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Math Help - comparisons and logarithms

  1. #1
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    comparisons and logarithms

    Hi. In comparisons, can work with logarithms in the usual manner? (Hoping, of course, my usual manner is the correct manner.)

    Is this valid:

    \frac{ln(n+2)}{ln(n+1)} \le \frac{n+2}{n+1}

    ln(n+2) \le \frac{ln(n+1)\,(n+2)}{n+1}

    n+2 \le e^{\frac{ln(n+1)\,(n+2)}{n+1}}

    n+2 \le (n+1)^{\frac{n+2}{n+1}}

    (Edit: assuming n \ge 1, if that matters.)
    Last edited by infraRed; March 28th 2013 at 02:14 PM.
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  2. #2
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    Re: comparisons and logarithms

    Hey infraRed.

    One approach you could try is showing d/dx [ln(x+2)] < d/dx (x+2) for all x >= 1 and show that ln(1+2) < 1+2.
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