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    Ok I get logs , but there is just one rule i dont get:

    How does a^\log_a x = X ?

    I get that if you log the LHS to get  \log_a a^(log_a x)

    then take the top log down and it becomes a sum of the 2 logs. Then  \log_a a =1 so i'm left with just  \log_a x but how does that = x?
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    You've forgotton to 'unlog' once you logged the LHS:

    \displaystyle \log_a a^{log_a x} = \log_a x \log_a a = \log_a x

    Now if log_a b_1 = log_a b_2 then b_1=b_2. Use this to prove the result.
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    Ooooo ok So im just doing a^loga x again but the a^ and the log cancel?

    Just like log and 10^ will cancel?
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    (Original post by Carlo08)
    Ooooo ok So im just doing a^loga x again but the a^ and the log cancel?

    Just like log and 10^ will cancel?
    Yes, exactly that. In fact, if you believe that, then go and read your original question again and it becomes obvious that a^{\log_a x} = x.
 
 
 
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