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# Triv

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Log4(x+2)+log4(x-2=2

Guest Jun 15, 2017
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I assume you want to find x

$$\log_4(x+2)+\log_4(x-2)= 2\\ \log_4((x+2)(x-2))=2 \\ (x+2)(x-2)=4^2\\x^2-4=16 \\ x^2= 20 \\ x=\pm2\sqrt5$$

However, the negative value is not valid in the original equation, so $$x = 2\sqrt5$$

Alan  Jun 15, 2017