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Thread: Trouble equating functions to check if one-to-one with natural log's

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    Trouble equating functions to check if one-to-one with natural log's

    $\displaystyle e^a/(1+4e^a)=e^b/(1+4e^b)$

    When I multiply each side by it's denominator I get:

    $\displaystyle e^a+4e^a^+^b=e^b+4e^a^+^b$

    hopefully I'm on the right track, then do I take the natural log of both sides? How does the natural log react in this situation with the addition? Do all the $\displaystyle e$'s cancel out?
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    Re: Trouble equating functions to check if one-to-one with natural log's

    Quote Originally Posted by Maskawisewin View Post
    $\displaystyle e^a/(1+4e^a)=e^b/(1+4e^b)$

    When I multiply each side by it's denominator I get:

    $\displaystyle e^a+4e^a^+^b=e^b+4e^a^+^b$

    hopefully I'm on the right track, then do I take the natural log of both sides? How does the natural log react in this situation with the addition? Do all the $\displaystyle e$'s cancel out?
    $\displaystyle e^a+4e^a^+^b=e^b+4e^a^+^b$

    start by subtracting $\displaystyle 4e^{a+b}$ from both sides ...
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    Re: Trouble equating functions to check if one-to-one with natural log's

    That was easy. Thank you. Always making it hard on myself.
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