1/LN(x) + 1/(2·LN(x) - 1) = 2
C.E. x > 0
LN(x) = t
1/t + 1/(2·t - 1) = 2
la porto alla forma intera:
t·(2·t - 1) ≠ 0---> t ≠ 1/2 ∧ t ≠ 0
(1/t + 1/(2·t - 1) = 2)·t·(2·t - 1)
3·t - 1 = 2·t·(2·t - 1)
2·t·(2·t - 1) - (3·t - 1) = 0
4·t^2 - 5·t + 1 = 0
(t - 1)·(4·t - 1) = 0
t = 1/4 ∨ t = 1
x = e ∨ x = e^(1/4)
1 / (ln x) + 1 /(2 ln x - 1) = 2;
lnx ≠ 0;
2 ln x - 1 ≠ 0; lnx ≠1/2; non si devono annullare i denominatori;
chiamiamo ln x = y;
1 / y + 1 / (2y - 1) = 2;
mcm = y (2y - 1);
2y - 1 + y = 2 y (2y - 1);
3y - 1 = 4y^2 - 2y;
4y^2 - 2y - 3y + 1 = 0;
4y^2 - 5y + 1 = 0;
y = [+ 5 +- radicequadrata(25 - 4 * 4 * 1)] /(2 * 4);
y = [+ 5 +- radicequadrata(25 - 16)] / 8;
y = [+ 5 +- radicequadrata(9)] / 8;
y1 = [+ 5 + 3] / 8 = 8/8;
y1 = 1;
y2 = [+ 5 - 3] / 8 = 2/8;
y2 = 1/4;
ln x = 1;
x1 = e^1 = e; e = 2,7182818...
lnx = 1/4;
x2 = e^(1/4) = radicequarta (e).
Ciao @alby