Estude! Com Prof. Caju

(IEZZI) log_x/2 8 + log_x/4 8 < (log₂ x⁴) / (log₂ x² - 4

gab: 0 < x < 2 ou x > 4

Felipe22,

BAst apassar tudo para a base 2 e depois resolver a inequação quociente..
Caso não consiga avise

Felipe22,

[tex3]\frac{log_28}{log_2\frac{x}{2}}+ \frac{log_28}{log_2\frac{x}{4}} < \frac{4log_2x}{2log_2x-4}\\

\frac{3log_22}{log_2x-log_2{2}}+ \frac{3}{log_2x- log_24} < \frac{4log_2x}{2log_2x-4}\\
\frac{3}{log_2x-1}+ \frac{3}{log_2x- 2} < \frac{4log_2x}{2log_2x-4}\\
log_2x = t \implies \frac{3}{t-1}+ \frac{3}{t- 2} < \frac{4t}{2t-4}\\
\frac{6t-12+6t-6-4t^2+4t}{2t^2-4t-2t+4} <0 \implies \frac{-4t^2+16t-18}{2t^2-6t+4} < 0 \\
\therefore \frac{-2t^2+8t-9}{t^2-3t+2} < 0 \\
Resolva~a~inequação...
[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac{log_28}{log_2\frac{x}{2}}+ \frac{log_28}{log_2\frac{x}{4}} < \frac{4log_2x}{2log_2x-4}\\

\frac{3log_22}{log_2x-log_2{2}}+ \frac{3}{log_2x- log_24} < \frac{4log_2x}{2log_2x-4}\\
\frac{3}{log_2x-1}+ \frac{3}{log_2x- 2} < \frac{4log_2x}{2log_2x-4}\\
log_2x = t \implies \frac{3}{t-1}+ \frac{3}{t- 2} < \frac{4t}{2t-4}\\
\frac{6t-12+6t-6-4t^2+4t}{2t^2-4t-2t+4} <0 \implies \frac{-4t^2+16t-18}{2t^2-6t+4} < 0 \\
\therefore \frac{-2t^2+8t-9}{t^2-3t+2} < 0 \\
Resolva~a~inequação...
[/tex3]