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BelleFlower

  • 2 years ago

Please wait for image! Logarithms question

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  1. BelleFlower
    • 2 years ago
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  2. BelleFlower
    • 2 years ago
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    Question 11! :)

  3. A.Avinash_Goutham
    • 2 years ago
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    loga - logb = log(a/b) loga + logb = log(a*b) x=log(e^x) logx=logy => x=y now help urself

  4. LogicalApple
    • 2 years ago
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    |dw:1356767087691:dw| Can you think of the next step?

  5. LogicalApple
    • 2 years ago
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    (That thing in the denominator of the parentheses is a 'b' -- sorry for sloppy artwork)

  6. BelleFlower
    • 2 years ago
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    I'm still not sure how to continue actually... :/ @LogicalApple

  7. LogicalApple
    • 2 years ago
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    Consider e and ln as inverse functions. I.e., ln(e^x) = x, and e^(ln(x)) = x. Try raising e to the power of both sides. For example, on the left side you would get e^(ln K) = K

  8. LogicalApple
    • 2 years ago
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    And whatever you do to one side you have to do to the other

  9. BelleFlower
    • 2 years ago
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    What are inverse functions? Sorry i don't really get it

  10. LogicalApple
    • 2 years ago
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    An inverse function cancels out the original function. Like square and square root functions are inverses of each other. Multiplication and division are inverses. Here, e and ln are inverses. That is why when we have e^(ln K), the e and ln cancel leaving only K.

  11. BelleFlower
    • 2 years ago
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    But I am supposed to solve this just using the power, product, quotient laws and the change-of-base law....

  12. LogicalApple
    • 2 years ago
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    But I see there is an 'e' in the answer, so you must be able to use the function.

  13. LogicalApple
    • 2 years ago
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    I will present a solution. Tell me if you have any questions about it. One sec.

  14. LogicalApple
    • 2 years ago
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