x+21 + √x } = 13^{0} = 1

√x+21 + √x = 21

By looking at the equation , we can see that x= 100.

Hence, option D is the correct answer.

**Problem 4 **: If 2 is the logarithm of a number to the base √3, then find out the logarithm of the same number for the base 3√3.

a.) 0

b.) ⅓

c.) ⅔

d.) 2

**Solution : **Option C is the correct answer.

Let the number be x.

log√3x = 2

3√3= (√3)^{3}

log3√3x = ⅓ * 2 = ⅔

**Problem 5 : **log_{a} (ab) = x , then log_{b}(ab) is equal to :

a.) x/(x+1)

b.) 1/x

c.) x/(x-1)

d.) x/(1-x)

**Solution: **Option C is the correct answer.

Log (ab) / log a = x

1+ (log b / log a ) = x

Log b / log a = x – 1

Log a / log b = 1 / (x-1)

Add 1 to both sides and solve

Log (ab) / log b = x / (x-1)

**Problem 6 : **log_{9} (3log_{2} (1 + log_{3} (1 + 2log_{2}x))) = 12. Find x.

a) 4

b) 12

c) 1

d) 2

**Solution **: Option D is the correct answer.

Log_{9 } (3log_{2} (1 + log_{3} (1 + 2log_{2}x)) = 12

3 log_{2}(1 + log_{3}(1 + 2log_{2}x)) = 9^{1/2}

log_{2}(1 + log_{3}(1 + 2log_{2}x) = 1

1 + log_{3}(1 + 2log_{2}x) = 2

log_{3}(1 + 2log_{2}x) = 1

log_{2}x = 1

Hence,the value of x = 2 .

**Problem 7 : **The domain of the function f(x) = log_{7} { log_{3}( log_{5}(20x – x^{2} – 91 ))} is

a.) (7,13)

b.) ( 8,12)

c.) (7,12)

d.) (12,13)

e.) None of these

(**XAT 2011**)

** **

**Solution : ** f(x) = log_{7} { log_{3}( log_{5}(20x – x^{2} – 91 ))}

For f(x) to be defined , log_{3}( log_{5}(20x – x^{2} – 91 )) > 0

log_{5}(20x – x^{2} – 91 ) > 1

(20x – x^{2} – 91 ) > 5

x^{2} – 20x + 96 < 0

( x – 8 ) (x – 12) < 0

8 < x < 12

Hence, option B is the correct answer.

**Problem 8**: If log { (x+y)/2 } = ½ (log x + log y), then which of the following is true ?

a.) x -y = 0

b.) x +y=0

c.) x +√y =0

d.) y +√x =0

**Solution : ** log { (x+y)/2 } = ½ ( log xy)

log { (x+y)/2 } = log (xy)^{½}

(x + y)/2 = √xy

Squaring both sides and simplifying,

(x-y)(x-y) = 0

Hence, a is the correct option.

**Problem 9 **: ( log x )/ y-z = ( log y )/ z-x = ( log z )/ x-y , then xyz = ?

a.) 0

b.) 1

c.) 2 ( x- y- z)

d.) None of the above

**Solution** : Option b is the correct answer.

Let ( log x )/ y-z = k

Log x = k (y – z)

x= 10^{k(y-z)}

Similarly , y = 10^{k(z-x)}

z= 10^{k(x-y)}

Hence , xyz = 1.

**Problem 10 **: If l = 1+ log_{a}bc, m = 1 + log_{b}ca , n = 1 + log_{c}ab, then lmn = ?

a.) l^{2} + m^{2} + n^{2}

b.) 1

c.) l + m+ n

d.) lm+ mn + nl

**Solution: **Option d is the correct answer.

l = 1+ log_{a}bc can be written as l = log_{a}abc

m = 1 + log_{b}ca can be written as m = log_{b}cab

n = 1 + log_{c}ab can be written as m = log_{c}abc

1/ l = log_{abc}a , 1/ l = log_{abc}a , 1/ l = log_{abc}a

Add above 3 expressions to get the correct answer.

**Q1**: If log (2^a × 3^b × 5^c) is the arithmetic mean of log (2^2 × 3^3 × 5), log (2^6 × 3 × 5^7), and log(2 × 3^2 × 5^4), then a equals

Check answer of logarithm Q1 from CAT

**Q2**: If x is a real number such that log(base 3)5 = log(base 5)(2 + x), then which of the following is true?

Check answer of logarithm Q2 from CAT

**Q3**: The value of log (base 0.008) √5 + log (base√3) 81 – 7 is equal to

Check answer of logarithm Q3 from CAT

**Q4**: Suppose, log(base3)x = log(base12)y = a, where x, y are positive numbers. If G is the geometric mean of x and y, and log(base6)G is equal to

Check answer of logarithm Q4 from CAT

I hope that you found this post on Logarithms for CAT helpful and you will now be able to solve these questions easily in the CAT exam. If you liked this post, please share it on Facebook and/or Whatsapp.

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Cheers,

Ravi Handa,

Founder, Handa Ka Funda

What a tremendous concept has given to us which is very helpful. Because they have provided us the peculiar concept about the logarithm.