Logarithm Questions

Logarithm questions with answers are provided for students to solve them and understand the concept elaborately. These questions are based on the logarithm chapter of Class 9, 10 and 11 syllabi. Practising these problems will not only help students to score good marks in academic exams but also participate in competitive exams conducted at the state or national level, such as Maths Olympiad.

The logarithmic function is an inverse of the exponential function. It is defined as:

y=logax, if and only if x=ay; for x>0, a>0, and a≠1.

Natural logarithmic function: The log function with base e is called natural logarithmic function and is denoted by loge.

f(x) = logex

The questions of logarithm could be solved based on the properties, given below:

  • Product rule: logb MN = logb M + logb N
  • Quotient rule: logb M/N = logb M – logb N
  • Power rule: logb Mp = P logb M
  • Zero Exponent Rule: loga 1 = 0
  • Change of Base Rule: logb (x) = ln x / ln b or logb (x) = log10 x / log10 b

Also, read:

Questions on Logarithm with Solutions

1. Express 53 = 125 in logarithm form.

Solution:

53 = 125

As we know,

ab = c ⇒ logac=b

Therefore;

Log5125 = 3

2. Express log101 = 0 in exponential form.

Solution:

Given, log101 = 0

By the rule, we know;

logac=b ⇒ ab = c

Hence,

100 = 1

3. Find the log of 32 to the base 4.

Solution: log432 = x

4x = 32

(22)x = 2x2x2x2x2

22x = 25

2x=5

x=5/2

Therefore,

log432 =5/2

4. Find x if log5(x-7)=1.

Solution: Given,

log5(x-7)=1

Using logarithm rules, we can write;

51 = x-7

5 = x-7

x=5+7

x=12

5. If logam=n, express an-1 in terms of a and m.

Solution:

logam=n

an=m

an/a=m/a

an-1=m/a

6. Solve for x if log(x-1)+log(x+1)=log21

Solution: log(x-1)+log(x+1)=log21

log(x-1)+log(x+1)=0

log[(x-1)(x+1)]=0

Since, log 1 = 0

(x-1)(x+1) = 1

x2-1=1

x2=2

x=± √2

Since, log of negative number is not defined.

Therefore, x=√2

7. Express log(75/16)-2log(5/9)+log(32/243) in terms of log 2 and log 3.

Solution: log(75/16)-2log(5/9)+log(32/243)

Since, nlogam=logamn

⇒log(75/16)-log(5/9)2+log(32/243)

⇒log(75/16)-log(25/81)+log(32/243)

Since, logam-logan=loga(m/n)

⇒log[(75/16)÷(25/81)]+log(32/243)

⇒log[(75/16)×(81/25)]+log(32/243)

⇒log(243/16)+log(32/243)

Since, logam+logan=logamn

⇒log(32/16)

⇒log2

8. Express 2logx+3logy=log a in logarithm free form.

Solution: 2logx+3logy=log a

logx2+logy3=log a   [By logarithm rule: logab = b log a]

log(x2y3)=log a  [By logarithm rule: log a + log b = log (ab) ]

x2y3 = a  [If logma = logmb, then a = b]

Video Lesson

Logarithmic Equations

9. Prove that: 2log(15/18)-log(25/162)+log(4/9)=log2

Solution: 2log(15/18)-log(25/162)+log(4/9)=log2

Taking L.H.S.:

⇒2log(15/18)-log(25/162)+log(4/9)

⇒log(15/18)2-log(25/162)+log(4/9)

⇒log(225/324)-log(25/162)+log(4/9)

⇒log[(225/324)(4/9)]-log(25/162)

⇒log[(225/324)(4/9)]/(25/162)

⇒log(72/36)

⇒log2 (R.H.S)

10. Express log10(2) + 1 in the form of log10x.

Solution: log10(2)+1

=log102+log1010  [Since, 1 = log1010 ]

=log10(2 x 10) [log10 a + log10 b = log10 ab]

=log1020

11. Find the value of x, if log10(x-10)=1.

Solution: Given, log10(x-10)=1.

log10(x-10) = log1010

x-10 = 10

x=10+10

x=20

12. Find the value of x, if log(x+5)+log(x-5)=4log2+2log3

Solution: Given,

log(x+5)+log(x-5)=4log2+2log3

log(x+5)(x-5) = 4log2+2log3 [log mn=log m+log n]

log(x2-25) = log24+log32

log(x2-25) = log16+log9

log(x2-25)=log(16×9)

log(x2-25)=log144

x2-25=144

x2=169

x=±√169

x=±13

13. Solve for x, if (log 225/log15) = log x

Solution: log x = (log 225/log15)

log x=[log(15×15)/log15]

log x = log 152/log 15

log x = 2log 15/log 15

log x = 2

Or

log10x=2

102=x

x=10×10

x=100

Practice Questions

  1. If log x = m+n and log y=m-n, express the value of log 10x/y2 in terms of m and n.
  2. Express 3-2=1/9 in logarithmic form.
  3. Express log100.01=-2 in exponential form.
  4. Find the logarithm of 1/81 to the base 27.
  5. Find x if log7(2x2-1)=2.

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7 Comments

  1. Wow! It was really helpful!

  2. Thanks you very much sir

  3. It was very helpful and fun
    Thanks so much

  4. Hello I hope it will be helpful

  5. yes I think it will be helpful because it had covered all the chapter in few examples

  6. It was very helpful to me tq

  7. Thank you so much,at least you are of help to me in mathematics, may God bless you for me.

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