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Question 1

Question 2

Решите задачу:

$(\frac{1}{49})^{-x}=\sqrt{\frac{1}{7}}\\ (\frac{1}{7})^{-2x}=(\frac{1}{7})^{\frac{1}{2}}\\ -2x=\frac{1}{2}\\ x=-\frac{1}{4}$

$4^x+7*2^{x-1}=4,5\\ 2^{2x}+7*2^x*2^{-1}=4,5\\ t=2^x; \ \ \ t\ \textgreater \ 0\\ t^2+3,5t=4,5\\ t^2+3,5t-4,5=0\\ D=3,5^2+4*4,5=30,25=5,5^2\\ t_1=\frac{-3,5+5,5}{2}=1; \ \ \ t_2=\frac{-3,5-5,5}{2}=-4,5\\ 2^x=1\\ x=0$

$\frac{1}{3^{5x-3}}\leq(\frac{1}{3})^{\frac{1}{5-3x}}\\ (\frac{1}{3})^{5x-3}\leq(\frac{1}{3})^{\frac{1}{5-3x}}\\ 5x-3\leq\frac{1}{5-3x}\\ (5x-3)(5-3x)\leq1\\ 25x-15-15x^2+9x-1\leq0\\ -15x^2+34x-16\leq0\\ \\ -15x^2+34x-16=0\\ D=34^2-4*(-15)*(-16)=196=14^2\\ x_1=\frac{-34+14}{30}=-\frac{2}3; \ \ \ x_2=\frac{-34-14}{30}=-1,6\\ x\in(-\infty;-1,6)\cup(-\frac{2}3;\infty)$

$log_2\frac{(\frac{1}{8})^3*2^{-0,5}}{(\frac{1}{4})^2*2^{\frac{1}{5}}}=log_2\frac{2^{-9}*2^{-0,5}}{2^{-\frac{1}{4}}*2^{\frac{1}{5}}}=log_2\frac{2^{-9,5}}{2^{-0,05}}}=log_22^{-9,55}=-9,55$

$a=log_{\frac{1}{5}}\frac{7}{5}=log_{\frac{1}{5}}7-log_{\frac{1}{5}}5^{-1}=log_{\frac{1}{5}}7-(-1)=log_{\frac{1}{5}}7+1\ \textgreater \ 1\\ b=(\frac{1}{3})^{\frac{2}{7}}=\frac{1^{\frac{2}{7}}}{3^{\frac{2}{7}}}=\frac{1}{3^{\frac{2}{7}}}\ \textless \ 1\\ a\ \textgreater \ b\\ \\ \\ a=\sqrt[4]{10000}=10\\ 3^{10}\ \textgreater \ 500 \ \ \ =\ \textgreater \ b=log_3500\ \textless \ 10\\ a\ \textless \ b \\ z$