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Thread: logarithm

  1. #1
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    Exclamation logarithm

    Without a calculator, show that:

    log 5 + log 9 + log 8 < 2
    log 19 log 19 log 19
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  2. #2
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    Quote Originally Posted by math123456 View Post
    Without a calculator, show that:

    log 5 + log 9 + log 8 < 2
    log 19 log 19 log 19
    $\displaystyle \frac{\log{5} + \log{9} + \log{8}}{\log{19}} = \frac{\log(5 \cdot 9 \cdot 8)}{\log{19}} = \frac{\log{360}}{\log{19}} < \frac{\log{361}}{\log{19}} = \frac{\log{19^2}}{\log{19}} = \frac{2\log{19}}{\log{19}} = 2$
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  3. #3
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    Hello, math123456!

    Without a calculator, show that: .$\displaystyle \frac{\log5}{\log19} + \frac{\log9}{\log19}+ \frac{\log 8}{\log19} \;<\;2$
    Base-Change Formula: .$\displaystyle \log_ba \:=\:\frac{\log a}{\log b}$


    The inequality becomes: .$\displaystyle \log_{19}5 + \log_{19}9 + \log_{19}8 \;=\;\log_{19}360$


    Since $\displaystyle 19^2 = 361$, then: .$\displaystyle \log_{19}361 \:=\:2$

    Then: .$\displaystyle 360 \,<\,361 \quad\Rightarrow\quad \log_{19}360 \,<\,\log_{19}361 $

    . . Therefore: .$\displaystyle \log_{19}360 \,<\,2$ . . . QED

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