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Math Help - Log rule confusion

  1. #1
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    Log rule confusion

    Looking for a bit of help on my log rules here, thought I had a relatively idea until trying to solve this:

    Prove
    \frac{3^{k - 1} - 1}{2} + 3^{k - 1} = \frac{3^k - 1}{2}

    So choosing to start with the left side and find the right:

    = \frac{3 * 3^{k - 1} - 1}{2}

    = (ln(3 * 3^{k - 1}) - \ln(2)) - \frac{1}{2} - can I just use ln on a portion of the equation like I have here

    = ((ln(3) + (k - 1)\ln(3)) - \ln(2)) - \frac{1}{2}

    = ((ln(3)(1 + (k - 1))) - \ln(2)) - \frac{1}{2}

    = ((k\ln(3)) - \ln(2)) - \frac{1}{2}

    = \frac{3^k - 1}{2}

    Really rusty on this so if anyone can have a quick look and let me know if there are errors in there that'd be great. (also first time using LaTeX)

    Thanks
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  2. #2
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    Re: Log rule confusion

    Skip the logarithms...

    \frac{3^{k-1} - 1}{2} + 3^{k-1} = \frac{3^{k-1} - 1}{2} + \frac{2(3^{k-1})}{2}

     = \frac{3(3^{k-1}) - 1}{2}

     = \frac{3^k - 1}{2}
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  3. #3
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    Re: Log rule confusion

    That's much easier, can you explain the basis of going from: 3(3^{k-1}) to 3^{k} ?
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  4. #4
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    Re: Log rule confusion

    Hint: a^b a^c = a^{b+c}.
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  5. #5
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    Re: Log rule confusion

    I've been doing too much math today, I was looking at that as ({3^{k-1}})^{3}

    Thanks for that!
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