Math-for-all-Grades.com

Remember the following important rule related to change of base in logarithms

for example:

log _{6} 7 = (log _{5} 7)/(log _{5} 6)

**How to change base of logarithms:**

log _{a} n = log _{b} n . log _{a} b

Without going into the proof and derivation of the rule, let us see this rule in a simple way
taking the help of the formula:

log _{a} b = (log b)/(log a)

**log _{b} n . log _{a} b **

= [(log n)/(log b)].[(log b)/(log a)]

= log _{a} n

So, log _{a} n = log _{b} n . log _{a} b

**Let us now solve a few problems based on the rule no. 3 and rule no. 4 on Base Change **

1. Prove the following

**1. log _{64} 3 = (log _{4} 3)/_{3} **

2. log _{pq} a = (log _{p} a)/(1 + log _{p} q)

*Solution1:*

We will convert all logs into base 4, as we see log to base 4 in the right side in question no.

1. So,

**log _{64} 3 = (log _{4} 3)/(log _{4} 64)**

from Law 3 on laws of logarithms, we have

**log _{a} (p)^{n} = n log _{a} p**

so, log _{4} 4^{ 3} = 3log 4^{4}

= 3 . 1 = 3, since log _{4}4 = 1

Therefore, (log _{4} 3)/ (log _{4} 64)

*Solution 2:*

we will convert all the logs into base p as we see base p on the right side.

log _{pq} a = (log _{p} a)/(log _{p} pq)

Now, log _{p} pq = log _{p} p + log _{p} q

= 1 + log _{p} q

Therefore,

log _{pq} a = (log _{p} a)/ (1 + log _{p} q)

Logarithms expressed or calculated to base 10 are called __Common Logarithms__

For eg, log _{10} 5, log _{10} p are examples of

common logarithms.

Common logarithms such as log _{10} x, log _{10}a

can also be merely expressed as log x or log a implying that the
base 10 is understood and therefore need not be shown.

This holds only for logarithms to base 10.

For logarithms to any base other than 10, the base must be clearly mentioned.
For example, if the base is 5, then it should be clearly mentioned as log_{5} x.