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Math Help - Solving for x

  1. #1
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    Solving for x

    \ln{x+2}={(x-3)^2}

    I think I'm forgetting an algebraic concept in solving for x in this problem. I have gone as far as square rooting both sides to yield: \sqrt{\ln{x+2}}={(x-3)}

    Is there anyway to detach the x from the natural log?
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  2. #2
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    Re: Solving for x

    Quote Originally Posted by AZach View Post
    \ln{x+2}={(x-3)^2}

    I think I'm forgetting an algebraic concept in solving for x in this problem. I have gone as far as square rooting both sides to yield: \sqrt{\ln{x+2}}={(x-3)}

    Is there anyway to detach the x from the natural log?
    You won't be able to solve this exactly. You'll have to use a numerical method like the bisection method.
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