Posted by **Kevin** on Monday, June 19, 2006 at 3:24am.

Can somebody solve this logarithm problem step by step for me please:

2(lnx)² + lnx-1 = 0

First let y = ln x be a new variable, , and solve the polynomial

2 y^2 + y - 1 = 0

2(y +1)(y - 1/2) = 0

Solutions are

y = -1 or +1/2

Thus x = e^y

and x= 1/e = 0.3679.. OR

e^0.5 = 1.6487..

- Math (Calculus) -
**Anonymous**, Friday, November 1, 2013 at 11:01pm
tg(x)sec^6(x)dx

## Answer This Question

## Related Questions

- math - how would you solve for y in this problem: ln(y-1)-ln2 = x +lnx ln(y-1)-...
- math - I am suppose to simplify the following problems: sqrt(x)/x (isn't that ...
- Calculus 1 - Find the derivative of y with respect to x. y=(x^6/6)(lnx)-(x^6/36...
- calculus - sorry to ask a second question so soon, but i'm just not getting this...
- Calculus - what is the limit as h approaches 0 of ((x+h)^pi-x^pi)/h? Multiple ...
- 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 - Find the derivatives, dy/dx for the following functions. a. x^2y^4 = ...
- Algebra 1--Step-by-Step - Can someone show me how to solve these step-by-step? ...
- Math logs - Solve the system of equations: y= (lnx)^2 + 2 (lnx^2) y = 3ln(1/x^2...

More Related Questions