Question

Решить неравенство

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Answer 1

$2x\geq log_5(29*10^{x-1}-4^x)\\29*10^{x-1}-4^x0\frac{29}{10}*10^x4^x(\frac{5}{2} )^x\frac{10}{29}=xlog_{\frac{5}{2} }(\frac{10}{29})\\5^{2x}\geq 29*10^{x-1}-4^x\\5^{2x}-\frac{29}{10} *2^x*5^x+2^{2x} \geq 0\\1-\frac{29}{10} *(\frac{2}{5} )^x+(\frac{2}{5} )^{2x}\geq 0\\10(\frac{2}{5} )^{2x}-29(\frac{2}{5} )^{x}+10\geq 0\\(\frac{2^x}{5^{x-1}}-2 )(\frac{2^{x+1}}{5^x}-5)\geq 0\\\\(\frac{2}{5} )^{x}\leq \frac{2}{5} =x\geq 1\\(\frac{2}{5} )^{2x}\geq \frac{5}{2}=x\leq -1$

Теперь нанесём наше ограничение и получаем:

$log_{\frac{5}{2} }(\frac{10}{29} )$