[tex]1. \ \lg 2 - \lg 3 - \lg 4 + \lg 6 = (\lg 2 + \lg 6) - (\lg 3 + \lg 4) = \lg (2 \cdot 6) - \lg (3 \cdot 4) = \lg 12 - \lg 12 = 0[/tex]
⁕⁕⁕
[tex]2. = \bigg[\log_{5} \bigg(\dfrac{1}{\not2} \cdot \dfrac{\not2}{\not3} \cdot ... \cdot \dfrac{\not23}{\not24} \cdot \dfrac{\not24}{25}\bigg) {\bf :} \log_{2} 2^2\bigg]^{2010}[/tex]
[tex]= \bigg(\log_{5} \dfrac{1}{25} : 2\log_{2}2\bigg)^{2010} = \bigg(\log_{5} 5^{-2} : 2\bigg)^{2010}[/tex]
[tex]= \Big(-2 : 2\Big)^{2010} = \Big(-1\Big)^{2010} = \bf 1[/tex]
⁕⁕⁕
3.a) x > 0, x ≠1 și x+4 > 0 ⇒ x > -4
⇒ x ∈ (0, +∞) \ {1}
b) x²-3x+2 > 0 ⇒ (x-1)(x-2) > 0 ⇒ x₁=1, x₂=2
⇒ x ∈ R \ [1; 2]
⁕⁕⁕
[tex]4. \ a = \log_{3} (3 \cdot 3^2 \cdot 3^3 \cdot 3^4) = \log_{3} 3^{1+2+3+4} = \log_{3} 3^{10} = 10 \log_{3} 3 = 10 \cdot 1 = \bf 10[/tex]
[tex]b = \log_{2} (2^2 \cdot 2^3 \cdot 2^4 \cdot 2^5) = \log_{2} 2^{2+3+4+5} = \log_{2} 2^{14} = 14 \log_{2} 2 = \bf 14[/tex]
[tex]\implies \bf a < b[/tex]
⁕⁕⁕
[tex]5. \ E = \dfrac{\log_{3} (2^2 \cdot 3^2) \cdot \log_{2^3} 3}{\log_{3} (2 \cdot 3^2) \cdot \log_{3}(2^3 \cdot 3^2)} = \dfrac{(\log_{3} 2^2 + \log_{3}3^2) \cdot \dfrac{1}{3} \log_{2} 3}{(\log_{3} 2 + \log_{3}3^2) \cdot (\log_{3}2^3 + \log_{3} 3^2)}[/tex]
[tex]= \dfrac{(2\log_{3} 2 + 2) \cdot \dfrac{1}{3} \log_{2} 3}{(\log_{3} 2 + 2) \cdot (3\log_{3} 2 + 2)} = \dfrac{\bigg(2\dfrac{1}{a} + 2\bigg) \cdot \dfrac{1}{3} a}{\bigg(\dfrac{1}{a} + 2\bigg) \cdot \bigg(3\dfrac{1}{a} + 2\bigg)} = \dfrac{\dfrac{2(a + 1)}{3}}{\dfrac{2a+1}{a} \cdot \dfrac{2a + 3}{a}}[/tex]
[tex]= \dfrac{2a^2(a+1)}{3(2a+1)(2a+3)}[/tex]
______
[tex]\boldsymbol{\red{ \log_{a} x^{n} = n \cdot \log_{a} x}; \ \ \log_{a} x = \dfrac{1}{\log_{x} a}}[/tex]
[tex]\boldsymbol{\log_{a} x \cdot y = \log_{a} x + \log_{a} y, \ \ \ \ \log_{a} \dfrac{x}{y} = \log_{a} x - \log_{a} y}\\[/tex]
[tex]\boldsymbol{\log_{ {a}^{n}} x = \dfrac{1}{n} \log_{a} x; \ \ \log_{a} a = 1} }[/tex]
