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Confucious

I'm not good with logs, can someone help me? log(x+2)+log(x-1)=1

  • 2 years ago
  • 2 years ago

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  1. henkjan
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    log(x+2)+log(x-1) = log ((x+2)(x-1)) maybe that helps ;P

    • 2 years ago
  2. henkjan
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    And perhaps another helpful rule: if log(a)=b than 10^b = a only in the case when 10 is the basenumber of the logaritm, but if it is not stated otherwise it is always 10.

    • 2 years ago
  3. henkjan
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    So (x+2)(x-1) = 10^1 x^2+x-2= 10 x^2+x-12 = 0 Solve for x and find x = -4 and x = 3

    • 2 years ago
  4. Confucious
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    but it's equal to one??, in the first line you have the equation equal to 10

    • 2 years ago
  5. henkjan
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    I applied the rule I explained in my second post here. because b is in this case 1, it means that 10^1 = a a = (x-1)(x+2) in this case

    • 2 years ago
  6. Confucious
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    isn't the b 10?

    • 2 years ago
  7. henkjan
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    no, the b is equal to the right side in this equation log ((x+2)(x-1)) =1 ...... so it's 1

    • 2 years ago
  8. Confucious
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    okay

    • 2 years ago
  9. Confucious
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    so if the equation was equal to 2 the b would be 10^2?

    • 2 years ago
  10. henkjan
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    log(x)+log(y) = log(x*y) if log(a)=b then 10^b = a only 2 rules I used

    • 2 years ago
  11. henkjan
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    yes, indeed than it would be a = 10^2 = 100

    • 2 years ago
  12. Confucious
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    :) alright, the answer at the back of the text is just 3, why isn't the -4 acceptable?

    • 2 years ago
  13. henkjan
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    oh, right.... a logaritmic can never have a negative number log(-x) is not possible... as 10^b = a ... a can only become positive in this formula

    • 2 years ago
  14. Confucious
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    oh okay, thank you very much

    • 2 years ago
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