Posted by **kristie** on Tuesday, July 11, 2006 at 10:04am.

how would you solve for y in this problem:

ln(y-1)-ln2 = x +lnx

ln(y-1)-ln2 = x +lnx Solve for y...

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

take the antilog of each side

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

solve for y.

## Answer This Question

## Related Questions

- math - I am suppose to simplify the following problems: sqrt(x)/x (isn't that ...
- calculus - (a) find the intervals on which f is incrs or decrs. (b) find the ...
- L'Hopital's rule - Find lim x->1+ of [(1/(x-1))-(1/lnx)]. Here is my work...
- calculus - sorry to ask a second question so soon, but i'm just not getting this...
- Calculus 1 - Find the derivative of y with respect to x. y=(x^6/6)(lnx)-(x^6/36...
- Calculus - what is the limit as h approaches 0 of ((x+h)^pi-x^pi)/h? Multiple ...
- Math (Calculus) - Can somebody solve this logarithm problem step by step for me ...
- Math logs - Solve the system of equations: y= (lnx)^2 + 2 (lnx^2) y = 3ln(1/x^2...
- calc - integral of 1 to e^4 dx/x(1+lnx) 1+lnx = a => dx/x = da dx/ x(1+lnx...
- Math - I cant figure this question out. I could use some help. Solve for x lnx-...

More Related Questions