# Base change of the logarithms

There is a relation that allows to convert logarithms into any other base: $$log_a x=\dfrac{log_b x}{log_b a}$$

Namely, if the logarithm of a number is divided by the logarithm of the base in which we want to express it, we obtain the value of the same logarithm in that base. For example:

$$log_3 7=\dfrac{log7}{log3}\simeq\dfrac{0,845}{0,477}\simeq1,771$$

Using this relation we can calculate logarithms that are not the decimal and the neperians calculated with a scientific calculator, since it is possible to apply both to do the conversion. For example:

$log_2 44$

$log_2 44=\dfrac{log44}{log2}\simeq5,459$

Or:

$log_2 44=\dfrac{ln44}{ln2}\simeq5,459$

In addition, it also allows us to simplify logarithms and express them in just one base, making the calculation easier. For example:

$log_2 \dfrac{13}{47^8}\cdot log_3\dfrac{\dfrac{1}{17}}{23^{12}}$

First, it is necessary to apply the properties of the logarithms to decompose the expression:

$$log_2 \dfrac{13}{47^8}\cdot log_3\dfrac{\dfrac{1}{17}}{23^{12}}=(log_2 13-log_2 47^{-8})\cdot (log_3 17^{-1}-log_3 23^{12})=$$ $$=(log_2 13+8\cdot log_2 47)\cdot (-1\cdot log_3 17-12\cdot log_3 23)$$

At this point, the base change is applied: $$=(log_2 13+8\cdot log_2 47)\cdot (-log_3 17-12\cdot log_3 23)=$$ $$=\Big(\dfrac{log13}{log2}+8\cdot\dfrac{log47}{log2}\Big)\cdot\Big(-\dfrac{log17}{log3}-12\cdot\dfrac{log23}{log3}\Big)\simeq$$ $$\simeq\Big(\dfrac{1,114}{0,301}+8\cdot\dfrac{1,672}{0,301}\Big)\cdot\Big(-\dfrac{1,230}{0,477}-12\cdot\dfrac{1,362}{0,477}\Big)\simeq$$ $$\simeq(3,701+8\cdot5,555)\cdot(-2,579-12\cdot2,855)\simeq$$ $$\simeq 48,141\cdot(-36,839)\simeq -1773,466$$