F.4maths對數函數

1.
㏒(3x+1) - ㏒(x-1) = ㏒5
㏒[(3x+1)/(x-1)] = ㏒5
(3x+1)/(x-1) =5
3x+1 =5x-5
2x=6
x=3

2.
12^(2x+1) = 8^3x
㏒12^(2x+1) = ㏒8^3x
(2x+1)㏒12= 3x ㏒8
2㏒12x+㏒12=3㏒8x
(2㏒12-3㏒8)x=-㏒12
x=-㏒12/(2㏒12-3㏒8)
x=-㏒12/㏒(12^2/8^3)
x=1.9589

3.
㏒(2x^2-19) = ㏒(5x-16)
2x^2-19= 5x-16
2x^2-5x-3=0
(x-3)(2x+1)=0
x=3 (捨去)or -1/2(捨去)
x无解