Logx (8x-3) - logx 4 = 2
logx [(8x-3)/4] = 2
x^2=(8x-3)/4
4x^2=4(8x-3)/4
4x^2=8x-3
4x^2-8x+3=8x-3-8x+3
4x^2-8x+3=0
4(4x^2-8x+3=0)
(4^2)(x^2)-8(4x)+12=0
(4x)^2-8(4x)+12=0
(4x-2)(4x-6)=0
2(4x/2-2/2)2(4x/2-6/2)=0
4(2x-1)(2x-3)=0
4(2x-1)(2x-3)/4=0/4
(2x-1)(2x-3)=0
Two options:
1) 2x-1=0
2x-1+1=0+1
2x=1
2x/2=1/2
x=1/2
2) 2x-3=0
2x-3+3=0+3
2x=3
2x/2=3/2
x=3/2
Answer: Two solutions: x=1/2 and x=3/2
It can be factored.
We need two numbers that multiply to -12 and add to -4, so are two terms would be -6 and 2.
(x - 6) (x + 2) = 0
x = 6
x = -2
