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    Logarithm

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    radioactive
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    Logarithm

    Post by radioactive on Mon Nov 08, 2010 10:27 pm

    In math research i got homework ( Freshman [I am in 9th grade, DONT ASK WHY I GOT THIS FOR HW])





    1- Log 4(base) ^64 = is the answer 3?



    2- Log2(base)^8+ Log 29(base)^32= Log2(base)^256









    Are my answers correct????

    I am a 9th grader ,so don't call me stupid!! Shocked Shocked
    I just have to do this for Math Research!


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    Ahmed
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    Re: Logarithm

    Post by Ahmed on Mon Nov 08, 2010 10:31 pm

    1-Yes
    2-I need to refresh my memory for logs. But Im busy atm.


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    Doc
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    Re: Logarithm

    Post by Doc on Tue Nov 09, 2010 12:02 am

    first one is right
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    The Banker
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    Re: Logarithm

    Post by The Banker on Wed Nov 10, 2010 6:56 pm

    1st is right
    2nd.. nope

    Log 2(base)^8+ Log 29(base)^32
    first thing u wanna do is change Log 29(base)^32 to a base of 2
    so that'll be (log 2(base)^32/log 2(base)^29)

    so what u've got in the end is Log 2(base)^8+(log 2(base)^32/log 2(base)^29)= 4.029 (3dp)

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    Re: Logarithm

    Post by radioactive on Wed Nov 10, 2010 8:18 pm

    Sorry! I wrote the question wronf for the 2nd one!!

    Its Log2(base)^8+ Log 2(base)^32= Log2(base)^256 NOT Log2(base)^8+ Log 29(base)^32= Log2(base)^256

    .... SORRY I WROTE 29 instead of 2!!! LOL SORRY!!


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    Coolmeia
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    Re: Logarithm

    Post by Coolmeia on Thu Nov 11, 2010 8:52 pm

    I'm in algebra 1 and I don't get it, sorry! We just did inequalities!

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