1) log1/9(x+2)+3log27(x+2)=1 2) log16x+log8x+log2x= 19/12
1) log1/9(x+2)+3log27(x+2)=1
2) log16x+log8x+log2x= 19/12
Ответ(ы) на вопрос:
1) [latex]log_{ \frac{1}{9}}(x+2)+3log_{27}(x+2)=1 [/latex]
[latex]log_{3^{-1}}(x+2)+3log_{3^{3}}(x+2)=1 [/latex]
[latex]-0.5*log_{3}(x+2)+log_{3}(x+2)=1[/latex]
[latex]0.5*log_{3}(x+2)=1[/latex]
[latex]log_{3}(x+2)=2[/latex]
[latex]x+2=9, x=7[/latex]
2) [latex]log_{16}x+log_{8}x+log_{2}x= \frac{19}{12} [/latex]
[latex]log_{2^{4}}x+log_{2^{3}}x+log_{2}x= \frac{19}{12} [/latex]
[latex] \frac{1}{4}*log_{2}x+ \frac{1}{3}*log_{2}x+log_{2}x= \frac{19}{12}[/latex]
[latex]3log_{2}x+ 4log_{2}x+12log_{2}x=19[/latex]
[latex]19log_{2}x=19[/latex]
[latex]log_{2}x=1[/latex]
[latex]x=2[/latex]
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