peux-tu résoudre cette équation avec les formules des puissances uniquement

[@ZuKoshooEdit]

Bien jouer

[@user-xv4ng9bh6c]

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[@bensaid50]

Merci

[@key_board_x]

4^(x) - 3^[x - (1/2)] = 3^[x + (1/2)] - 2^(2x - 1) 4^(x) + 2^(2x - 1) = 3^[x + (1/2)] + 3^[x - (1/2)] 4^(x) + [2^(2x) * 2^(- 1)] = [3^(x) * 3^(1/2)] + [3^(x) * 3^(- 1/2)] 4^(x) + [{2^(2)}^(x) / 2^(1)] = [3^(x) * 3^(1/2)] + [3^(x) / 3^(1/2)] 4^(x) + [4^(x) / 2] = [3^(x) * √3] + [3^(x) / √3] 4^(x) + [4^(x) * (1/2)] = [3^(x) * √3] + [3^(x) * (1/√3)] 4^(x) * [1 + (1/2)] = 3^(x) * [√3 + (1/√3)] 4^(x) * (3/2) = 3^(x) * [(3 + 1)/√3] 4^(x) * (3/2) = 3^(x) * (4/√3) 4^(x) / 3^(x) = (4/√3) / (3/2) (4/3)^(x) = (4/√3) * (2/3) (4/3)^(x) = 8/(3√3) (4/3)^(x) = [2^(3)]/(3√3) (4/3)^(x) = [2^(3)]/(√3 * √3 * √3) (4/3)^(x) = [2^(3)]/[(√3)^(3)] (4/3)^(x) = (2/√3)^(3) (4/3)^(x) = (2/√3)^(2 + 1) (4/3)^(x) = [(2/√3)^(2)] * (2/√3)^(1) (4/3)^(x) = (4/3) * 2/√3 (4/3)^(x) / (4/3) = 2/√3 (4/3)^(x - 1) = 2/√3 [(2/√3)²]^(x - 1) = 2/√3 (2/√3)^[2.(x - 1)] = (2/√3)^(1) 2.(x - 1) = 1 2x - 2 = 1 2x = 3 → x = 3/2