Post a New Question

Calculus

posted by .

Solve for x in the equation ln(16-12x)-2lnx)= ln4

  • Calculus -

    ln [ (16-12x)/x^2 ] = ln 4
    (16-12x)/x^2 = 4
    4x^2 = 16-12x
    x^2 + 3x - 4 = 0
    (x+4)(x-1) = 0
    x = -4 or x=1

    but if x = -4 , 2lnx in the original equation is not defined, so

    x = 1

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. algebra2

    1/3 - 5/6 = 1/X SOLVE 1/3 - 5/6 = 1/X Let's remove the denominators by multiplying by 12X 12X*(1/3) - 12X*(5/6) = 12X*(1/X) Clear the fractions. 4X - 10X = 12 -6X=12 multipy by -1 6X=-12 Divide both sides by 6 6X/6 = -12/6 X = -2 Check …
  2. calculus

    I'm feeling pretty stupid right now because I can't remember how to do these. if f(x) = (2x-5)/(x^2-4), write an equation of the tangent line to f at point x=0. *How do I find the slope of the tangent line?
  3. math

    2^(x^2 - 35)= 4^x How do I solve this? The x^2 is messing me up. So far I have: ln(2^(x^2-35))=ln(4^x) (x^2-35)ln 2= xln4 x^2ln 2 - 35ln2= xln4 x^2ln 2- xln4= 35ln2 x(xln2 - ln4)= 35ln2 I would have taken the log base 2 of each side,
  4. Calculus

    Solve for x: lnx^3 - 2lnx = 1 Please try and show how you got the answer best as you can thank you :)
  5. Calculus

    Solve for x in the equation ln(16-12x)-2lnx = ln4
  6. precalc

    solve the following logarithmic equation: 2lnx= log base e^3 343
  7. Algebra II

    Is this correct? Solve: 2lnx-2=0 2lnx=2 Lnex=1 E^1=x
  8. Pre-Calculus

    Assume that x, y, and z are positive numbers. Use the properties of logarithms to write the expression 2lnx - 6lny + 1/3ln e^12 as a simplified logarithm. Is this 2lnx - 6lny + 4, or can it be simplified further?
  9. Calculus!! HELP!!

    1) The sides of the triangle shown increase in such a way that (dz/dt=1) and (dx/dt=(3dy/dx)) At the instant when x = 12 and y = 5, what is the value of dx/dt?
  10. Algebra

    I came across a problem and noticed that (ln4)/2 = ln√4. How do I rewrite (ln4)/2 into ln√4 or verify that the two are equal without a calculator?

More Similar Questions

Post a New Question