Question

encontrei A=18.... queria ter certeza e tambem saber a ultima parte do exercicio

Answer 1

$\text{A} = \sqrt[n]{\frac{90}{9^{n + 2} + 3 ^{2n + 2}}} \\\\ \text{A} = \sqrt[n]{\frac{90}{(3^2)^{n + 2} + 3 ^{2n} \cdot 3^2}} \\\\ \text{A} = \sqrt[n]{\frac{90}{3^{2n + 4} + 3 ^{2n} \cdot 3^2}} \\\\ \text{A} = \sqrt[n]{\frac{90}{3^{2n} \cdot 3^4 + 3 ^{2n} \cdot 3^2}} \\\\ \text{A} = \sqrt[n]{\frac{90}{3^{2n} \cdot 81 + 3 ^{2n} \cdot 9}} \\\\ \text{Considerando temporariamente 3}^n = x, \text{temos:} \\\\ \text{A} = \sqrt[n]{\frac{90}{(3^n)^2 \cdot 81 + (3^n)^2 \cdot 9}}$
$\text{A} = \sqrt[n]{\frac{90}{x^2 \cdot 81 + x^2 \cdot 9}} \\\\ \text{A} = \sqrt[n]{\frac{90}{81x^2 + 9x^2}} \\\\ \text{A} = \sqrt[n]{\frac{\not{90}}{\not{90}x^2}} \\\\ \text{A} = \sqrt[n]{\frac{1}{x^2}} \rightarrow \sqrt[n]{\frac{1}{3^n}} = {\frac{\sqrt[n]{1}}{\sqrt[n]{3^n}}} = \boxed{\frac{1}{3}} \\ --------------------- \\ \bullet \text{log}_{\frac{1}{9}} \text{A} \rightarrow \text{log}_{\frac{1}{9}} \frac{1}{3} \rightarrow (\frac{1}{9})^x = \frac{1}{3} \\\\ (\frac{1}{9})^x = \frac{1}{3}$
$(9^{-1})^x = 3^{-1} \\\\ 9^{-x} = 3^{-1} \\\\ (3^2)^{-x} = 3^{-1} \\\\ 3^{-2x} = 3^{-1} \\\\ -2x = -1 \\\\ \boxed{x = \frac{1}{2}} \rightarrow \boxed{\text{log}_{\frac{1}{9}} \text{A} = \frac{1}{2}}$