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marcelie

  • one year ago

help please !!!! i attached the picture below

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  1. marcelie
    • one year ago
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    |dw:1444196703724:dw|

  2. marcelie
    • one year ago
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    |dw:1444196743208:dw|

  3. jojokiw3
    • one year ago
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    ooo, so logs are like exponents, when you multiply exponents, what do you do with them actually?

  4. marcelie
    • one year ago
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    yea

  5. jojokiw3
    • one year ago
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    What's the answer to this question? x^2(x^2)?

  6. marcelie
    • one year ago
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    the answer in my sheet says x cant equal -3 and x = 2

  7. jojokiw3
    • one year ago
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    Hm, I don't think you can get an numerical answer to this question, because it doesn't have an equal sign.

  8. jojokiw3
    • one year ago
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    But you can simplify the equation more

  9. marcelie
    • one year ago
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    yeh one sec ..

  10. freckles
    • one year ago
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    |dw:1444197015545:dw|

  11. marcelie
    • one year ago
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    1 Attachment
  12. freckles
    • one year ago
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    \[\log_2(x+2)+\log_2(x-1)=2?\]

  13. marcelie
    • one year ago
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    yes.

  14. jojokiw3
    • one year ago
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    ooo, that makes more sense.

  15. freckles
    • one year ago
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    that first step you have there looks great little confused on your third line you have you should do 2^ on both sides for your second line

  16. marcelie
    • one year ago
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    the other one should be log base 2 (2) ?

  17. marcelie
    • one year ago
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    k

  18. freckles
    • one year ago
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    \[2^{\log_2((x-1)(x+2))}=2^2\] see we do 2^ on both sides

  19. freckles
    • one year ago
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    this gives us: \[(x-1)(x+2)=4\]

  20. jojokiw3
    • one year ago
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    Oh, I'm sorry.

  21. freckles
    • one year ago
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    you will need to expand and get everything on one side like you are going to want 0 on that right hand side there

  22. marcelie
    • one year ago
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    hmmm. im confused

  23. freckles
    • one year ago
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    on what I did?

  24. freckles
    • one year ago
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    or what I'm asking you to do?

  25. marcelie
    • one year ago
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    yes. why 2 on both sides

  26. freckles
    • one year ago
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    \[a^{\log_a(y)}=y \\ \text{ for } y>0 \text{ and } a \in (0,1) \cup (1,\infty)\]

  27. freckles
    • one year ago
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    \[\text{ that is I knew } 2^{\log_2((x+2)(x-1))}=(x+2)(x-1)\]

  28. freckles
    • one year ago
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    for example if we had: \[\log_{4}((x+2)(x-1))=2 \\ \text{ I would do 4^ on both sides } \\ 4^{(\log_4(x+2)(x-1))}=4^2 \\ \text{ which the whole point in doing the 4^ } \\ \text{ is written below } \\ (x+2)(x-1)=4^2 \\ \text{ we get \to write in its equivalent exponent form } \]

  29. freckles
    • one year ago
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    do you understand that the following two are inverses: \[y=a^x \text{ and } y=\log_a(x)\]

  30. freckles
    • one year ago
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    \[f(x)=a^x \text{ and } f^{-1}(x)=\log_a(x) \text{ are inverses iff } \\ f(f^{-1}(x))=x \text{ and } f^{-1}(f(x))=x\]

  31. freckles
    • one year ago
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    that is: \[a^{\log_a(x)}=x \text{ and } \log_a(a^x)=x\] there are some domain restrictions but yeah

  32. freckles
    • one year ago
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    \[\text{ so we had } \log_2(stuff)=2 \\ \text{ and I did } \color{red}{2}^{\log_\color{green}{2}(stuff)}=\color{red}{2}^2 \\ stuff=4\]

  33. freckles
    • one year ago
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    I did 2^ on both sides because of the green 2 there

  34. freckles
    • one year ago
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    if the green would have been 5 I would have did 5^ on both sides

  35. marcelie
    • one year ago
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    woahhh o.o

  36. freckles
    • one year ago
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    if that doesn't make sense to you I'm sure you have written log equations in their equivalent exponential form right?

  37. freckles
    • one year ago
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    \[\log_a(y)=x \implies a^{x}=y \\ \log_2([(x-1)(x+2)]=2 \implies 2^{2}=(x-1)(x+2)\]

  38. marcelie
    • one year ago
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    oh got it xD

  39. marcelie
    • one year ago
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    thank you :)

  40. freckles
    • one year ago
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    np

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