Normal line to the curve y=ln(2x+a) at x=1 is perpendicular to x-4y+2 = 4ln3. find the value of a.

Hello,

The normal perpendicular to x-4y+2=4ln(3) or y=(x+2-4ln(3))/4
has as slope -4

So y'=(ln(2x+a))'=1/(2x+a) *2 if x=1 , y'=1/(2+a)
Thus
1/(2+a)=-4 ==>2+a=-1/4==>a=-1/4-2==>a=-9/4