Răspuns:
Explicație pas cu pas:
1) 5ˣ⁺¹ + 5ˣ⁺² = 3ˣ⁺¹ + 3ˣ⁺² + 3ˣ⁺³ <=>
5ˣ⁺¹·(1+5) = 3ˣ⁺¹·(1+3+9) =>
6·5·5ˣ = 3·13·3ˣ =>
10·5ˣ = 13·3ˣ =>
5ˣ/3ˣ = 13/10 <=>
(5/3)ˣ = 13/10 =>
x = log₅₍₃(13/10)
2) 3ˣ⁻¹ - 5ˣ⁺¹ = 3ˣ⁺³ - 5ˣ⁺² <=>
3ˣ/3 - 5·5ˣ = 3³·3ˣ - 5²·5ˣ I·3 =>
3ˣ - 3⁴·3ˣ = 15·5ˣ-5²·5ˣ =>
3ˣ·(1-81) = 5ˣ·(15-25) <=>
80·3ˣ = 10·5ˣ I:10 =>
8·3ˣ = 5ˣ =>
5ˣ/3ˣ = 8 =>
(5/3)ˣ = 8 =>
x = log₅₍₃8
3) 32·√[(0,125)³⁻ˣ⁽²] = 2^√(x+2) <=> conditie x+2 ≥ 0
2⁵·√(2⁻³·⁽⁶⁻ˣ⁾²) = 2^√(x+2) <=>
2⁵·√(2⁻⁹⁺³ˣ) = 2^√(x+2) =>
(2³ˣ⁻⁹⁺¹⁰)¹⁾² = 2^√(x+2) =>
(2³ˣ⁺¹)¹⁾² = 2^√(x+2) I² =>
2³ˣ⁺¹ = 2^2√(x+2) =>
3x+1 = 2√(x+2) I² =>
9x²+6x+1 = 4x+8 =>
9x²+2x-7 = 0 => x₁,₂ = [-2±√(4+252)]/18 = >
x₁,₂ = (-2±16)/18 => x₁ = -18/18 => x₁ = -1 ;
x₂ = 14/18 => x₂ = 7/9
