Bài 6.12 trang 10 SBT Toán 11 - Kết nối tri thức: Chứng minh rằng: \({\rm{lo}}{{\rm{g}}_a}\left( {x + \sqrt {{x^2} - 1} } \right) + {\rm{lo}}{{\rm{g}}_a}\left( {x - \sqrt {{x^2} - 1}...

Chứng minh rằng:

a) ${\rm{lo}}{{\rm{g}}_a}\left( {x + \sqrt {{x^2} - 1} } \right) + {\rm{lo}}{{\rm{g}}_a}\left( {x - \sqrt {{x^2} - 1} } \right) = 0$;

b) ${\rm{ln}}\left( {1 + {e^{2x}}} \right) = 2x + {\rm{ln}}\left( {1 + {e^{ - 2x}}} \right)$.

Áp dụng quy tắc tính logarit

${\log _a}(MN) = {\log _a}M + {\log _a}N;$

Biến đổi $1 + {e^{2x}}{e^{2x}} = \left( {1 + {e^{ - 2x}}} \right)$

a) ${\rm{lo}}{{\rm{g}}_a}\left( {x + \sqrt {{x^2} - 1} } \right) + {\rm{lo}}{{\rm{g}}_a}\left( {x - \sqrt {{x^2} - 1} } \right) = {\rm{lo}}{{\rm{g}}_a}\left[ {\left( {x + \sqrt {{x^2} - 1} } \right)\left( {x - \sqrt {{x^2} - 1} } \right)} \right]$

${\rm{ = lo}}{{\rm{g}}_a}\left( {{x^2} - \left( {{x^2} - 1} \right)} \right) =$$= {\rm{lo}}{{\rm{g}}_a}1 = 0$.

b) ${\rm{ln}}\left( {1 + {e^{2x}}} \right) = {\rm{ln}}\left[ {{e^{2x}}\left( {1 + {e^{ - 2x}}} \right)} \right] = {\rm{ln}}{e^{2x}} + {\rm{ln}}\left( {1 + {e^{ - 2x}}} \right)$$= 2x + {\rm{ln}}\left( {1 + {e^{ - 2x}}} \right){\rm{.\;}}$