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Laws of Logarithms

Last Updated : 23 Jul, 2025
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The logarithm is the exponent or power to which a base is raised to get a particular number. For example, 'a' is the logarithm of 'm' to the base of 'x' if xm = a, then we can write it as m = logxa. Logarithms are invented to speed up the calculations and time will be reduced when we are multiplying many digits using logarithms. Now, Let's discuss the laws of logarithms below.

Three Basic Laws of Logarithms

There are three common laws of logarithms that are derived using the basic rules of exponents. These three laws are the product rule law, quotient rule law, power rule law. Let's take a look at the laws in detail.

First Law of logarithm or Product Rule

Let a = xn and b = xm where base x should be greater than zero and x is not equal to zero. i.e., x > 0 and x ≠ 0. from this we can write them as
n = logxa and m = logxb ⇢ (1)

By using the first law of exponents we know that xn × xm = xn + m ⇢ (2)
ab = xn × xm = xn + m (From equation 2)

Now apply the logarithm to the above equation we get as below,
logxab = n + m

From equation 1 we can write as logxab = logxa + logxb

This is the first law of Logarithms/ Product Rule Law.

logxab = logxa + logxb

For more than two numbers : logxabc = logxa + logxb + logxc.

Second Law of Logarithm or Quotient Rule

Let a = xn and b = xm where base x should be greater than zero and x is not equal to zero. i.e., x > 0 and x ≠ 0. from this we can write them as,
n = logxa and m = logxb ⇢ (1)

By using the first law of exponents we know that xn / xm = xn - m ⇢ (2)
a/b = xn / xm = xn - m ⇢ (From equation 2)

Now apply the logarithm to the above equation we get as below,
logx(a/b) = n - m

From equation 1 we can write as logx(a/b) = logxa - logxb

This is the second law of Logarithms/ quotient Rule Law.

logx(a/b) = logxa - logxb

Third Law of Logarithm or Power Rule

Let a = xn ⇢ (i),    
Where base x should be greater than zero and x is not equal to zero. i.e., x > 0 and x ≠ 0. from this we can write them as, n = logxa ⇢ (i)

If we raise both sides of the equation (i) with the power of 'm' then we get it as follows,
am = (xn)m = xnm

Let am be a single quantity and apply logarithm to the above equation then,
logxam = nm

logxam = m.logxa

This is the third law of logarithms. It states that the logarithm of a power number can be obtained by multiplying the logarithm of the number by that number.

Also Read,

Sample Problems on Laws of Logarithms

Problem 1: Expand log 21.
Solution: 

As we know that logxab = logxa + logxb (From first law of logarithm)

So, log 21 = log (3 × 7) 
= log 3 + log 7

Problem 2: Expand log (125/64).
Solution: 

As we know that logx(a/b) = logxa - logxb (From second law of logarithm)

So, log (125/64) = log 125 - log 64
= log 53 - log 43

logxam = m.logxa (From third law of logarithm), we can write it as,
= 3 log 5 - 3 log 4
= 3(log 5 - log 4)

Problem 3: Write 3log 2 + 5 log3 - 5log 2 as a single logarithm.
Solution: 

3log 2 + 5 log3 - 5log 2

= log 23 + log 35 - log 25
= log 8 + log 243 - log 32
= log(8 × 243) - log 32
= log 1944 - log 32
= log (1944/32)

Problem 4: Write log 16 - log 2 as a single logarithm.
Solution: 

log(16/2)

= log(8) = log(23)
= 3 log 2

Problem 5: write 3 log 4 as a single logarithm
Solution: 

From the power rule law, we can write it as,

= log 43 = log 64

Problem 6: Write 2 log 3- 3 log 2 as a single logarithm
Solution: 

log 32 - log 23 

= log 9 - log 8 
= log (9/8)


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