Pré-Vestibular ⇒ (FEI) Inequação Logarítmica Tópico resolvido

[ariadni04]

Resolva a inequação:

[tex3]log_{\frac{1}{2}} (x - 1) - log_{\frac{1}{2}} (x + 1) \lt log_{\frac{1}{2}} + 1[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]log_{\frac{1}{2}} (x - 1) - log_{\frac{1}{2}} (x + 1) \lt log_{\frac{1}{2}} + 1[/tex3]

[petras]

O correto seria:
[tex3]log_{\frac{1}{2}} (x - 1) - log_{\frac{1}{2}} (x + 1) \lt log_{\frac{1}{2}}(x-2) + 1\\
CE: x-1 > 0,x+1 > 0,x-2 > 0\therefore \boxed{x>2}\\
log_\frac{1}{2}(\frac{x-1}{x+1}) < log_\frac{1}{2} (x-2) + log_\frac{1}{2}\frac{1}{2}\\
log_\frac{1}{2}(\frac{x-1}{x+1}) < log_\frac{1}{2} (x-2)(\frac{1}{2}) \\ \text{base entre 0 e 1 inverte a desigualdade}\\
(\frac{x-1}{x+1}) > \frac{x-2}{2}\rightarrow (\frac{x-1}{x+1}) -\frac{x-2}{2}> 0\rightarrow\frac{2x-2-x^2+2x-x+2}{2(x+1)} \\
\frac{-x^2+3 x}{2x+2}> 0\rightarrow x< -1(não~atende~CE)~ou~0 < x < 3\\
mas ~x > 2\therefore \boxed{\color{red}{2 < x < 3}}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]log_{\frac{1}{2}} (x - 1) - log_{\frac{1}{2}} (x + 1) \lt log_{\frac{1}{2}}(x-2) + 1\\
CE: x-1 > 0,x+1 > 0,x-2 > 0\therefore \boxed{x>2}\\
log_\frac{1}{2}(\frac{x-1}{x+1}) < log_\frac{1}{2} (x-2) + log_\frac{1}{2}\frac{1}{2}\\
log_\frac{1}{2}(\frac{x-1}{x+1}) < log_\frac{1}{2} (x-2)(\frac{1}{2}) \\ \text{base entre 0 e 1 inverte a desigualdade}\\
(\frac{x-1}{x+1}) > \frac{x-2}{2}\rightarrow (\frac{x-1}{x+1}) -\frac{x-2}{2}> 0\rightarrow\frac{2x-2-x^2+2x-x+2}{2(x+1)} \\
\frac{-x^2+3 x}{2x+2}> 0\rightarrow x< -1(não~atende~CE)~ou~0 < x < 3\\
mas ~x > 2\therefore \boxed{\color{red}{2 < x < 3}}[/tex3]