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 Logarithm

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radioactive
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radioactive


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PostSubject: Logarithm    Logarithm  I_icon_minitimeMon 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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PostSubject: Re: Logarithm    Logarithm  I_icon_minitimeMon 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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PostSubject: Re: Logarithm    Logarithm  I_icon_minitimeTue Nov 09, 2010 12:02 am

first one is right
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The Banker
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PostSubject: Re: Logarithm    Logarithm  I_icon_minitimeWed Nov 10, 2010 6:56 pm

1st is right
2nd.. nope

Quote :
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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radioactive
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PostSubject: Re: Logarithm    Logarithm  I_icon_minitimeWed 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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PostSubject: Re: Logarithm    Logarithm  I_icon_minitimeThu 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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