Math

Solve for x:

e^4x-3 = 2e^2x1



ln2x+6) - ln(3x-1) = ln(2)

Then, simply the radicals

asked by Jake
  1. How about some parentheses?

    Just guessing here, I read

    e^(4x-3) = 2e^(2x)
    No idea what the final 1 is. Anyway, I'll work this, and then you can fix it if necessary.

    Taking logs of both sides, we get

    4x-3 = ln2 + 2x
    2x = 3+ln2
    x = (3+ln2)/2

    ln(2x+6) - ln(3x-1) = ln(2)
    (2x+6)(3x-1) = 2
    Now just solve the quadratic

    posted by Steve

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