Question

(FGV-SP) se 9^x= 72+3^x entao x^2+ 5 vale

Answer 1

$9^x=72+3^x\\\\ \big(3^2\big)^{\!x}=72+3^x\\\\ 3^{2x}=72+3^x\\\\ 3^{2x}-3^x-72=0\\\\ \big(3^{x}\big)^{\!2}-3^x-72=0$


Faça a seguinte mudança de variável:

$3^x=t~~~~(t>0)$


E a equação fica

$t^2-t-72=0~~~\Rightarrow~~\left\{ \!\begin{array}{l} a=1\\b=-1\\c=-72 \end{array} \right.\\\\\\ \Delta=b^2-4ac\\\\ \Delta=(-1)^2-4\cdot 1\cdot (-72)\\\\ \Delta=1+288\\\\ \Delta=289\\\\ \Delta=17^2$


$t=\dfrac{-b\pm \sqrt{\Delta}}{2a}\\\\\\ t=\dfrac{-(-1)\pm \sqrt{17^2}}{2\cdot 1}\\\\\\ t=\dfrac{1\pm 17}{2}\\\\\\ \begin{array}{rcl} t=\dfrac{1+17}{2}&~\text{ ou }~&t=\dfrac{1-17}{2}\\\\ t=\dfrac{18}{2}&~\text{ ou }~&t=\dfrac{-16}{2}\\\\ t=9&~\text{ ou }~&t=-8~~(\text{n\~ao serve}) \end{array}$


Então,

$t=9$


Voltando à variável $x,$ temos

$3^x=9\\\\ 3^x=3^2\\\\ \therefore~~\boxed{\begin{array}{c}x=2 \end{array}}$


Então,

$x^2+5\\\\ =2^2+5\\\\ =4+5\\\\\\ \therefore~~\boxed{\begin{array}{c}x^2+5=9 \end{array}}$


Bons estudos! :-)