College Algebra

posted by .

How would I go about solving for x in the equation:
ln(4x-2)-ln(4)=-ln(x-2)
Thanks in advance

  • College Algebra -

    ln[(4x-2)/4] = ln[1/(x-2)]
    (4x -2)/4 = 1/(x-2)
    x - 1/2 = 1/(x-2)
    x^2 -x/2 -2x +1 = 1
    x^2 -(5/2)x = 0
    x = 0 or 5/2
    0 is not allowed since ln of a negative number is undefined
    x = 5/2

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Algebra

    Explain the difference between solving a linear equation graphically and solving a system of equations graphically. How would I go about doing this?
  2. Alg2/Trig

    How would I go about solving this equation algebraically?
  3. College Intermediate Algebra

    Explain the reason to use tradional solving with a quadratic equation with one variable.
  4. College Algebra

    Solve the following equation in the complex number system x^4+x^3+4x^2+10x-60=0 College Algebra - Reiny, Monday, January 2, 2012 at 11:49pm hint x = 2 and x = -3 are roots so either do long division by x-2 and x+3 to get a quadratic …
  5. calculus

    find the limit. lim (5-e^x)/(5+3e^x) (x-> infinity) how would i go about solving this?
  6. algebra

    How do go about solving a equation symbolically 3x+7 + 3x7 =-8 4. 5. 5
  7. college algebra

    I need help solving this equation. 3√z+1=-3 How do I do this problem?
  8. college algebra

    compare and contrast the ways in which solving a linear inequality is similar / dissimilar to solving a linear equation please explain
  9. Precalculus

    10x^2=2^x Hi! How do I go about solving an equation like this?
  10. college Algebra

    solving rational equation trying to figure out where i did wrong x/x-4 -4/x-5 = 4/x^2-9x +20 x/x-4 -4/x-5= 4/(x-4)(x-5) domain x not equal +- 4,5 x(x-4)(x-5)/x-4 - 4(x-4)(x-5)/x-5= 4(x-4)(x-5)/(x-4)(x-5) x(x-5)-4(x-4)=4 x^2-5x-4x+16=4 …

More Similar Questions