Pentru a logaritma o expresie putem să alegem orice logaritm, în orice bază. Vom alege logaritmul natural: ln
a) condiție: a,b,c>0
[tex]\ln \Big(5a^4b^{\frac{5}{6}} c^3\Big) = \ln 5 + \ln a^4 + \ln b^{\frac{5}{6}} + \ln c^3 = \ln 5 + 4\ln a + \dfrac{5}{6}\ln b + 3\ln c[/tex]
b) condiție: a,b,x,y>0
[tex]\ln \Big[(8a^2b^4):(3x\sqrt{y})\Big] = \ln \Big(2^3a^2b^4\Big) - \ln \Big(3xy^{\frac{1}{2}} \Big) = \Big(\ln 2^3 + \ln a^2 + \ln b^4 \Big) - \Big(\ln 3 + \ln x + \ln y^{\frac{1}{2}} \Big) = 3\ln 2 + 2\ln a + 4 \ln b - \ln 3 - \ln x - \dfrac{1}{2}\ln y[/tex]
c) condiție: a,b>0
[tex]\ln \Big(8\sqrt[3]{4a^2b^3} \Big) = \ln \Big(2^3\sqrt[3]{2^2a^2b^3} \Big) = \ln \Big(2^3 \cdot 2^{\frac{2}{3} } a^{\frac{2}{3} } b^{\frac{3}{3} }} \Big) = \ln 2^3 + \ln 2^{\frac{2}{3} } + \ln a^{\frac{2}{3} } + \ln b^{\frac{3}{3} }} = 3\ln 2 + \dfrac{2}{3}\ln 2 + \dfrac{2}{3}\ln a + \ln b = 3\dfrac{2}{3}\ln 2 + \dfrac{2}{3}\ln a + \ln b[/tex]
d) condiție: a,b>0
[tex]\ln \Big[(a^2\sqrt[4]{b^2}):(b^2\sqrt[3]{a})\Big] = \ln \Big(a^2b^{\frac{2}{4} }\Big) - \ln \Big(b^2a^{\frac{1}{3} }\Big) = \Big(\ln a^2 + \ln b^{\frac{1}{2} }\Big) - \Big(\ln b^2 + \ln a^{\frac{1}{3} }\Big) = 2\ln a + \dfrac{1}{2}\ln b - 2\ln b - \dfrac{1}{3}\ln a = \dfrac{5}{3}\ln a - \dfrac{3}{2}\ln b[/tex]
e) condiție: a,b>0
[tex]\ln \dfrac{4a^3\sqrt[3]{8a^7b^3} }{3\sqrt[2]{a^5b} } = \ln \Big( 4a^3\sqrt[3]{2^3a^7b^3}\Big) - \ln \Big(3\sqrt[2]{a^5b} \Big) = \ln \Big( 4 a^3 2^{\frac{3}{3}} a^{\frac{7}{3}} b^{\frac{3}{3} }\Big) - \ln \Big(3 a^{\frac{5}{2}} b^{\frac{1}{2}} \Big) = \Big( \ln 2^2 + \ln a^3 + \ln 2^{\frac{3}{3}} + \ln a^{\frac{7}{3}} + \ln b^{\frac{3}{3} }\Big) - \Big(\ln 3 + \ln a^{\frac{5}{2}} + \ln b^{\frac{1}{2}} \Big) =[/tex]
[tex]= 2\ln 2 + 3\ln a + \ln 2 + \dfrac{7}{3}\ln a + \ln b - \ln 3 - \dfrac{5}{2}\ln a - \dfrac{1}{2}\ln b = 3\ln 2 + \dfrac{18+14-15}{6}\ln a - \ln 3 + \dfrac{2-1}{2}\ln b = 3\ln 2 + \dfrac{17}{6}\ln a - \ln 3 + \dfrac{1}{2}\ln b[/tex]
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Am utilizat proprietățile:
[tex]\boldsymbol{\log_{a} x \cdot y = \log_{a} x + \log_{a} y, \ \ \ \ \log_{a} \bigg(\dfrac{x}{y} \bigg) = \log_{a} x - \log_{a} y}\\[/tex]
[tex]\boldsymbol{\red{ \log_{a} x^{n} = n \cdot \log_{a} x}}[/tex]
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Pentru a logaritma o expresie putem să alegem orice logaritm, în orice bază. Vom alege logaritmul natural: ln
a) condiție: a,b,c>0
[tex]\ln \Big(5a^4b^{\frac{5}{6}} c^3\Big) = \ln 5 + \ln a^4 + \ln b^{\frac{5}{6}} + \ln c^3 = \ln 5 + 4\ln a + \dfrac{5}{6}\ln b + 3\ln c[/tex]
b) condiție: a,b,x,y>0
[tex]\ln \Big[(8a^2b^4):(3x\sqrt{y})\Big] = \ln \Big(2^3a^2b^4\Big) - \ln \Big(3xy^{\frac{1}{2}} \Big) = \Big(\ln 2^3 + \ln a^2 + \ln b^4 \Big) - \Big(\ln 3 + \ln x + \ln y^{\frac{1}{2}} \Big) = 3\ln 2 + 2\ln a + 4 \ln b - \ln 3 - \ln x - \dfrac{1}{2}\ln y[/tex]
c) condiție: a,b>0
[tex]\ln \Big(8\sqrt[3]{4a^2b^3} \Big) = \ln \Big(2^3\sqrt[3]{2^2a^2b^3} \Big) = \ln \Big(2^3 \cdot 2^{\frac{2}{3} } a^{\frac{2}{3} } b^{\frac{3}{3} }} \Big) = \ln 2^3 + \ln 2^{\frac{2}{3} } + \ln a^{\frac{2}{3} } + \ln b^{\frac{3}{3} }} = 3\ln 2 + \dfrac{2}{3}\ln 2 + \dfrac{2}{3}\ln a + \ln b = 3\dfrac{2}{3}\ln 2 + \dfrac{2}{3}\ln a + \ln b[/tex]
d) condiție: a,b>0
[tex]\ln \Big[(a^2\sqrt[4]{b^2}):(b^2\sqrt[3]{a})\Big] = \ln \Big(a^2b^{\frac{2}{4} }\Big) - \ln \Big(b^2a^{\frac{1}{3} }\Big) = \Big(\ln a^2 + \ln b^{\frac{1}{2} }\Big) - \Big(\ln b^2 + \ln a^{\frac{1}{3} }\Big) = 2\ln a + \dfrac{1}{2}\ln b - 2\ln b - \dfrac{1}{3}\ln a = \dfrac{5}{3}\ln a - \dfrac{3}{2}\ln b[/tex]
e) condiție: a,b>0
[tex]\ln \dfrac{4a^3\sqrt[3]{8a^7b^3} }{3\sqrt[2]{a^5b} } = \ln \Big( 4a^3\sqrt[3]{2^3a^7b^3}\Big) - \ln \Big(3\sqrt[2]{a^5b} \Big) = \ln \Big( 4 a^3 2^{\frac{3}{3}} a^{\frac{7}{3}} b^{\frac{3}{3} }\Big) - \ln \Big(3 a^{\frac{5}{2}} b^{\frac{1}{2}} \Big) = \Big( \ln 2^2 + \ln a^3 + \ln 2^{\frac{3}{3}} + \ln a^{\frac{7}{3}} + \ln b^{\frac{3}{3} }\Big) - \Big(\ln 3 + \ln a^{\frac{5}{2}} + \ln b^{\frac{1}{2}} \Big) =[/tex]
[tex]= 2\ln 2 + 3\ln a + \ln 2 + \dfrac{7}{3}\ln a + \ln b - \ln 3 - \dfrac{5}{2}\ln a - \dfrac{1}{2}\ln b = 3\ln 2 + \dfrac{18+14-15}{6}\ln a - \ln 3 + \dfrac{2-1}{2}\ln b = 3\ln 2 + \dfrac{17}{6}\ln a - \ln 3 + \dfrac{1}{2}\ln b[/tex]
______
Am utilizat proprietățile:
[tex]\boldsymbol{\log_{a} x \cdot y = \log_{a} x + \log_{a} y, \ \ \ \ \log_{a} \bigg(\dfrac{x}{y} \bigg) = \log_{a} x - \log_{a} y}\\[/tex]
[tex]\boldsymbol{\red{ \log_{a} x^{n} = n \cdot \log_{a} x}}[/tex]