Ensino Médio ⇒ inequação logaritmica Tópico resolvido

[Felipe22]

(IEZZI) log (base (2x+5)/2 de [(x-5)/2x-3)]^2 > 0

gab:-5/2 < x < -2 ou -3/2 < x < 8/3 e x[tex3]\neq 3/2[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\neq 3/2[/tex3]

[petras]

Felipe22,

[tex3]log_{\frac{2x+5}{2}} \left(\frac{x-5}{2x-3}\right)^2 > 0\\
C.E.: \\\boxed{2x+5 >0 \implies x > -\frac{5}{2} \wedge \frac{2x+5}{2} \neq 1 \implies x \neq -\frac{3}{2}\\
2x-3 \neq 0 \implies x \neq \frac{3}{2} \wedge (\frac{x-5}{2x-3})^2>0 \implies x-5 \neq 0 \therefore x \neq 5}(i)\\
Caso1: \frac{2x+5}{2} >1 \implies x >-\frac{3}{2}(ii)\\
(\frac{x-5}{2x-3})^2 > 1 \implies -2 < x < \frac{8}{3} (iii)\\
S1=(i) \cap(ii) \cap(iii) = -\frac{3}{2} < x < \frac{8}{3} \wedge x \ne \frac{3}{2}\\
Caso2:0 < \frac{2x+5}{2} < 1 \implies -\frac{5}{2} < x < -\frac{3}{2}(iv)\\
(\frac{x-5}{2x-3})^2 {\color{red} <} 1 \implies x < -2 \vee x > \frac{8}{3}(v)\\
S2= (i) \cap (iv)\cap(v)= -\frac{5}{2} < x < -2\\
S = S1 \cup S2 = \boxed{ -\frac{3}{2} < x < \frac{8}{2} \wedge x \neq \frac{3}{2} \vee -\frac{5}{2} < x < -2}



[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]log_{\frac{2x+5}{2}} \left(\frac{x-5}{2x-3}\right)^2 > 0\\
C.E.: \\\boxed{2x+5 >0 \implies x > -\frac{5}{2} \wedge \frac{2x+5}{2} \neq 1 \implies x \neq -\frac{3}{2}\\
2x-3 \neq 0 \implies x \neq \frac{3}{2} \wedge (\frac{x-5}{2x-3})^2>0 \implies x-5 \neq 0 \therefore x \neq 5}(i)\\
Caso1: \frac{2x+5}{2} >1 \implies x >-\frac{3}{2}(ii)\\
(\frac{x-5}{2x-3})^2 > 1 \implies -2 < x < \frac{8}{3} (iii)\\
S1=(i) \cap(ii) \cap(iii) = -\frac{3}{2} < x < \frac{8}{3} \wedge x \ne \frac{3}{2}\\
Caso2:0 < \frac{2x+5}{2} < 1 \implies -\frac{5}{2} < x < -\frac{3}{2}(iv)\\
(\frac{x-5}{2x-3})^2 {\color{red} <} 1 \implies x < -2 \vee x > \frac{8}{3}(v)\\
S2= (i) \cap (iv)\cap(v)= -\frac{5}{2} < x < -2\\
S = S1 \cup S2 = \boxed{ -\frac{3}{2} < x < \frac{8}{2} \wedge x \neq \frac{3}{2} \vee -\frac{5}{2} < x < -2}



[/tex3]