How do you solve $log_6(b^2+2) + log_6 2=2$ ?

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Answer 1 · 2015-06-20T04:45:05

$b=+-4$

Start by usig the following property of logs:

$log_ab+log_ac=log_a(b*c)$
so you get:

$log_6[2(b^2+2)]=2$

now use the definition of log:
$log_ax=b ->x=a^b$

in your case:
$2(b^2+2)=6^2$
$b^2+2=18$
$b^2=16$
$b=+-sqrt(16)=+-4$

How do you solve log_6(b^2+2) + log_6 2=2log_6(b^2+2) + log_6 2=2?