10-log-3-6-15-log-3-2-3-6-log-3-2-3-5-log-3-4-3-

Question Number 209991 by efronzo1 last updated on 28/Jul/24

\frac{10^{\log_{3} (6)} . 15^{\log_{3} (\frac{2}{3})}}{6^{\log_{3} (\frac{2}{3})} . 5^{\log_{3} (\frac{4}{3})}} = ?

Answered by Rasheed.Sindhi last updated on 28/Jul/24

\frac{10^{\log_{3} (6)} . 15^{\log_{3} (\frac{2}{3})}}{6^{\log_{3} (\frac{2}{3})} . 5^{\log_{3} (\frac{4}{3})}}

= \frac{2^{\log_{3} 6} . 5^{\log_{3} 6} . 3^{\log_{3} (\frac{2}{3})} . 5^{\log_{3} (\frac{2}{3})}}{2^{\log_{3} (\frac{2}{3})} . 3^{\log_{3} (\frac{2}{3})} . 5^{\log_{3} (\frac{4}{3})}}

= \frac{2^{\log_{3} 6} . 5^{\log_{3} 6} . 5^{\log_{3} (\frac{2}{3})}}{2^{\log_{3} (\frac{2}{3})} . 5^{\log_{3} (\frac{4}{3})}}

= 2^{\log_{3} 2 + \log_{3} 3 - (\log_{3} 2 - \log_{3} 3)} . 5^{\log_{3} 2 - \log_{3} 3 - (\log_{3} 4 - \log_{3} 3)}

= 2^{\log_{3} 2 + \log_{3} 3 - \log_{3} 2 + \log_{3} 3} . 5^{\log_{3} 6 + \log_{3} 2 - \log_{3} 3 - {2 \log}_{3} 2 + \log_{3} 3}

= 2^{\log_{3} 3 + \log_{3} 3} . 5^{\log_{3} 2 + \log_{3} 3 + \log_{3} 2 - {2 \log}_{3} 2}

= 2^{{2 \log}_{3} 3} . 5^{\log_{3} 3}

= 2^{2 (1)} . 5^{1} = 4.5 = 20

Answered by alusto22 last updated on 28/Jul/24

= 6^{\log_{3} (10) - \log_{3} (\frac{2}{3})} . \frac{15^{\log_{3} (\frac{2}{3})}}{5^{\log_{3} (\frac{4}{3})}}

= 6^{\log_{3} (15)} . \frac{15^{\log_{3} (\frac{2}{3})}}{5^{\log_{3} (\frac{4}{3})}}

= 15^{\log_{3} (6) + \log_{3} (\frac{2}{3})} . \frac{1}{5^{\log_{3} (\frac{4}{3})}}

= 15^{\log_{3} (4)} . 5^{- \log_{3} (\frac{4}{3})}

= 5^{\log_{3} (3)} . 3^{\log_{3} (4)}

= 20

Facebook Tweet Pin