Question

Sabendo que log base 20 de 2 = a e log base 20 de 3 = b, calcule:

log base 12 de 25

Answer 1

$\log_{12}{25} = \frac{\log_{20}{25}}{ \log_{20}{12} } = \frac{\log_{20}{ {5}^{2} }}{\log_{20}{(4 \cdot 3)}} \\ = \frac{2 \cdot \log_{20}{5} }{\log_{20}{4} + \log_{20}{3}} = \frac{2 \cdot \log_{20}{( \frac{20}{4} )}}{\log_{20}{ {2}^{2} + \log_{20}{3} }} \\ = \frac{2 \cdot ( \log_{20}{20} - \log_{20}{4} )}{2 \cdot \log_{20}{2} + \log_{20}{3} } \\ = \frac{2 \cdot ( \log_{20}{20} - \log_{20}{ {2}^{2} } )}{2 \cdot \log_{20}{2} + \log_{20}{3} } \\ = \frac{2 \cdot ( \log_{20}{20} - 2 \cdot \log_{20}{2 } )}{2 \cdot \log_{20}{2} + \log_{20}{3} } \\ = \frac{2 \cdot ( 1 - 2a )}{2a + b } \\ \\ \log_{12}{25} = \frac{2 - 4a}{2a + b}$