Question

O conjunto solução da inequação: 3.2^x+2 -2^2x>32

Answer 1

$\mathrm{3.2^{x+2}-2^{2x}\ \textgreater \ 32\ \to\ 3.2^x.2^2-2^{2x}\ \textgreater \ 32}\\\\ \mathrm{-(2^x)^2+12.2^x-32\ \textgreater \ 0\ \to\ (2^x)^2-12.2^x+32\ \textless \ 0}\\\\ \mathrm{*\ Se\ 2^x=m\ \to\ m^2-12m+32\ \textless \ 0}\\\\ \textbf{Pela f\'ormula quadr\'atica, teremos que:}\\\\ \mathrm{m=\dfrac{-(-12)\pm\sqrt{(-12)^2-4.(1).(32)}}{2.1}}\\\\ \mathrm{m=\dfrac{12\pm\sqrt{144-128}}{2}=\dfrac{12\pm\sqrt{16}}{2}=\dfrac{12\pm4}{2}=6\pm2}\\\\ \mathrm{m\ \textless \ 8\ \ \|\ \ m\ \textgreater \ 4\ \to\ \boxed{\mathrm{4\ \textless \ m\ \textless \ 8}}}$

$\mathrm{*\ Como\ m=2^x\ \to\ 4\ \textless \ 2^x\ \textless \ 8}\\\\ \mathrm{\log{2^2}\ \textless \ \log{2^x}\ \textless \ \log{2^3}\ \to\ 2.\log{2}\ \textless \ x.\log{2}\ \textless \ 3.\log{2}}\\\\ \mathrm{\dfrac{2.\log{2}}{\log{2}}\ \textless \ \dfrac{x.\log{2}}{\log{2}}\ \textless \ \dfrac{3.\log{2}}{\log{2}}\ \to\ 2\ \textless \ x\ \textless \ 3}\\\\\\ \boxed{\boxed{{\mathbf{S=\{x\in\mathbb{R}\ |\ 2\ \textless \ x\ \textless \ 3\}}}}}$