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Let g(x)=log4(x^2-4x-21). Find the domain of g(x).
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\[g(x)=\frac{\ln(x^2-4x-21)}{\ln(4)} \text{ domain is } x^2-4x-21>0\]
make sure that the mess inside the log is positive, because you cannot take the log of a negative number. in other words solve \[x^2-4x-21>0\] for x
you do this via \[(x-7)(x+3)>0\] and this is a parabola that opens up, so it will be negative between the zeros and positive outside of them. your answer therefore will be \[(-\infty,-3)\cup (7,\infty)\]
THANKS!!!
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