How do you evaluate ${log}_{3} 18$ $log_3 18$ ?

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Answer 1 · 2016-05-19T01:44:00

$= {log}_{3} (9 \times 2)$ $= log_3(9 xx 2)$

$= {log}_{3} (9) + {log}_{3} (2)$ $= log_3(9) + log_3(2)$

$= \frac{log 9}{log 3} + \frac{log 2}{log 3}$ $= log9/log3 + log2/log3$

$= \frac{{log 3}^{2}}{log 3} + \frac{log 2}{log 3}$ $= log3^2/log3 + log2/log3$

$= 2cancel(log3)/cancel(log3) + log2/log3$

$= 2 + log_3(2)$

Hopefully this helps!

How do you evaluate log_3 18log318log_3 18?