㏒10(4-2x^2) = ㏒10(a-x)+1 = ㏒10(10a-10x);
则有:4-2x^2 = 10a-10x ,
整理得:x^2-5x+(5a-2) = 0 ;
方程有实根,则有:判别式 = 25-4(5a-2) ≥ 0 ,
解得:a ≤ 33/20 ;且已知,a∈N,可得:a = 1 .
将 a = 1 代入 x^2-5x+(5a-2) = 0 ,可得:x^2-5x+3 = 0 ,解得:x = (5±√13)/2 ;
考虑对数有意义,必须:4-2x2 > 0 ,a-x = 1-x > 0 ,即需要:-√2 < x < 1 ;
所以,方程的根只有一个:x = (5-√13)/2 .