Salve, qualcuno saprebbe dirmi come risolvere queste espressioni esponenziali, anche solo qualcuno di loro. Vi ringrazio in anticipo.
82)
ln(e^4x) = 4x;
1/e^2 = e^-2;
ln(e^-2) = - 2.
4x = -2;
x = - 2/4 = - 1/2.
1/2 = 2^-1; 4 = 2^2;
4^x = 2^-1;
(2^2)^x = 2^-1;
2 ^ 2x = 2 ^ -1;
2x = - 1;
x = - 1/2;
infatti:
4 ^ (-1/2) = (1/4)^1/2 = radice(1/4) = 1/2.
9^x = 1/ (3^3),
(3^2)^x = 3 ^-3;
3^2x = 3 ^ -3;
2x = - 3;
x = - 3/2;
9^(-3/2) = (1/9) ^ 3/2 = [radicequadrata(1/9)]^3 = (1/3)^3 = 1/27.
Sono equazioni elementari.
Si possono ricondurre alla stessa base : a^x = a^q => x = q.
e^(4x) = 1/e^2 => e^(4x) = e^(-2) => 4x = -2 => x = -2/4 = -1/2
(1/2) = 4^x => (2^2)^x = 2^(-1) => 2^(2x) = 2^(-1) => 2x = - 1 => x = -1/2
9^x = 1/27 => (3^2)^x = 1/(3^3) => 3^(2x) = 3^(-3) => 2x = -3 => x = -3/2.
83)
0.3^x = 0.09 => 0.3^x = 0.3^2 => x = 2
(1/5)^x = 25 => 5^(-x) = 5^2 => - x = 2 => x = -2
[(2/3)^x]^2 = 8/27 => (2/3)^(2x) = (2/3)^3 => 2x = 3 => x = 3/2
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Answers & Comments
82)
ln(e^4x) = 4x;
1/e^2 = e^-2;
ln(e^-2) = - 2.
4x = -2;
x = - 2/4 = - 1/2.
1/2 = 2^-1; 4 = 2^2;
4^x = 2^-1;
(2^2)^x = 2^-1;
2 ^ 2x = 2 ^ -1;
2x = - 1;
x = - 1/2;
infatti:
4 ^ (-1/2) = (1/4)^1/2 = radice(1/4) = 1/2.
9^x = 1/ (3^3),
(3^2)^x = 3 ^-3;
3^2x = 3 ^ -3;
2x = - 3;
x = - 3/2;
9^(-3/2) = (1/9) ^ 3/2 = [radicequadrata(1/9)]^3 = (1/3)^3 = 1/27.
Sono equazioni elementari.
Si possono ricondurre alla stessa base : a^x = a^q => x = q.
82)
e^(4x) = 1/e^2 => e^(4x) = e^(-2) => 4x = -2 => x = -2/4 = -1/2
(1/2) = 4^x => (2^2)^x = 2^(-1) => 2^(2x) = 2^(-1) => 2x = - 1 => x = -1/2
9^x = 1/27 => (3^2)^x = 1/(3^3) => 3^(2x) = 3^(-3) => 2x = -3 => x = -3/2.
83)
0.3^x = 0.09 => 0.3^x = 0.3^2 => x = 2
(1/5)^x = 25 => 5^(-x) = 5^2 => - x = 2 => x = -2
[(2/3)^x]^2 = 8/27 => (2/3)^(2x) = (2/3)^3 => 2x = 3 => x = 3/2