Question

Помогите пожалуйста

Answer 1

1)
$(8^{log_8 \sqrt[3]{5} })^3=(\sqrt[3]{5} })^3=5$

2)
$log_3 125* log_{25} 27 =log_3 5^3*log_{5^2} 27 = 3log_3 5* \frac{3log_5 3}{2} = \frac{9log_5 3}{2log_5 3}=4.5$

3)
$-log_2 log_2 \sqrt{ \sqrt[4]{2} } =-log_2 log_2 (2^{ \frac{1}{4} })^{ \frac{1}{2} }=-log_2 \frac{1}{16} log_2 2=-log_2 \frac{1}{16} =log_2 16\\=4$

4)
$25^{ \frac{1}{2} log_{ \sqrt{3} }4}=25^{log_3 4}$

5)
$log_6 9=log_6 x-log_6 4 \\\\ log_6 36=log_6 x \\\\ x=36$

6)
$log_4 (x^2+2x+49)=3 \\\\ log_4 (x^2+2x+49)=log_4 64 \\\\ x^2+2x-15=0 \\\\ (x+5)(x-3)=0 \\\\ x_1+x_2=-5+3=-2$

7)
$log_9 (20x-16)-log_9 4=log_9 18 \\\\ log_9 (20x-16)=log_9 72 \\\\ 20x-16=72 \\\\ 20x=88 \\\\ x=4.4$

8)
$5^{log_5 (2x^2)}=13x-21 \\\\ 2x^2=13x-21 \\\\ 2x^2-13x+21=0 \\\\ (x-3)(2x-7)=0 \\\\ x_1+x_2=3+3.5=6.5$

9)
$\sqrt{1+log_{ \frac{1}{8} }x} \\\\ \left \{ {{1+log_{ \frac{1}{8} }x \geq 0} \atop {x\ \textgreater \ 0}} \right. \\\\ \left \{ {{log_{ \frac{1}{8} }x \geq -1} \atop {x\ \textgreater \ 0}} \right. \\\\ \left \{ {{log_{ \frac{1}{8} }x \geq log_{ \frac{1}{8} } 8} \atop {x\ \textgreater \ 0}} \right. \\\\ \left \{ {{}x \leq 8} \atop {x\ \textgreater \ 0}} \right. \\\\ x\in(0;8]$

10)
$log^2 _2 x-4log_2 x+3 \geq 0 \\\\ log_2 x=t \\\\ t^2-4t+3 \geq 0 \\\\ (t-3)(t-1) \geq 0 \\\\ \left \{ {{t \geq 3} \atop {t \leq 1}} \right. \\\\ \left \{ {{log_2 x \geq 3} \atop {log_2 x \leq 1}} \right. \\\\ \left \{ {{log_2 x \geq log_2 8} \atop {log_2 x \leq log_2 2}} \right. \\\\ \left \{ {{x \geq8} \atop {x \leq 2}} \right. \\\\ x=-1; 0; 1; 2\\ (4)$