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=log_ax, if and only if x=a^y; 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 log_e.

f(x) = log_ex

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

Also, read:

• Logarithms
• Logarithm Table

Questions on Logarithm with Solutions

1. Express 5³ = 125 in logarithm form.

Solution:

5³ = 125

As we know,

a^b = c ⇒ log_ac=b

Therefore;

Log₅125 = 3

2. Express log₁₀1 = 0 in exponential form.

Solution:

Given, log₁₀1 = 0

By the rule, we know;

log_ac=b ⇒ a^b = c

Hence,

10⁰ = 1

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

Solution: log₄32 = x

4^x = 32

(2²)^x = 2x2x2x2x2

2^2x = 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 log_am=n, express a^n-1 in terms of a and m.

Solution:

log_am=n

aⁿ=m

aⁿ/a=m/a

a^n-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, nlog_am=log_amⁿ

⇒log(75/16)-log(5/9)²+log(32/243)

⇒log(75/16)-log(25/81)+log(32/243)

Since, log_am-log_an=log_a(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, log_am+log_an=log_amn

⇒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: loga^b = b log a]

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

x²y³ = a  [If log_ma = log_mb, then a = b]

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