How do you solve ${log}_{2} (x + 2) - {log}_{2} (x - 2) = 1$ $Log_2(x+2)-Log_2(x-2)=1$ ?

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How do you solve ${log}_{2} (x + 2) - {log}_{2} (x - 2) = 1$ $Log_2(x+2)-Log_2(x-2)=1$ ?

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Answer 1 · 2016-01-03T16:33:39

It is $x = 6$ $x=6$

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It is

${log}_{2} (x + 2) - {log}_{2} (x - 2) = 1 \Rightarrow {log}_{2} (\frac{x + 2}{x - 2}) = 1 \Rightarrow \frac{x + 2}{x - 2} = 2 \Rightarrow x + 2 = 2 \cdot (x - 2) \Rightarrow x + 2 = 2 x - 4 \Rightarrow x = 6$ $log_2(x+2)-log_2(x-2)=1=>log_2((x+2)/(x-2))=1=> (x+2)/(x-2)=2=>x+2=2*(x-2)=>x+2=2x-4=>x=6$

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How do you solve ${log}_{2} (x + 2) - {log}_{2} (x - 2) = 1$ $Log_2(x+2)-Log_2(x-2)=1$ ?

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How do you solve Log_2(x+2)-Log_2(x-2)=1log2(x+2)−log2(x−2)=1Log_2(x+2)-Log_2(x-2)=1?

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How do you solve ${log}_{2} (x + 2) - {log}_{2} (x - 2) = 1$ $Log_2(x+2)-Log_2(x-2)=1$ ?