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Math Help - Stuck on how to solve simultaneous exponential equations

  1. #1
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    Stuck on how to solve simultaneous exponential equations

    Hello

    I have the equations:

    6 = a2^n and 18 = a3^n

    I am really stuck on this. This is what I tried:

    2^n = \frac{6}{a}

    3^n = \frac{18}{a}

    n log2 = log \frac{6}{a} = log(6) - log(a)

    n log3 = log \frac{18}{a} = log(18) - log(a)

    n = \frac{log(6) - log(a)}{log2}

    n = \frac{log(18) - log(a)}{log3}

    then

    \frac{log(6) - log(a)}{log2} =  \frac{log(18) - log(a)}{log3}

    Is this the way to solve this sort of problem? Or am I missing an easier way?

    If this is the way, it looks complicated. How would I proceed from here?

    Angus
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  2. #2
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    Crna Gora
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    Re: Stuck on how to solve simultaneous exponential equations

    Quote Originally Posted by angypangy View Post
    Hello

    I have the equations:

    6 = a2^n and 18 = a3^n

    I am really stuck on this. This is what I tried:

    2^n = \frac{6}{a}

    3^n = \frac{18}{a}

    n log2 = log \frac{6}{a} = log(6) - log(a)

    n log3 = log \frac{18}{a} = log(18) - log(a)

    n = \frac{log(6) - log(a)}{log2}

    n = \frac{log(18) - log(a)}{log3}

    then

    \frac{log(6) - log(a)}{log2} =  \frac{log(18) - log(a)}{log3}

    Is this the way to solve this sort of problem? Or am I missing an easier way?

    If this is the way, it looks complicated. How would I proceed from here?

    Angus
    a \cdot 3^n=6 \cdot 3 \Rightarrow a \cdot 3^n=a \cdot 2^n \cdot 3 \Rightarrow 3^n = 2^n \cdot 3 \Rightarrow \left(\frac{3}{2}\right)^n=3

    \Rightarrow n=\log_{\frac{3}{2}} 3 ; a \neq 0
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