вправа 4.37 гдз 11 клас алгебра Істер 2019
Вправа 4.37
Умова:
Обчисліть:
1) log2√31/12; 2) log√6/32,25; 3) log√8(4√2); 4) log0,4√2625/64.
1) log2√31/12; 2) log√6/32,25; 3) log√8(4√2); 4) log0,4√2625/64.
Відповідь:
1) log2√31/12 = log√1212-1 = 2 • (-1) • log1212 = -2;
2) log√6/32,25 = log√6/91,52 = log(2/3)1/2(2/3)2 = 2 • (-2) = -4;
3) log√8(4√2) = log21/3221/2 = 3 • 2 1/2 = 15/2 = 7,5;
4) log0,4√2625/64 = log√2•16/100(25/8)2 = log(0,32)1/20,32-2 = 2 • (-2) = -4.
2) log√6/32,25 = log√6/91,52 = log(2/3)1/2(2/3)2 = 2 • (-2) = -4;
3) log√8(4√2) = log21/3221/2 = 3 • 2 1/2 = 15/2 = 7,5;
4) log0,4√2625/64 = log√2•16/100(25/8)2 = log(0,32)1/20,32-2 = 2 • (-2) = -4.