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

$$\frac{{10}^{\log {}_3\left(6\right)}.{15}^{\log {}_3\left(\frac{2}{3}\right)}}{6^{\log {}_3\left(\frac{2}{3}\right)}.5^{\log {}_3\left(\frac{4}{3}\right)}}=?$$

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

$$\frac{{10}^{\log {}_3\left(6\right)}.{15}^{\log {}_3\left(\frac{2}{3}\right)}}{6^{\log {}_3\left(\frac{2}{3}\right)}.5^{\log {}_3\left(\frac{4}{3}\right)}}$$$$=\frac{2^{{\log }_{3}6}.5^{{\log }_{3}6}.\cancel{3^{{\log }_3\left(\frac{2}{3}\right)}}.5^{{\log }_3\left(\frac{2}{3}\right)}}{2^{{\log }_3\left(\frac{2}{3}\right)}.\cancel{3^{{\log }_3\left(\frac{2}{3}\right)}}.5^{{\log }_3\left(\frac{4}{3}\right)}}$$$$=\frac{2^{{\log }_{3}6}.5^{{\log }_{3}6}.5^{{\log }_3\left(\frac{2}{3}\right)}}{2^{{\log }_3\left(\frac{2}{3}\right)}.5^{{\log }_3\left(\frac{4}{3}\right)}}$$$$=2^{{\log }_{3}2+{\log }_{3}3-({\log }_{3}2-{\log }_{3}3)}.5^{{\log }_{3}2-{\log }_{3}3-({\log }_{3}4-{\log }_{3}3)}$$$$=2^{\cancel{{\log }_{3}2}+{\log }_{3}3-\cancel{{\log }_{3}2}+{\log }_{3}3}.5^{{\log }_{3}6+{\log }_{3}2-\cancel{{\log }_{3}3}-{2\log }_{3}2+\cancel{{\log }_{3}3}}$$$$=2^{{\log }_{3}3+{\log }_{3}3}.5^{{\log }_{3}2+{\log }_{3}3+{\log }_{3}2-{2\log }_{3}2}$$$$=2^{{2\log }_{3}3}.5^{{\log }_{3}3}$$$$=2^{2\left(1\right)}.5^1=4.5=20$$

Answered by alusto22 last updated on 28/Jul/24

$$=6^{\log {}_3\left(10\right)-\log {}_3\left(\frac{2}{3}\right)}.\frac{{15}^{\log {}_3\left(\frac{2}{3}\right)}}{5^{\log {}_3\left(\frac{4}{3}\right)}}$$$$=6^{\log {}_3\left(15\right)}.\frac{{15}^{\log {}_3\left(\frac{2}{3}\right)}}{5^{\log {}_3\left(\frac{4}{3}\right)}}$$$$={15}^{\log {}_3\left(6\right)+\log {}_3\left(\frac{2}{3}\right)}.\frac{1}{5^{\log {}_3\left(\frac{4}{3}\right)}}$$$$={15}^{\log {}_3\left(4\right)}.5^{-\log {}_3\left(\frac{4}{3}\right)}$$$$=5^{\log {}_3\left(3\right)}.3^{\log {}_3\left(4\right)}$$$$=20$$

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