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Math Help - A logaritm problem from an old trig textbook

  1. #1
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    A logaritm problem from an old trig textbook

    The equation is a logarithm problem that reads:

    (a^2)(e^-ax)-(b^2)(e^-bx)=0 where a>b

    The answer is:

    (2(log[base e]a-log[base e]b))/(a-b)

    How do I get from the problem to the solution.

    Thank you
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  2. #2
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    Re: A logaritm problem from an old trig textbook

    Quote Originally Posted by Crisor2431 View Post
    The equation is a logarithm problem that reads:

    (a^2)(e^-ax)-(b^2)(e^-bx)=0 where a>b

    The answer is:

    (2(log[base e]a-log[base e]b))/(a-b)

    How do I get from the problem to the solution.

    Thank you
    a^2 e^{-ax} = b^2 e^{-bx}

    Take the natural log of both sides:
    ln \left ( a^2 e^{-ax} \right ) = ln \left ( b^2 e^{-bx} \right )

    ln(a^2) - ax = ln(b^2) - bx

    Can you finish from here?

    -Dan
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  3. #3
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    Re: A logaritm problem from an old trig textbook

    I believe so. Thank you.
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  4. #4
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    Re: A logaritm problem from an old trig textbook

    Hello, Crisor2431!

    I assume that you know that \log_e(x) is written \ln(x).


    (a^2)(e^-ax)-(b^2)(e^-bx)=0 where a>b

    The answer is: (2(log[base e]a - log[base e]b)) / (a - b)

    How do I get from the problem to the solution?

    We have: . a^2e^{-ax} - b^2e^{-bx} \:=\:0 \quad\Rightarrow\quad b^2e^{-bx} \:=\:a^2e^{-ax}

    . . . . . . . . \frac{e^{-bx}}{e^{-ax}} \:=\:\frac{a^2}{b^2} \quad\Rightarrow\quad e^{ax-bx} \:=\:\tfrac{a^2}{b^2} \quad\Rightarrow\quad e^{(a-b)x} \:=\:\left(\tfrac{a}{b}\right)^2

    \text{Take logs: }\:\ln\left(e^{(a-b)x}\right) \;=\;\ln\left(\tfrac{a}{b}\right)^2 \quad\Rightarrow\quad (a-b)x\underbrace{\ln(e)}_{\text{This is 1}} \:=\:2\ln\left(\tfrac{a}{b}\right)

    Therefore: . x \;=\;\dfrac{2\ln(\frac{a}{b})}{a-b}
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