How do you solve $2 log (x - 4) = 4 log 2$ $2 Log (x - 4) = 4 log 2$ ?

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How do you solve $2 log (x - 4) = 4 log 2$ $2 Log (x - 4) = 4 log 2$ ?

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Answer 1 · 2015-11-25T03:07:18

We must have $x - 4 > 0 \Rightarrow x > 4$ $x-4>0=>x>4$ hence

$2 log (x - 4) = 4 log 2 \Rightarrow {log (x - 4)}^{2} = {log 2}^{4} \Rightarrow {(x - 4)}^{2} = 2^{4} \Rightarrow x - 4 = \pm 4 \Rightarrow x = 8 or x = 0$ $2log(x-4)=4log2=>log(x-4)^2=log2^4=>(x-4)^2=2^4=>x-4=+-4=>x=8 or x=0$

Finally $x = 8$ $x=8$

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How do you solve $2 log (x - 4) = 4 log 2$ $2 Log (x - 4) = 4 log 2$ ?

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