Question

Sa se rezolve ecuatia:

log₂x +log₄x+log₈x=4

Answer 1

$\displaystyle log_2x+log_4x+log_8x=4 \\ \\ log_4x= \frac{log_2x}{log_24} = \frac{log_2x}{log_22^2} = \frac{log_2x}{2log_22} = \frac{log_2x}{2} \\ \\ log_8x= \frac{log_2x}{log_28} = \frac{log_2x}{log_22^3} = \frac{log_2x}{3log_22} = \frac{log_2x}{3} \\ \\ log_2x+log_4x+log_8x=4 \\ \\ log_2x+ \frac{log_2x}{2} + \frac{log_2x}{3} =4 \\ \\ 6log_2x+3log_2x+2log_2x=6 \cdot 4 \\ \\ 11log_2x=24 \Rightarrow log_2x= \frac{24}{11} \Rightarrow x=2^ \frac{24}{11} \Rightarrow \boxed{x=4 \cdot 2^ \frac{2}{11} }$