Question

Сравните числа (подробно) 1) log(1/2)1/3 и log(1/3)1/2 2) 2^(2log(2)5+log(1/9)9) и √8 решите уравнение log(3)x+log(√3)x+log(1/3)x=6 решить неравенство log(3√2)(x-5)+log(3√2)(x+12)≤2

Answer 1

$1)\; \; log_{\frac{1}{2}}\frac{1}{3}=log_{2^{-1}}3^{-1}=+log_23\\\\log_23\ \textgreater \ log_22=1\\\\log_{\frac{1}{3}}\frac{1}{2}=log_{3^{-1}}2^{-1}=log_32\\\\log_32\ \textless \ log_33=1\\\\\Rightarrow \; \; log_23\ \textgreater \ log_32\; \; \Rightarrow \underline {log_{\frac{1}{2}}\frac{1}{3}\ \textgreater \ log_{\frac{1}{3}}\frac{1}{2}}$

$2)\; \; 2^{2log_25}+log_{\frac{1}{9}}9=2^{log_25^2}+log_{9^{-1}}}9=5^2-log_99=25-1=24\\\\\sqrt8\ \textless \ \sqrt9,\; \; \sqrt9=3\; \; \Rightarrow \; \; \; \sqrt8\ \textless \ 3\\\\24\ \textgreater \ 3\; \; \Rightarrow \; \; (2^{2log_25}+log_{\frac{1}{9}}9)\ \textgreater \ \sqrt8$

$3)\; \; log_3x+log_{\sqrt3}x+log_{\frac{1}{3}}x=6\; ,\; \; \; \; ODZ:\; \; x\ \textgreater \ 0\\\\log_3x+log_{3^{\frac{1}{2}}}x+log_{3^{-1}}x=6\; \; \; [\, log_{a^{k}}x=\frac{1}{k}log_{a}x\, ]\\\\log_3x-2log_3x-log_3x=6\\\\-2log_3x=6\; \; \to \; \; log_3x=-3\; ,\; \; x=3^{-3}=\frac{1}{27}$

$4)\; \; log_{3\sqrt2}\, (x-5)+log_{3\sqrt2}(x+12) \leq 2\; ,\\\\ODZ:\; ,x\ \textgreater \ 5,x\ \textgreater \ -12\; \; \; \to \; \; x\ \textgreater \ 5\\\\log_{3\sqrt2}( (x-5)\cdot (x+12)) \leq log_{3\sqrt2}(3\sqrt2)^2\\\\(x-5)\cdot (x+12) \leq 18$

$x^2+7x-78 \leq 0\\\\x_1=-13\\\\x_2=6\\\\Otvet:x=6.$