Question

Aratati ca log(2)(radical din 7+radical din 3)+log(2)(radical din 7+radical din 3)=2.

Answer 1

Ai scris gresit textul dar mi-am dat seama.Se aplica formula: $log a+ log b= loga*b$

$log2( \sqrt{7} + \sqrt{3} )+log( \sqrt{7} - \sqrt{3} )=log2( \sqrt{7} + \sqrt{3} )( \sqrt{7} - \sqrt{3} )$= log2(7-3)=log2(4)=2
am mai aplicat formula (a-b)*(a+b)=a^2-b^2

Answer 2

$\displaystyle log_2( \sqrt{7} + \sqrt{3} )+log_2( \sqrt{7} - \sqrt{3} )=log_2[( \sqrt{7} + \sqrt{3} )( \sqrt{7} - \sqrt{3} )]= \\ =log_2(7-3)=log_24=log_22^2=2log_22= 2$