Question

(MACK) "Seja A = $\sqrt[n]{ \frac{90}{9^{n+2} + 3^{2n+2}}}$ , onde n > 1 é um número natural. Então ㏒1/9 A vale" :

a) -2;
b) -1;
c) 0;
d) 1;
e) 2.

Gabarito ⇒ Letra D

Answer 1

$\begin{array}{lcl} A&\!\!\!=\!\!\!&\,^n\!\!\!\!\sqrt{\dfrac{90}{9^{n+2}+3^{2n+2}}}\\\\ &\!\!\!=\!\!\!&\,^n\!\!\!\!\sqrt{\dfrac{90}{(3^2)^{n+2}+3^{2n+2}}}\\\\ &\!\!\!=\!\!\!&\,^n\!\!\!\!\sqrt{\dfrac{90}{3^{2(n+2)}+3^{2n+2}}}\\\\ &\!\!\!=\!\!\!&\,^n\!\!\!\!\sqrt{\dfrac{90}{3^{2n+4}+3^{2n+2}}}\\\\ &\!\!\!=\!\!\!&\,^n\!\!\!\!\sqrt{\dfrac{90}{3^{2n+2+2}+3^{2n+2}}}\\\\ &\!\!\!=\!\!\!&\,^n\!\!\!\!\sqrt{\dfrac{90}{3^{2n+2}\cdot 3^2+3^{2n+2}}} \end{array}$

   $\begin{array}{lcl} &\!\!\!=\!\!\!&\,^n\!\!\!\!\sqrt{\dfrac{90}{3^{2n+2}\cdot (3^2+1)}}\\\\ &\!\!\!=\!\!\!&\,^n\!\!\!\!\sqrt{\dfrac{90}{3^{2n+2}\cdot (9+1)}}\\\\ &\!\!\!=\!\!\!&\,^n\!\!\!\!\sqrt{\dfrac{90}{3^{2n+2}\cdot 10}}\\\\ &\!\!\!=\!\!\!&\,^n\!\!\!\!\sqrt{\dfrac{9}{3^{2n+2}}}\\\\ &\!\!\!=\!\!\!&\,^n\!\!\!\!\sqrt{\dfrac{3^2}{3^{2n+2}}} \end{array}$

$\begin{array}{lcl} &\!\!\!=\!\!\!&\,^n\!\!\!\!\sqrt{3^{2-(2n+2)}}\\\\ &\!\!\!=\!\!\!&\,^n\!\!\!\!\sqrt{3^{-2n}}\\\\ &\!\!\!=\!\!\!&(3^{-2n})^{\frac{1}{n}}\\\\ &\!\!\!=\!\!\!&3^{-2n\cdot \frac{1}{n}}\\\\ &\!\!\!=\!\!\!&3^{-2}\\\\ &\!\!\!=\!\!\!&\dfrac{1}{3^2}\\\\\\ A&\!\!\!=\!\!\!&\dfrac{1}{9} \end{array}$


Portanto,

$\begin{array}{lcl} \mathrm{\ell og}_{\frac{1}{9}}\,A&\!\!\!=\!\!\!&\mathrm{\ell og}_{\frac{1}{9}}\,\dfrac{1}{9}\\\\\\ \mathrm{\ell og}_{\frac{1}{9}}\,A&\!\!\!=\!\!\!&1\quad\quad\checkmark \end{array}$


Resposta: alternativa d) 1.


Dúvidas? Comente.


Bons estudos! :-)