How do you solve ${log}_{6} {(x - 4)}^{2} = 2$ $log_6 (x-4) ^2=2$ ?

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How do you solve ${log}_{6} {(x - 4)}^{2} = 2$ $log_6 (x-4) ^2=2$ ?

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Answer 1 · 2018-05-09T14:26:57

$x_{1} = 4 + 6^{\sqrt{2}}$ $x_"1"=4+6^sqrt2$
$x_{2} = 4 + \frac{1}{6^{\sqrt{2}}}$ $x_"2"=4+1/(6^sqrt2)$

Explanation:

$log_"6"(x-4)²=2$
$(ln(x-4)/ln(6))²=2$
$ln(x-4)/ln(6)=±sqrt2$
$ln(x-4)=±ln(6)sqrt2$
$x-4=6^(±sqrt2)$
$x=4+6^(±sqrt2)$
$x_"1"=4+6^sqrt2$
$x_"2"=4+1/(6^sqrt2)$
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How do you solve ${log}_{6} {(x - 4)}^{2} = 2$ $log_6 (x-4) ^2=2$ ?

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How do you solve ${log}_{6} {(x - 4)}^{2} = 2$ $log_6 (x-4) ^2=2$ ?