Question

a) 1/27

b) 1/16

c) 1/9

d) 1/8

e) 1/4

Answer 1

Resposta:

$\textsf{Leia abaixo}$

Explicação passo a passo:

$\sf A = \begin{bmatrix}\cancel\sf 1&\cancel\sf 8\\\cancel\sf 2^{x + 1}&\cancel\sf log_2\:(y + 5)\end{bmatrix}$

$\sf A^{-1} = \begin{bmatrix}\cancel\sf 3&\cancel\sf -8\\\cancel\sf -\dfrac{1}{4}&\cancel\sf 1\end{bmatrix}$

$\sf A\:.\:A^{-1} = I$

$\sf \begin{bmatrix}\cancel\sf 1&\cancel\sf 8\\\cancel\sf 2^{x + 1}&\cancel\sf log_2\:(y + 5)\end{bmatrix}.\begin{bmatrix}\cancel\sf 3&\cancel\sf -8\\\cancel\sf -\dfrac{1}{4}&\cancel\sf 1\end{bmatrix} = \begin{bmatrix}\cancel\sf 1&\cancel\sf 0\\\cancel\sf 0&\cancel\sf 1\end{bmatrix}$

$\begin{cases}\sf 3\:.\:2^{x + 1} - \dfrac{1}{4}\:.\:log_2\:(y + 5) = 0\\\sf -8\:.\:2^{x + 1} + log_2\:(y + 5) = 1\end{cases}$

$\sf 12\:.\:2^{x + 1} -1 - 8\:.\:2^{x + 1} = 0$

$\sf 4\:.\:2^{x + 1} = 1$

$\sf 2^{x + 1} = 2^{-2}$

$\sf x + 1 = -2$

$\sf x = -3$

$\sf -8\:.\:2^{-3 + 1} + log_2\:(y + 5) = 1$

$\sf -\dfrac{8}{4} + log_2\:(y + 5) = 1$

$\sf log_2\:(y + 5) = 3$

$\sf log_2\:(y + 5) = log_2\:2^3$

$\sf y + 5 = 8$

$\sf y = 3$

$\sf y^x = 3^{-3}$

$\boxed{\boxed{\sf y^x = \dfrac{1}{27}}}$