Base change of the logarithms

There is a relation that allows to convert logarithms into any other base: $l o g_{a} x = \frac{l o g_{b} x}{l o g_{b} a}$

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:

Example

$l o g_{3} 7 = \frac{l o g 7}{l o g 3} ≃ \frac{0, 845}{0, 477} ≃ 1, 771$

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:

Example

$l o g_{2} 44$

$l o g_{2} 44 = \frac{l o g 44}{l o g 2} ≃ 5, 459$

Or:

$l o g_{2} 44 = \frac{l n 44}{l n 2} ≃ 5, 459$

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

Example

$l o g_{2} \frac{13}{47^{8}} \cdot l o g_{3} \frac{\frac{1}{17}}{23^{12}}$

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

$l o g_{2} \frac{13}{47^{8}} \cdot l o g_{3} \frac{\frac{1}{17}}{23^{12}} = (l o g_{2} 13 - l o g_{2} 47^{- 8}) \cdot (l o g_{3} 17^{- 1} - l o g_{3} 23^{12}) =$ $= (l o g_{2} 13 + 8 \cdot l o g_{2} 47) \cdot (- 1 \cdot l o g_{3} 17 - 12 \cdot l o g_{3} 23)$

At this point, the base change is applied: $= (l o g_{2} 13 + 8 \cdot l o g_{2} 47) \cdot (- l o g_{3} 17 - 12 \cdot l o g_{3} 23) =$ $= (\frac{l o g 13}{l o g 2} + 8 \cdot \frac{l o g 47}{l o g 2}) \cdot (- \frac{l o g 17}{l o g 3} - 12 \cdot \frac{l o g 23}{l o g 3}) ≃$ $≃ (\frac{1, 114}{0, 301} + 8 \cdot \frac{1, 672}{0, 301}) \cdot (- \frac{1, 230}{0, 477} - 12 \cdot \frac{1, 362}{0, 477}) ≃$ $≃ (3, 701 + 8 \cdot 5, 555) \cdot (- 2, 579 - 12 \cdot 2, 855) ≃$ $≃ 48, 141 \cdot (- 36, 839) ≃ - 1773, 466$