Ex 3 va rog mult! Acela cu logaritmii

[tex]log_{2}(x) + log_{4}(x) + log_{8}(x) = \frac{11}{6} \\ log_{2}(x) + log_{ {2}^{2} }(x) + log_{ {2}^{3} }(x) = \frac{11}{6} \\ log_{2}(x) + \frac{1}{2} \cdot log_{2}(x) + \frac{1}{3} \cdot log_{2}(x) = \frac{11}{6} \\ \frac{6 \cdot log_{2}(x) + 3 \cdot log_{2}(x) + 2 \cdot log_{2}(x)}{6} = \frac{11}{6} \\ \frac{11 \cdot log_{2}(x)}{6} = \frac{11}{6} \iff log_{2}(x) = 1 \\ \implies \bf x = 2[/tex]

Seethh Seethh · Answer 2 · 2022-07-24T23:31:24+03:00

[tex]log_2x+log_4x+log_8x=\dfrac{11}{6} ~~~~~~~~~~~~~~~~C.E.~x > 0\\\\log_4x=\dfrac{log_2x}{log_24}=\dfrac{log_2x}{log_22^2}=\dfrac{log_2x}{2log_22}=\dfrac{log_2x}{2} \\\\log_8x=\dfrac{log_2x}{log_28} =\dfrac{log_2x}{log_22^3} =\dfrac{log_2x}{3log_33}=\dfrac{log_2x}{3} \\\\log_2x+\dfrac{log_2x}{2} +\dfrac{log_2x}{3}=\dfrac{11}{6} \\\\6log_2x+3log_2x+2log_2x=11\\\\11log_2x=11\Rightarrow log_2x=1 \Rightarrow log_2x=log_22\Rightarrow \boxed{x=2}[/tex]