Question

A soma das raizes da equaçao:
2 elevado a 2×-1-2elevado a ×+4=2 elevado a ×+2-32... me ajudem por favor e urgente

Answer 1

$\mathrm{2^{2x-1}-2^{x+4}=2^{x+2}-32}\\\\ \mathrm{\dfrac{2^{2x}}{2}-2^x.2^4=2^x.2^2-32}\ \to\ \mathrm{2^{2x}-2^x.2^5=2^x.2^3-64}\\\\ \mathrm{\big(2^x\big)^2-32.2^x=8.2^x-64}\ \to\ \mathrm{\big(2^x\big)^2-40.2^x+64=0}\\\\ \mathrm{*\ Se\ 2^x=w\ \to\ w^2-40w+64=0}\\\\ \textbf{Pela f\'ormula quadr\'atica, teremos que:}\\\\ \mathrm{w=\dfrac{-(-40)\pm\sqrt{(-40)^2-4(1)(64)}}{2.1}}\\\\ \mathrm{w=\dfrac{40\pm\sqrt{1344}}{2}=\dfrac{40\pm8\sqrt{21}}{2}=4(5\pm\sqrt{21})}$

$\mathrm{2^x=4(5\pm\sqrt{21})\ \to\ \log{2^x}=\log{4(5\pm\sqrt{21})}}\\\\ \mathrm{x.\log{2}=\log{4(5\pm\sqrt{21})}\ \to\ x=\dfrac{\log{4(5\pm\sqrt{21})}}{\log{2}}}\\\\ \mathrm{x=\log_2{4(5\pm\sqrt{21})}\ \to\ x=log_24+\log_2{(5\pm\sqrt{21})}}\\\\ \mathrm{x=2+\log_2{(5\pm\sqrt{21})}}\\\\ \boxed{\mathrm{x=\{2+\log_2{(5+\sqrt{21})},2+\log_2{(5-\sqrt{21})}\}}}$

$\mathbf{Soma\ das\ ra\'izes:}}\\\\ \mathrm{2+\log_2{(5+\sqrt{21})}+2+log_2{(5-\sqrt{21})}=}\\\\ \mathrm{=4+\log_2{\big((5+\sqrt{21})(5-\sqrt{21})\big)}=}\\\\ \mathrm{=4+\log_2{\big(5^2-(\sqrt{21})^2\big)}=4+\log_2{(25-21)}=}\\\\ \mathrm{=4+\log_2{4}=4+2=\boxed{\mathbf{6}}}$