How do you evaluate ${log}_{18} (9) + {log}_{18} (2)$ $log_18(9) + log_18(2)$ ?

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How do you evaluate ${log}_{18} (9) + {log}_{18} (2)$ $log_18(9) + log_18(2)$ ?

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Answer 1 · 2016-10-29T13:12:35

$1$ $1$

Explanation:

Using the $laws of logarithms$ $color(blue)"laws of logarithms"$

$color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(logx+logy=log(xy)" and " log_b b=1)color(white)(2/2)|)))$

$rArrlog_(18)9+log_(18)2=log_(18)18=1$

Answer 2 · 2016-10-29T17:03:15

The answer is $log_18 (9)+log_18 (2)$ $=1$ .

Explanation:

$log_18 (9)+log_18 (2)$
$=log_18 (9*2)$
$=log_18 (18)$
$=log18/log18$
$=1$ . (answer).

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How do you evaluate ${log}_{18} (9) + {log}_{18} (2)$ $log_18(9) + log_18(2)$ ?

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How do you evaluate ${log}_{18} (9) + {log}_{18} (2)$ $log_18(9) + log_18(2)$ ?