Question

Resolva a seguinte equação exponencial:
$\displaystyle4^{\,x}-3^{\,x-\frac{1}{2}}=3^{\,x+\frac{1}{2}}-2^{\,2x-1}$

Answer 1

Resposta:

$\textsf{Leia abaixo}$

Explicação passo a passo:

$\sf 4^x - 3^{x - \frac{1}{2}} = 3^{x + \frac{1}{2}} - 2^{2x - 1}}}$

$\sf 2^{2x} - \dfrac{3^{x}}{\sqrt{3}} = 3^{x}\:.\:\sqrt{3} - \dfrac{2^{2x}}{2}}}}$

$\sf 2\sqrt{3}\:.\:2^{2x} - 2\:.\:3^{x} = 2\sqrt{3}\:.\:3^{x}\:.\:\sqrt{3} - 2^{2x}\:.\:\sqrt{3}$

$\sf 2\sqrt{3}\:.\:2^{2x} +2^{2x}\:.\:\sqrt{3} = 6\:.\:3^{x} + 2\:.\:3^{x}$

$\sf 2^{2x}\:.\:(2\sqrt{3} +\sqrt{3}) = 3^x\:.\:(6 + 2)$

$\sf 2^{2x}\:.\:3\sqrt{3} = 8\:.\:3^x$

$\sf \dfrac{3^x}{2^{2x}} = \dfrac{3\sqrt{3}}{8}$

$\sf \left(\dfrac{3}{4}\right)^x = \dfrac{3\sqrt{3}}{8}$

$\Large \boxed{\sf x = log_{\left(\frac{3}{4}\right)}\left(\dfrac{3\sqrt{3}}{8}\right)}}$

$\Large \boxed{\sf x = log_{\left(\frac{3}{4}\right)}\left(\dfrac{3}{4}\:.\:\dfrac{\sqrt{3}}{2}\right)}}$

$\Large \boxed{\sf x = log_{\left(\frac{3}{4}\right)}\left(\dfrac{3}{4}\right) + log_{\left(\frac{3}{4}\right)}\left(\dfrac{3}{4}\right)^\frac{1}{2}}}$

$\Large \boxed{\sf x = 1 + \dfrac{1}{2}}$

$\Large \boxed{\sf x = \dfrac{3}{2}}$