Question

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Answer 1

Answer:   x = 6

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Work Shown:

$\log_{4}(x+10)+\log_{4}(x-2)=\log_{4}(64)\\\\\log_{4}\left((x+10)(x-2)\right)=\log_{4}(64)\\\\(x+10)(x-2)=64\\\\x^2-2x+10x-20=64\\\\x^2-2x+10x-20-64=0$

$x^2+8x-84=0\\\\(x+14)(x-6)=0\\\\x+14=0 \ \text{ or } \ x-6=0\\\\x=-14 \ \text{ or } \ x=6$

Those are the possible solutions, but plugging x = -14 back into the original equation will lead to an error. So we rule x = -14 out

x = 6 works as a solution however