Question

Решите, пожалуйста, уравнение, желательно с подробным решением.

Answer 1

Ответ:

$36\, log_{\frac{1}{8}}^2\, x+4\, log_{\frac{1}{4}}\, x-5=0\ \ \ ,\ \ \ \ ODZ:\ x>0\ ,\\\\36\cdot \dfrac{1}{9}\, log^2_{\frac{1}{2}}\, x+2\, log_{\frac{1}{2}}\, x-5=0\\\\t=log_{\frac{1}{2}}\, x\ \ ,\ \ \ 4t^2+2t-5=0\ \ ,\ \ D/4=21\ ,\\\\t_1=\dfrac{-1-\sqrt{21}}{4}\approx -1,395\ \ \ ,\ \ \ t_2=\dfrac{-1+\sqrt{21}}{4}\approx 0,896\\\\\\a)\ \ log_{\frac{1}{2}}\, x=\dfrac{-1-\sqrt{21}}{4}\ \ ,\ \ log_{2^{-1}}\, x=\dfrac{-1-\sqrt{21}}{4}\ \ ,\ \ -log_2\, x=\dfrac{-1-\sqrt{21}}{4}\ \ ,$

$log_2x=\dfrac{1+\sqrt{21}}{4}\approx 1,396\ \ \ ,\ \ \ x=2^{\frac{1+\sqrt{21}}{4}}\approx 2^{1,396}\approx 2,632\\\\\\b)\ \ log_{\frac{1}{2}}\, x=\dfrac{-1+\sqrt{21}}{4}\ \ ,\ \ log_{2^{-1}}\, x=\dfrac{-1+\sqrt{21}}{4}\ \ ,\ \ -log_2\, x=\dfrac{-1+\sqrt{21}}{4}\ ,\\\\\\log_2x=\dfrac{1-\sqrt{21}}{4}\approx -0,896\ \ ,\ \ \ x=2^{\frac{1-\sqrt{21}}{4}}\approx 2^{-0,896}\approx 0,537\\\\\\c)\ \ x\in [\ 0,5\ ;\ 5\ ]\, :\ \ x_1=2^{\frac{1+\sqrt{21}}{4}}\ \ ,\ \ x_2=2^{\frac{1-\sqrt{21}}{4}}\ .$