Question

4 elevado a x + 6 elevado a x = 9 elevado a x
Qual o valor de x?

Answer 1

Resposta:

$\textsf{Leia abaixo}$

Explicação passo a passo:

$\sf{4^x + 6^x = 9^x}$

$\sf{\dfrac{4^x}{4^x} + \dfrac{6^x}{4^x} = \dfrac{9^x}{4^x}}$

$\sf{1 + \left(\dfrac{6}{4}\right)^x = \left(\dfrac{9}{4}\right)^x}$

$\sf{1 + \left(\dfrac{3}{2}\right)^x = \left(\dfrac{3}{2}\right)^{2x}}$

$\sf{\left(\dfrac{3}{2}\right)^{2x} - \left(\dfrac{3}{2}\right)^x - 1 = 0}$

$\sf{y = \left(\dfrac{3}{2}\right)^x}$

$\sf{y^2 - y - 1 = 0}$

$\sf{\Delta = b^2 - 4.a.c}$

$\sf{\Delta = (-1)^2 - 4.1.(-1)}$

$\sf{\Delta = 1 + 4}$

$\sf{\Delta = 5}$

$\sf{y = \dfrac{-b \pm \sqrt{\Delta}}{2a} = \dfrac{1 \pm \sqrt{5}}{2} \rightarrow \begin{cases}\sf{y' = \dfrac{1 + \sqrt{5}}{2}}\\\\\sf{y'' = \dfrac{1 - \sqrt{5}}{2}}\end{cases}}$

$\mathsf{\left(\dfrac{3}{2}\right)^x = \dfrac{1 + \sqrt{5}}{2}}$

$\Large \boxed{\boxed{\sf{x = log_{\frac{3}{2}}\left(\dfrac{1 + \sqrt{5}}{2}\right)}}}$