Помогите пожалуйста хоть до черты(((

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мохинсан_zn · Answer 1 · 2018-03-17T08:00:39+0000

1.
$36^{\log_{6}(5)+\log_{9}(81)}=36^{\log_{6}(5)+2}=36^{\log_{6}(5)+\log_{6}(6^2)}=\\ =36^{\log_{6}(5\cdot36)}=6^{2\log_{6}(180)}=6^{\log_{6}(108^2)}=\\ =180^2=(9\cdot20)^2=9^2\cdot20^2=81\cdot400=32400$
2.a)
0;x+1>0=>x>0;\\ \lg( \frac{x}{12})= -\lg(1+x)- \frac{1}{2}\lg(4);\\ \lg( \frac{x}{12} )=\lg((1+x)^{-1})-\lg((4)^{ \frac{1}{2} });\\ \lg( \frac{x}{12} )=\lg( \frac{1}{2(x+1)} );\\ \frac{x}{6}= \frac{1}{x+1}\\ \frac{x^2+x-6}{6(x+1)}=0\\ x^2+x-6=0;\\ D=1+24=25;\\ x_{1}= \frac{-1-5}{2}=-3\notin(0;+\infty);\\ x_{2}= \frac{-1+5}{2} =2\notin(0;+\infty);\\ x=2;" alt="\lg x-\lg12=\log_{0,1}(x+1)-log_{100}(4);\\ x>0;x+1>0=>x>0;\\ \lg( \frac{x}{12})= -\lg(1+x)- \frac{1}{2}\lg(4);\\ \lg( \frac{x}{12} )=\lg((1+x)^{-1})-\lg((4)^{ \frac{1}{2} });\\ \lg( \frac{x}{12} )=\lg( \frac{1}{2(x+1)} );\\ \frac{x}{6}= \frac{1}{x+1}\\ \frac{x^2+x-6}{6(x+1)}=0\\ x^2+x-6=0;\\ D=1+24=25;\\ x_{1}= \frac{-1-5}{2}=-3\notin(0;+\infty);\\ x_{2}= \frac{-1+5}{2} =2\notin(0;+\infty);\\ x=2;" align="absmiddle" class="latex-formula">
b)
0} \atop { \frac{9}{x-1}>0 }} \right. =>x-1>0; x>1; x\in(1;+\infty);\\ \log_{3}^2(x-1)+\log_{3}(9^2)-\log_{3}((x-1)^2)=7;\\ \log_{3}^2(x-1)-2\log_{3}(x-1)=7-4=3;\\ log_{3}(x-1)=t;\\ t^2-2t-3=0; D=4+12=16;\\ t_{1}= \frac{2-4}{2}=-1=>\log_{3}(x-1)=\log_{3}(3^{-1})=> x-1= \frac{1}{3}=>x=- \frac{2}{3}\\\notin(1;+\infty);\\ t_{2}= \frac{2+4}{2}=3;=>\log_{3}(x-1)=\log_{3}(3^3);=>x-1=27=>\\ x\in(0;+\infty);\\ " alt="\log_{3}^2(x-1)-2\log_{ \frac{1}{3}}( \frac{9}{x-1} )=2^{\log_{2}(7)};\\ \left \{ {{x-1>0} \atop { \frac{9}{x-1}>0 }} \right. =>x-1>0; x>1; x\in(1;+\infty);\\ \log_{3}^2(x-1)+\log_{3}(9^2)-\log_{3}((x-1)^2)=7;\\ \log_{3}^2(x-1)-2\log_{3}(x-1)=7-4=3;\\ log_{3}(x-1)=t;\\ t^2-2t-3=0; D=4+12=16;\\ t_{1}= \frac{2-4}{2}=-1=>\log_{3}(x-1)=\log_{3}(3^{-1})=> x-1= \frac{1}{3}=>x=- \frac{2}{3}\\\notin(1;+\infty);\\ t_{2}= \frac{2+4}{2}=3;=>\log_{3}(x-1)=\log_{3}(3^3);=>x-1=27=>\\ x\in(0;+\infty);\\ " align="absmiddle" class="latex-formula">
x=28;

c)
0;\\ e^{\ln(x^{\ln(x)})}=e^2x;\\ e^{\ln^2(x)}=e^2x;\\ \ln(e^{\ln^2(x)})=\ln(e^2)+\ln(x);\\ ln^2(x)-\ln(x)-2=0;\\ D=1+8=9;\\ \frac{1-3}{2}=-1; x=e^{(-1)}\in(0;+\infty)\\ \frac{1+3}{2}=2; x=e^2\in(0;+\infty)\\ x=e^{-1}; x=e^2;\\ " alt="x^{\ln(x)}=e^2\cdot x;\\ x>0;\\ e^{\ln(x^{\ln(x)})}=e^2x;\\ e^{\ln^2(x)}=e^2x;\\ \ln(e^{\ln^2(x)})=\ln(e^2)+\ln(x);\\ ln^2(x)-\ln(x)-2=0;\\ D=1+8=9;\\ \frac{1-3}{2}=-1; x=e^{(-1)}\in(0;+\infty)\\ \frac{1+3}{2}=2; x=e^2\in(0;+\infty)\\ x=e^{-1}; x=e^2;\\ " align="absmiddle" class="latex-formula">

3.
a) -3\log_{ \frac{1}{5}} ( \sqrt[3]{ \frac{1}{5}} );\\ x>2; x\in(2;+\infty);\\ \log_{ \frac{1}{3}}(x-2)>-1;\\ \log_{ \frac{1}{3}}(x-2)>\log_{ \frac{1}{3}}((\frac{1}{3})^-1) ;\\ x-2<3; x<5; \\ x\in(2;5)" alt="\log_{ \frac{1}{3}}(x-2)>-3\log_{ \frac{1}{5}} ( \sqrt[3]{ \frac{1}{5}} );\\ x>2; x\in(2;+\infty);\\ \log_{ \frac{1}{3}}(x-2)>-1;\\ \log_{ \frac{1}{3}}(x-2)>\log_{ \frac{1}{3}}((\frac{1}{3})^-1) ;\\ x-2<3; x<5; \\ x\in(2;5)" align="absmiddle" class="latex-formula">
b) [tex](1 \frac{11}{25})^{\log_{9}(x)}>( \frac{5}{6} )^{\log_{ \frac{1}{9}}(6-5x)} ; \left \{ {{x>0;} \atop {x< \frac{6}{5} }} \right.\\ ( \frac{36}{25} )^{\log_{9(x)}}>( \frac{6}{5} )^{\log_{9}(6-5x)};\\ 2\log_{9}(x)>\log_{9}(6-5x); x^2+5x-6>0;\\ D=25+24=49;\\ x_{1}= \frac{-5-7}{2}=-6\\ x_{2}= \frac{-5+7}{2}=1\\ \left \{ {{01}} \right. }} \right.\\ 1<>