Logs question, help plz?

Remember the laws of exponents
& I'm assumming that ln is a natural (rather than a common) logarithm/

(1/2) ln ( y^2+1) = ln | 1+x | + ln A
ln ( y^2+1)^(1/2) = ln | 1+x | + ln A
ln (( y^2+1)^(1/2)) - ln | 1+x | = ln A
ln ((( y^2+1)^(1/2)/(| 1+x |)=ln A
((( y^2+1)^(1/2)/(| 1+x |)=exp(ln A)=A
(( y^2+1)^(1/2)=A| 1+x |
y^2+1=(A|x+1|)^2
y^2= (A|x+1|)^2-1
y=sqrt[(A|x+1|)^2-1]

This one is easier to put into y in terms of x
y + ln (y) = ln ( x- 1 ) + 1
y-1+ln(y)=ln(x-1)
y-1=ln(x-1)-ln(y)
y-1=ln((x-1)/y)
ln((x-1)/y)=y-1
(x-1)/y=exp(y-1)
x-1=y*exp(y-1)
x=y*exp(y-1)+1

Rap

thnx very much Rap