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algebra2sucks Group Title

Show why the log100 + log10000 = 2log(3)27

  • one year ago
  • one year ago

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  1. satellite73 Group Title
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    compute all the numbers and see that they are equal

    • one year ago
  2. satellite73 Group Title
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    you need \(\log(100)\) and also \(\log(10000)\) first

    • one year ago
  3. satellite73 Group Title
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    if you have a power of ten , the log counts the zeros

    • one year ago
  4. algebra2sucks Group Title
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    lol i have no idea how to do this

    • one year ago
  5. ZeHanz Group Title
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    When log 100 = a, this means: \[10^a=100\]Now, because 100 is a power of 10, you can see very quickly what log 100 is... Same for log 10000 >> what power of 10 is it?

    • one year ago
  6. ZeHanz Group Title
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    If you have problems understanding logarithms, always remember one thing: LOGARITHMS are EXPONENTS

    • one year ago
  7. ZeHanz Group Title
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    In your problem, I also see a logarithm with base 3. I will call it b: \[\log_{3} 27=b\]So b is that logarithm, base 3 of 27. This means, because logarithms are exponents, \[3^b=27\]Now, 27, doesn't it sound familiar? Yes! It is a power of 3! In fact, it is 3³:\[3^b=3^3\]so b = 3. The right hand side of the problem doubles that number, so we have 6 there. Can you see now, why the left hand side is 6 as well?

    • one year ago
  8. algebra2sucks Group Title
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    Yea i guess so how do we answer this

    • one year ago
  9. ZeHanz Group Title
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    Begin with the left hand side, replace log(100) and log(10000) with the numbers we've got. These will add up to 6. Right hand side is 2*3=6 also. Now you have shown that they are equal...

    • one year ago
  10. algebra2sucks Group Title
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    so 6 is the answeR?

    • one year ago
  11. ZeHanz Group Title
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    If you re-read the question, you'll see that "6" ids not the answer. You have to show that the left hand side and right hand side are equal. Although they seem very different, both sides give you 6 if you calculate them. Therefore they are equal. That is what you had to prove.

    • one year ago
  12. algebra2sucks Group Title
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    oh so how to i prove that with the equation right? i have no idea how to do this at all i know one side equals 6

    • one year ago
  13. algebra2sucks Group Title
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    how do we get 6 for the left hand side?

    • one year ago
  14. satellite73 Group Title
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    \(\log(100)\) measn \(\log_{10}(100)\) and since \(100=10^2\) we know \(\log(100)=2\)

    • one year ago
  15. satellite73 Group Title
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    last job is to solve \(\log(10000)\) i.e. solve \(10^y=10000\) for \(y\)

    • one year ago
  16. algebra2sucks Group Title
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    3 would be the y?

    • one year ago
  17. algebra2sucks Group Title
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    @satellite73 ?

    • one year ago
  18. ZeHanz Group Title
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    What makes you think it is 3?

    • one year ago
  19. algebra2sucks Group Title
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    because it has to equal to six idk i'm confused

    • one year ago
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