Solve for y (natural logs)....?

Question

ln (y-1) - ln 2 = x + ln x

Answer 1

ln(y - 1) - ln 2 = x + ln x

=> ln(y - 1) - ln 2 - ln x = x

=> ln{(y - 1)/2x} = x

=> (y - 1)/2x = e^x

=> y - 1 = 2x.e^x

=> y = 1 + 2x.e^x <==ANSWER

Hope this help u :)

Answer 2

nothing differ between the other log

ln (y-1) - ln 2 = x ln e + ln x

ln { (y-1) /2 }= ln {e^x (x) }

cut out the ln then u'll get

y-1/2=xe^x

y=2xe^x+1

sorry if i wrong

Answer 3

Ln(y-1)/2x =x,e^x=y-1/2x,2xe^x=y-1

then  

        y=1 + 2xe ^x

Answer 4

ln[(y-1)/2]=x+lnx

e^(ln[(y-1)/2])=e^(x+lnx)

(y-1)/2=xe^x

y-1=2xe^x

y=2xe^x+1