How do you solve ${log}_{4} {(x - 6)}^{3} = 6$ $log_4 (x-6)^3= 6$ ?

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Answer 1 · 2015-11-07T21:29:29

$x = 22$ $x=22$

By using laws of logs we can write this equation as :

$3 {log}_{4} (x - 6) = 6$ $3log_4(x-6)=6$

Now divide throughout by 3 to get

${log}_{4} (x - 6) = 2$ $log_4(x-6)=2$

Now applying the definition of log, we get

$x - 6 = 4^{2}$ $x-6=4^2$

$therefore x=22$

How do you solve log_4 (x-6)^3= 6log4(x−6)3=6log_4 (x-6)^3= 6?