Bonjour
Yacine931
Exercice 58
[tex]a)\ \ln(2+\sqrt{3})+\ln(2-\sqrt{3})=\ln[(2+\sqrt{3})(2+\sqrt{3})]\\\\\ln(2+\sqrt{3})+\ln(2-\sqrt{3})=\ln[2^2-(\sqrt{3})^2]\\\\\ln(2+\sqrt{3})+\ln(2-\sqrt{3})=\ln(4-3)\\\\\ln(2+\sqrt{3})+\ln(2-\sqrt{3})=\ln1\\\\\boxed{\ln(2+\sqrt{3})+\ln(2-\sqrt{3})=0}[/tex]
[tex]b)\ \ln\sqrt{\sqrt{3}+1}+\ln\sqrt{\sqrt{3}-1}=\ln(\sqrt{\sqrt{3}+1}\times\sqrt{\sqrt{3}-1})\\\\\ln\sqrt{\sqrt{3}+1}+\ln\sqrt{\sqrt{3}-1}=\ln(\sqrt{(\sqrt{3}+1)(\sqrt{3}-1)}\\\\\ln\sqrt{\sqrt{3}+1}+\ln\sqrt{\sqrt{3}-1}=\ln(\sqrt{(\sqrt{3})^2-1^2}\\\\\ln\sqrt{\sqrt{3}+1}+\ln\sqrt{\sqrt{3}-1}=\ln\sqrt{3-1}\\\\\ln\sqrt{\sqrt{3}+1}+\ln\sqrt{\sqrt{3}-1}=\ln\sqrt{2}\\\\\boxed{\ln\sqrt{\sqrt{3}+1}+\ln\sqrt{\sqrt{3}-1}=\dfrac{1}{2}\ln{2}}[/tex]
Exercice 59
[tex]a)\ \ln(e\sqrt{e})=\ln e+\ln\sqrt{e}\\\\\ln(e\sqrt{e})=\ln e+\dfrac{1}{2}\ln e\\\\\ln(e\sqrt{e})=1+\dfrac{1}{2}\times1\\\\\boxed{\ln(e\sqrt{e})=\dfrac{3}{2}}\\\\\\b)\ \ln(e^2)+\ln(e^{-2})=2-2\\\\\boxed{\ln(e^2)+\ln(e^{-2})=0}[/tex]
[tex]c)\ \ln e-\ln(\dfrac{1}{e})^2=1-\ln(\dfrac{1}{e^2})\\\\\ln e-\ln(\dfrac{1}{e})^2=1-[\ln 1-\ln(e^2)]\\\\\ln e-\ln(\dfrac{1}{e})^2=1-[0-2]\\\\\ln e-\ln(\dfrac{1}{e})^2=1+2\\\\\boxed{\ln e-\ln(\dfrac{1}{e})^2=3}[/tex]
