Question

Na igualdade $3.2 ^{x+1}= \sqrt{72}$, o valor de $x^{-1}$ é

(A) 4.
(B) -1/2
(C) -2.
(D) 2.
(E) 1/2.

Answer 1

$3\cdot2^{x+1}=\sqrt{72}~\Rightarrow~3^2\cdot(2^{x+1})^2=(\sqrt{72})^2~\Rightarrow~9\cdot2^{2x+2}=72$

$2^{2x+2}=\dfrac{72}{9}$

$2^{2x+2}=8$

$\not2^{2x+2}=\not2^3$

$2x+2=3$

$2x=1$

$x=\dfrac{1}{2}$

$x^{-1}=\left(\dfrac{1}{2}\right)^{-1}=2$

Letra D