Question

Salut, am nevoie de ajutor la ex. 1.255C( explicatii pas cu pas, va rog)

Answer 1

Răspuns:

a

Explicație pas cu pas:

$\sqrt{log_a(ax) + log_x(ax)} + \sqrt{log_a(\frac{x}{a}) + log_x(\frac{a}{x})} = 2\sqrt{a}\\\textrm{Aplicam } log_k(ab) = log_k(a) + log_k(b) \textrm{ si } log_k(a/b) = log_k (a) - log_k(b)\\\sqrt{log_a(a) + log_a(x) + log_x(a) + log_x(x)} + \sqrt{log_a(x) - log_a(a) + log_x(a) - log_x(x)} = 2\sqrt{a}$

$\sqrt{1 + log_a(x) + log_x(a) + 1} + \sqrt{log_a(x) + log_x(a) - 1 - 1} = 2\sqrt{a}\\\\\sqrt{log_a(x) + log_x(a) + 2} + \sqrt{log_a(x) + log_x(a) - 2} = 2\sqrt{a}\\\\\textrm{Notam b = }log_a(x) + log_x(a)\\\\\sqrt{b+2} + \sqrt{b-2} = 2\sqrt{a}\\\\\sqrt{b+2}\sqrt{b-2} = \sqrt{(b+2)(b-2)} = \sqrt{b^2 - 4}\\\\\sqrt{b+2} + \sqrt{b-2} = 2\sqrt{a}\Bigg | \hat{}\,\, 2\\\\ b \cancel{+ 2} + b \cancel{- 2} + 2\sqrt{b^2 - 4} = 4a\\\\2b+2\sqrt{b^2 - 4} = 4a\Bigg | \div 2\\\\ b + \sqrt{b^2 - 4} = 2a\\\\ log_a(x) + log_x(a) + \sqrt{(log_a(x) + log_x(a))^2 - 4} = 2a\\\\ log_a(x) + log_x(a) + \sqrt{log_a^2(x) + log_x^2(a) + 2log_x(a)\cdot log_a(x) - 4} = 2a$

$\textrm{Aplicam } a\cdot log_b(c) = log_b(c^a)\textrm{ astfel:}\\\\log_a(x)\cdot log_x(a) = log_x(a^{log_a(x)})\\\\\textrm{Acum, }a^{log_a(b)} = b\\\\ \implies log_x(a^{log_a(x)}) = log_x(x) = 1 = log_a(x) \cdot log_x(a)$

$log_a(x) + log_x(a) + \sqrt{log_a^2(x) + log_x^2(a) + 2\cdot 1 - 4} = 2a\\\\log_a(x) + log_x(a) + \sqrt{log_a^2(x) + log_x^2(a) - 2} = 2a\\\\log_a(x) + log_x(a) + \sqrt{log_a^2(x) + log_x^2(a) - 2\cdot log_a(x)\cdot log_x(a)} = 2a\\\\log_a(x) + log_x(a) + \sqrt{(log_a(x) - log_x(a))^2} = 2a\\\\log_a(x) + log_x(a) + | log_a(x) - log_x(a) | = 2a\\\\\textrm{Daca } log_a(x) - log_x(a) \geq 0:\\\\2log_a(x) = 2a\\\\a = log_a(x)\\\\ log_a(a^a) = log_a(x)\Rightarrow x = a^a\\\\\textrm{Daca }log_a(x) - log_x(a) < 0: \\\\ 2log_x(a) = 2a\\\\ log_x(a) = a\\\\ log_x(a) = log_x(x^a)\\\\ a = x^a\Big | \hat{}\,\,\frac{1}{a}\\\\ x = a^\frac{1}{a}\\\\\implies \boxed{x = a^{a^{\pm 1}}}$