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Math Help - similar logarithm problem!

  1. #1
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    similar logarithm problem!

    e^2x-7e^x+10

    bit confused on how to find the x?
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  2. #2
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     e^{2x}-7e^x+10=0

     (e^{x})^2-7e^x+10=0

    Lets make e^{x} = a

    Now we have

     a^2-7a+10=0

    which can be factored as follows

    (a-5)(a-2)=0

    by the null factor law

    a=2,5

    and bring our original unknown back into the picture

    e^x=2,5

    x=\ln(2),\ln(5)

    Does this make sense?
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  3. #3
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    Quote Originally Posted by frozenflames View Post
    e^2x-7e^x+10

    bit confused on how to find the x?
    This can't be solved because it doesn't equal anything. Also, use parentheses to avoid confusion. You could mean e^(2)*x or e^(2x).
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  4. #4
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    Thanks Jameson, my solution only assumes the expression is equal to zero.
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  5. #5
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    okay, that makes a lot of sense so there are two answers technically and I forgot to take the ln of the 2 and 5. that is where i was going wrong..thanks so much!
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  6. #6
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    Quote Originally Posted by pickslides View Post
    Thanks Jameson, my solution only assumes the expression is equal to zero.
    Given the way the problem worked out, it seems that was a good assumption. It's just not good to see people solve for things that can't be solved.
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