f(x)=(\frac{1}{3})^{x+3}-1
f(x)=0\\\\ (\frac{1}{3})^{x+3}=1\\\\ (\frac{1}{3}^{x+3}+(\frac{1}{3})^0\\\\ x+3=0\\\\ x=-3
,
g(x)=log_{\frac{1}{2}}(19+x)+p
g(-3)=0\\\\ log_{\frac{1}{2}}[19+(-3)]+p=0\\\\log_{\frac{1}{2}}16+p=0\\\\ p=-log_{\frac{1}{2}}16^{-1}\\\\ p= log_{\frac{1}{2}}\frac{1}{16}\\\\ p=log_{\frac{1}{2}}(\frac{1}{2})^4\\\\ p=4