Надо во всех уравнениях привести логарифмы к одному основанию.1) log5(x) - 3log5(5)/log5(x) = 2 ОДЗ: х >0 log5(x) -3/log5(x) = 2 | ·log5(x) log²5(x) -3 = 2log5(x) log²5(x) -2log5(x) -3 = 0log5(x) = t t² -2t -3 = 0по т. Виета t1 = 3 и t2 = -1a) t=3 б) t = -1log5(x) = 3 log5(x) = -1x = 5^3 x = 5^-1x = 125 x = 1/52)log4(3x+7) + 1/log4(3x +7) = 2,5 |·log4(3x +7) ОДЗ: 3х +7>0 ⇒x > -7/3log²4(3x+7) +1 = 2,5log4(3x+7)log4(3x +7) = tt² +1 = 2,5tt² - 2,5t +1 = 0по т. Виета t1 = 2 и t2 = 1/2а) t=2 б) t = 1/2log4(3x+7) = 2 log4(3x+7)=1/23x+7 = 4² 3x +7 = 4^1/23x +7 = 16 3x +7 = 23x = 9 3x = -5x = 3 (в ОДЗ входит) x = -5/3 ( в ОДЗ входит)Ответ: 3; -5/33) log1/3(x +2 = 3logx(1/3) ОДЗ: х>0 log1/3(x) +2 = 3/log1/3(x) |·log1/3(x) log²1/3(x) +2log1/3(x) = 3 log1/3(x) = tt² +2t -3 = 0по т. виета t1 = -3 и t = 1a) t = -3 б) t=1log1/3(x) = -3 log1/3(x) = 1x = 27 x = 1/34)log8(27x -1) + 1/log8(27x-1) = 10/3 |· 3log8(27x -1) 3 log²8(27x -1) +3 = 10 log8(27x -1) ОДЗ: 27х-1>0log8(27x -1) = t x > 1/27 3t² -10t + 3 = 0t1 = 3 и t2 = 1/3a) t=3 б) t = 1/3log8(27x -1) = 3 log8(27x -1) = 1/327x -1 = 8³ 27x - 1= 8^1/3 27x -1 = 512 27x -1 = 227x = 513 27x = 3 x = 513/27 x = 3/27 х = 19 (в ОДЗ входит) х = 1/9 (в ОДЗ входит)Ответ: 19; 1/9