Pre-Calculus

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I need help solving this logarithmic equation: ln x + ln (x+3) = 1.

I know ln x + ln (x+3)=ln [x(x+3)] and that is ln x^2 + 3x when broken down. Then you have to get rid of ln by taking the natural base e of both sides so in the end you have x^2+3x = e^(1). That's when I got stuck.

  • Pre-Calculus -

    You are on the right track. Remember that e^1 = e = 2.718..
    You just need to use the quadratric equation to solve
    x^2 + 3x -2.718 = 0
    since it cannot be factored easily.

    x = [-3 +/- sqrt (19.87)]/2
    = 1.458/2 or -7.458/2 = 0.729 or -3.729

  • Pre-Calculus -

    x+5=4

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