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Math Help - Need a few logarithm formulas...

  1. #1
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    Need a few logarithm formulas...

    Or at least I think I do...


    These are the two problems I have:

    #1: 4^{4x}=2000; x=?

    #2: 4log(x-4)=16; x=?


    ...so I basically just need to know what x equals, and just plugging in numbers isn't very practical Can anyone help?
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  2. #2
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    4^{4x}=2000 take the log of both sides
    log(4^{4x})=log(2000) but now, one of the log rules says that log(a^b)=b*log(a)

    so 4x*log4=log2000 and 4x=\frac{log2000}{log4} so just divide by 4 and x= \frac{log2000}{4log4}

    in general though, people seem to like to reduce logs for no reason that I can see... so log4= log2^2=2log2 so x= \frac{log2000}{8log2}



    4log(x-4)=16 divide by 4 and log(x-4)=4, now raise each side to a power of 10, so 10^{log(x-4)}=10^4

    so x-4=10000, and x=10004
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  3. #3
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    Thanks man, I understand the second one, but for the first one, is x= \frac{log2000}{8log2} the answer? Because when I put that into a calculator, I get .1242136307, and then when I add that value into the equation

    4^{4x}=2000 as 4^{4 * .1242136307}

    I get 1.991297872, which isn't close to 2,000...
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  4. #4
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    x=\frac{log2000}{8log2} when you put it in your calculator, you didn't use order of operations correctly so your calculator did this: \frac{log2000}{8}*log2
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  5. #5
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    Ahhhh ok. I went back and entered it in as:
    Code:
    (log(2000))/(8log(2))
    And got 1.370723036, which worked perfectly...thanks for all of your help
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