How do you solve ${log}_{6} (b^{2} + 2) + {log}_{6} 2 = 2$ $log_6(b^2+2) + log_6 2=2$ ?

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Answer 1 · 2015-06-20T04:45:05

$b = \pm 4$ $b=+-4$

Start by usig the following property of logs:

${log}_{a} b + {log}_{a} c = {log}_{a} (b \cdot c)$ $log_ab+log_ac=log_a(b*c)$
so you get:

${log}_{6} [2 (b^{2} + 2)] = 2$ $log_6[2(b^2+2)]=2$

now use the definition of log:
${log}_{a} x = b \to x = a^{b}$ $log_ax=b ->x=a^b$

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

How do you solve log_6(b^2+2) + log_6 2=2log6(b2+2)+log62=2log_6(b^2+2) + log_6 2=2?