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    If  |a^x| < b ,

    How do you prove that  x > log_{|a|} b ?

    Any hints? Thanks.
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    Um, take logs to base |a|?
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    (Original post by DFranklin)
    Um, take logs to base |a|?

    I forgot the fact that  |a^x| = |a|^x .
    However, why is the inequality sign reversed?

    Thanks.
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    It shouldn't be, I think.
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    (Original post by melbss01)
    If  |a^x| < b ,

    How do you prove that  x > log_{|a|} b ?

    Any hints? Thanks.
    If 10 squared is less than 1000, how can 2 be greater than Log 1000, which is 3?
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    (Original post by stevencarrwork)
    If 10 squared is less than 1000, how can 2 be greater than Log 1000, which is 3?
    i was thinking that

    i think the question has been written wrong and the inequality should be the same way round as the initial inequality sign

    EDIT: i was trying to think of a good example but you beat me to it :p:
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    (Original post by Anon the 7th)
    i was thinking that

    i think the question has been written wrong and the inequality should be the same way round as the initial inequality sign

    EDIT: i was trying to think of a good example but you beat me to it :p:
    Yup. Since \dfrac{\mbox{d}}{\mbox{d}x} \log_k x = \dfrac{1}{x\ln k} > 0, it is an increasing function, it's true that p > q \iff \log p > \log q. If you let k = |a| (just to avoid confusion), then:
    \newline

k^x > b > 0\newline

\Rightarrow x > \log_k b\newline

\Rightarrow x > \log_{|a|} b

    (I'm sure you can accept that if b \le 0, it doesn't work.)
 
 
 

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