How do you condense $2 {log}_{3} 6 - {log}_{3} 4$ $2log_3 6 - log_3 4$ ?

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Answer 1 · 2015-09-20T23:35:14

$2 {log}_{3} 6 - {log}_{3} 4 = 2$ $2 log_3 6 - log_3 4 = 2$

Use ${log}_{a} b + {log}_{a} c = {log}_{a} (b c)$ $log_a b + log_a c = log_a (bc)$

and ${log}_{a} (\frac{1}{b}) = - {log}_{a} b$ $log_a (1/b) = -log_a b$

$2 {log}_{3} 6 - {log}_{3} 4 = {log}_{3} 6^{2} - {log}_{3} 4 = {log}_{3} (\frac{36}{4})$ $2 log_3 6 - log_3 4 = log_3 6^2 - log_3 4 = log_3 (36 / 4)$

$= {log}_{3} 9 = {log}_{3} 3^{2} = 2$ $= log_3 9 = log_3 3^2 = 2$

How do you condense 2log_3 6 - log_3 4 2log36−log342log_3 6 - log_3 4 ?