How do you condense $(1/2)log_6 9 +log_6 5$ ?

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Answer 1 · 2016-06-02T01:19:14

I tried this:

We can write:
$log_6(9)^(1/2)+log_6(5)=$
$=log_6(sqrt(9))+log_6(5)=$
$=log_6(3)+log_6(5)=$
$=log_6(3*5)=log_6(15)$

Answer 2 · 2016-06-02T01:25:09

$log_6 15=log 15 /log 6=1.5114$ , nearly.

Use $n log_b a = log_b (a^n), log_b(mn)=log_b m+log_b n,and log _b a = log_c a/log_c b$ .

So, $(1/2)log_6 9 + log_6 5$

$=log_6(9^(1/2)) + log_6 5$

$=log_6 (3X5)=log_6 15$

$=log 15/log 6=1.5114$ , nearly

How do you condense (1/2)log_6 9 +log_6 5(1/2)log_6 9 +log_6 5?