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Thread: Why can't use natural log for this equation?

  1. #1
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    Why can't use natural log for this equation?

    So I have solved the equation below the proper way but cannot understand why applying the natural logarithm this way is wrong when it does work for other equations, what rule I am I not following here ?

    https://imgur.com/a/cbOVa
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  2. #2
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    Re: Why can't use natural log for this equation?

    egad...

    logs don't work at all like that

    what you could do is

    $e^{2n} - 2e^n = 6$

    $\ln\left(e^{2n} - 2e^n \right) = \ln(6)$

    but this is as far as you can go using this strategy.

    The actual way to solve this problem is with the substitution

    $u = e^n$

    Then we can rewrite the original equation as

    $u^2 - 2u - 6 = 0$

    Solve this for $u$

    and then $x = \ln(u),~u>0$

    If there are any solutions for $u$ such that $u\leq 0$ they can be ignored.
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  3. #3
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    Re: Why can't use natural log for this equation?

    Quote Originally Posted by BadLuchaThings View Post
    So I have solved the equation below the proper way but cannot understand why applying the natural logarithm this way is wrong when it does work for other equations, what rule I am I not following here ?
    https://imgur.com/a/cbOVa
    Well the logarithm is not an additive function.
    $\Large\log(e^{2n}-2e^{n}+6)\ne\log(e^{2n})-\log(e^{n})-\log(6)$
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  4. #4
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    Re: Why can't use natural log for this equation?

    Wow I went through like 50 questions without a problem and I'm surprised how not knowing this didn't bite me. So if I'm understanding correctly, I can't take the log of every single term separately in an equation.
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    Re: Why can't use natural log for this equation?

    Quote Originally Posted by BadLuchaThings View Post
    Wow I went through like 50 questions without a problem and I'm surprised how not knowing this didn't bite me. So if I'm understanding correctly, I can't take the log of every single term separately in an equation.
    $\ln(a+b) \neq \ln(a) + \ln(b)$
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  6. #6
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    Re: Why can't use natural log for this equation?

    Instead, let $y= e^x$ so the equation becomes $y^2- 2y- 6= 0$. Solve that by completing the square or the quadratic formula then take the logarithm.
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