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solve for x: log3(x^2-4)-log3(x+2)=2
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\[\log_{3} (x^2-4)/(x+2) =2 \]
\[\log_{3}(x-2)=2 \]
this is just simplfying in steps and than from there it's 3^2=(x-2) and you can solve for x
if this helped please become a fan it's basically just logarhithimic rules the subtraction sign tells you the two binomials were being divided and than you can simplify after factoring the top, and than cancel and than i'm not sure what the method or rule i used what it was called but you just raise the base of the log to the 2 and than set it equal to the x-2
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