Question

o valor da expressão 4elevado a x+1 + 4elevado ao cubo -x = 257

Answer 1

$Propriedades\ utilizadas\ : \\\\ ---------------------- \\\\ \cdot \ a^x\ =\ a^y\ \Rightarrow\ x\ =\ y \\\\ \cdot\ Produto\ de\ pot\hat{e}ncia\ na\ mesma\ base \ : \ a^{x+y}\ =\ a^x\ .\ a^y\\\\ \cdot\ Quociente\ de\ pot\hat{e}ncia\ na\ mesma\ base\ :\ a^{x-y}\ =\ a^x\ .\ \frac{1}{a^y}\\\\ \cdot Mudan\c{c}a\ de\ base\ : \ a^x\ =\ y$

$\cdot\ Dada\ uma\ equa\c{c}\tilde{a}o\ do\ tipo\ \ ax^2\ +\ bx\ +\ c \ =\ 0 \ \ temos\\ que\ o\ discriminante\ (\Delta)\ e\ a\ F\acute{o}rmula\ de\ Bhaskara\ s\tilde{a}o\ ,\\ respectivamente\ ,\\\\ \Delta\ =\ b^2\ -\ 4\ .\ a\ .\ c\ \ \ \ \ \ \ e \ \ \ \ \ \ \ x\ =\ \frac{-b\ ^+_-\ \sqrt{\Delta}}{2a} \\\\ ----------------------$

$Dada\ a\ express\tilde{a}o\ :\\\\ 4^{x+1}\ +\ 4^{3-x}\ =\ 257\\\\ 4^x\ .\ 4^1\ +\ 4^3\ .\ \frac{1}{4^x}\ =\ 257\\\\ 4\ .\ 4^x\ +\ 64\ .\ \frac{1}{4^x}\ =\ 257\\\\ 4\ .\ 4^x\ .\ \frac{4^x}{4^x}\ +\ 64\ .\ \frac{1}{4^x}\ = \ 257\ .\ \frac{4^x}{4^x}\\\\ 4\ .\ 4^x\ .\ 4^x\ +\ 64\ =\ 257\ .\ 4^x\\\\ \boxed{\boxed{Mudan\c{c}a\ de\ base\ :\ \ \ 4^x\ =\ y\ \ }}\\\\\\ 4\ .\ y\ .\ y\ +\ 64\ =\ 257\ .\ y\\\\ 4\ .\ y^2\ -\ 257\ .\ y\ +\ 64\ =\ 0$

$\Delta\ =\ b^2\ -\ 4\ .\ a\ .\ c\\\\ \Delta\ =\ (-257)^2\ -\ 4\ .\ (4)\ .\ (64)\\\\ \Delta\ =\ 650025\\\\ y\ =\ \frac{-\ b\ ^+_-\ \sqrt{\Delta}}{2\ .\ a}\\\\ y\ =\ \frac{-(-257)\ ^+_- \sqrt{650025}}{2\ .\ (4)}\\\\ y\ =\ \frac{257\ ^+_-\ 255}{8}\\\\\\ y'\ =\ 64\\\\ y''\ = \frac{1}{4}$


$Voltando\ a\ mudan\c{c}a\ de\ base\ : \\\\ 4^x\ =\ y'\\\\ 4^x\ =\ 64\\\\ 4^x\ =\ 4^3\\\\ x\ =\ 3\\\\\\ 4^x\ =\ y''\\\\ 4^x\ =\ \frac{1}{4}\\\\ 4^x\ =\ 4^-^1 \\\\ x\ =\ -1$


$Assim\ ,\\\\ S\ =\ \begin{Bmatrix} -1 & , & 3 \end{Bmatrix}$