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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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 · 2015-07-30T13:37:01

I found: $b = \pm 4$ $b=+-4$

Explanation:

You can start using a property of logs:
$log x + log y = log x y$ $logx+logy=logxy$
so you get:
${log}_{6} [2 (b^{2} + 2)] = 2$ $log_6[2(b^2+2)]=2$
using the definition of log:
${log}_{a} x = b \to x = a^{b}$ $log_ax=b -> x=a^b$ you get:
$2 (b^{2} + 2) = 6^{2} = 36$ $2(b^2+2)=6^2=36$
$b^{2} + 2 = \frac{36}{2} = 18$ $b^2+2=36/2=18$
$b^{2} = 18 - 2 = 16$ $b^2=18-2=16$
so that $b = \pm \sqrt{16} = \pm 4$ $b=+-sqrt(16)=+-4$

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