Logs Change of Base Formula | Logarithm

Here you will learn logs change of base formula in logarithm and logarithmic inequalities with example.

Let’s begin –

Logs Change of Base Formula – Base Changing Theorem

It can be stated as “quotient of the algorithm of two numbers is independent of their common base.”

Symbolically, \(log_b m\) = \(log_a m \over log_a b\), where a > 0 , \(a\neq1\), b > 0, \(b\neq1\).

NOTE :

(i)   \(log_b a\) . \(log_a b\) = \(log a \over log b\) . \(log b \over log a\) = 1 ;   Hence   \(log_b a\)=\(1 \over log_a b\)

(ii)   \(a^{log_b c} \) = \(c^{log_b a} \)

(iii)   \(log_{a^k} m\) = \(1\over k\)\(log_a m\)          It is known as Base Power Formula.

(iv)   The base of the logarithm can be any positive number other than 1, but in normal practice, only two bases are popular, these are 10 and e(=2.718 approx.). Logarithms of numbers to the base 10 are named as ‘common logarithm’ and the logarithms of numbers to the base e are called Natural or Napierian logarithm. We will consider logx as \(log_e x\) or lnx.

Example : Evaluate the given log : \(81^{l\over {log_5 3}}\) + \(27^{log_9 36}\) + \(3^{4\over {log_7 9}}\).

Solution : \(81^{log_3 5}\) + \(3^{3log_9 36}\) + \(3^{4log_9 7}\)
\(\implies\) \(3^{4log_3 5}\) + \(3^{log_3 {(36)}^{3/2}}\) + \(3^{log_3 {7}^2}\)
= 625 + 216 + 49
= 890.

Note :

(i)  for a given value of N,\(log_aN\) will give us a unique value.

(ii)  Logarithm of zero does not exist.

(iii)  Logarithm of negative reals are not defined in the system of real numbers.

Logarithmic Inequalities

(i)   \(log_a x\) < \(log_a y\)    <=>   [ x < y if a > 1 && x > y if 0 < a < 1 ]

(ii)   If a > 1, then     (a) \(log_a x\) < p => 0 < x < \(a^p\)             (b) \(log_a x\) > p => x > \(a^p\)

(iii)   If 0 < a < 1, then     (a) \(log_a x\) < p => x > \(a^p\)             (b) \(log_a x\) > p => 0 < x < \(a^p\)

Hope you learnt logs change of base formula in logarithm. To learn more practice more questions and get ahead in competition. Good Luck!

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