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Cutiepo0

How to write as a simple logarithm: log a + log (a+b) - log c - log d

  • one year ago
  • one year ago

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  1. AnimalAin
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    The rules of logarithms are important here:\[\log(AB)=logA+logB\]\[\log(\frac{A}{B})=logA-logB\]

    • one year ago
  2. Cutiepo0
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    I got as far as log (((a^2)+ab)/(c/d))

    • one year ago
  3. Cutiepo0
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    but in the answer, the bottom of the equation is cd not c/d and I don't get why :(

    • one year ago
  4. AnimalAin
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    Working from that,\[\log a + \log (a+b) - \log c - \log d=\log(\frac{a(a+b)}{cd})\]

    • one year ago
  5. AnimalAin
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    OK, this time step-by-step. From the product rule\[\log a + \log (a+b) - \log c - \log d=\log[a(a+b)]-\log c-\log d\]

    • one year ago
  6. Cutiepo0
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    ok following so far

    • one year ago
  7. AnimalAin
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    Now, from the quotient rule\[\log[a(a+b)]−logc−logd=\log[\frac{a(a+b)}{c}]-logd\]

    • one year ago
  8. AnimalAin
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    Apply the quotient rule again, to get\[\log[\frac{a(a+b)}{c}]−logd=\log[\frac{a(a+b)}{cd}]\]

    • one year ago
  9. AnimalAin
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    Got it now?

    • one year ago
  10. Cutiepo0
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    and then with log d you do the same step as the last time!! Oh! How did you know to do the c first and then d though, I just did the first two parts with the multiplication law, and thelast two with the division law, and then both together with the division law

    • one year ago
  11. AnimalAin
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    You eat an elephant one bite at a time. Do math the same way.

    • one year ago
  12. Cutiepo0
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    haha okay thanks so much :)

    • one year ago
  13. AnimalAin
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    Do math every day.

    • one year ago
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