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anonymous

  • one year ago

Evalute log 2(x-2) + log 2 (x+5)=3

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  1. anonymous
    • one year ago
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    @Zarkon

  2. Nnesha
    • one year ago
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    familiar with the log properties ?

  3. anonymous
    • one year ago
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    yes

  4. Nnesha
    • one year ago
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    okay so apply one of them at left side \[\huge\rm \log_2(x-2) \color{reD}{+} \log_2 (x+5)=3\] there is plus sign whichnoe one would you use ? quotient or product property ?

  5. anonymous
    • one year ago
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    product

  6. Nnesha
    • one year ago
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    yes \[\huge\rm log_b x + \log_b y = \log_b(x \times y)\] try to apply this on ur original question pretty sure you can do it!

  7. anonymous
    • one year ago
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    x^2+3x-10

  8. Nnesha
    • one year ago
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    yes \[\huge\rm log_2 (x^2+3x-10)=3\] now we have to convert log to exponential form

  9. anonymous
    • one year ago
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    x^2+3x=13

  10. Nnesha
    • one year ago
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    how did you get 13 ?

  11. Nnesha
    • one year ago
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    ohh well log_2 is still there \[\huge\rm log_b x + \log_b y = \color{ReD}{\log_b}(x \times y)\] \[\huge\rm\color{reD}{ log_2} (x^2+3x-10)=3\] log_b is a common factor

  12. anonymous
    • one year ago
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    now what do i do?

  13. UnkleRhaukus
    • one year ago
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    now, exponentiate both sides of the equation, with the base 2

  14. anonymous
    • one year ago
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    What does that mean?

  15. Nnesha
    • one year ago
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    |dw:1439613747571:dw| convert log to exponential form

  16. UnkleRhaukus
    • one year ago
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    from\[LHS = RHS\] \[2^{LHS} = 2^{RHS}\]

  17. anonymous
    • one year ago
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    ok hold on let me try..

  18. anonymous
    • one year ago
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    y=2^x ?? Im so confused.

  19. UnkleRhaukus
    • one year ago
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    \[\begin{align}\log_2(x^2+3x-10)&=3\\ 2^{\log_2(x^2+3x-10)}&=2^3\\ (x^2+3x-10)&=2^3\end{align} \]

  20. Nnesha
    • one year ago
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    that was an example |dw:1439613949291:dw|

  21. anonymous
    • one year ago
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    Yay! :) Would that be the final answer?

  22. UnkleRhaukus
    • one year ago
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    solve the quadratic

  23. anonymous
    • one year ago
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    I can do that :) Please hold on

  24. anonymous
    • one year ago
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    (x-5)(x+2)

  25. anonymous
    • one year ago
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    wait! I factored wrong! Ughhh hold again

  26. anonymous
    • one year ago
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    (x+5)(x-2)

  27. UnkleRhaukus
    • one year ago
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    \[\begin{align}\log_2(x^2+3x-10)&=3\\ 2^{\log_2(x^2+3x-10)}&=2^3\\ (x^2+3x-10)&=2^3\\ x^2+3x-18&=0\\ &= \end{align}\]

  28. anonymous
    • one year ago
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    (x+6)(x-3)

  29. UnkleRhaukus
    • one year ago
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    \[x^2+3x-18=0\\ (x+6)(x-3)=0\] good, so what values of \(x\) do you get?

  30. anonymous
    • one year ago
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    x=-6,3

  31. UnkleRhaukus
    • one year ago
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    If we try plugging the first of these (\(x=-6\)) back into the original equation \[\log _2(x-2) +\log_2 (x+5)=3\] we get\[\log _2(-6-2) +\log_2 (-6+5)=3\\\log _2(-8) +\log_2 (-1)=3\] .... logs of negative numbers... hmm doesn't make much sense. If we plugging the second of these (\(x=3\)) back into the original equation \[\log _2(x-2) +\log_2 (x+5)=3\] we get\[\log _2(3-2) +\log_2 (3+5)=3\\\log _2(1) +\log_2 (8)=3\\0+\log_2 (2^3)=3\\3\log_2 (2)=3 \\3=3\] which is always true. so this is our solution

  32. anonymous
    • one year ago
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    Thanks :)!! I understand it now!!!

  33. anonymous
    • one year ago
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    If I post another question, will u answer it?

  34. anonymous
    • one year ago
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    solve 7-7x=(3x+2)(x-1) and check for extraneous solutions

  35. anonymous
    • one year ago
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    7 - 7x = 3x² - x - 2 3x² + 6x - 9 = 0 3(x + 3)(x - 1) = 0 x = -3, 1

  36. anonymous
    • one year ago
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    Would it be extraneous? Thanks geny55!!! :D

  37. anonymous
    • one year ago
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    medal please?

  38. anonymous
    • one year ago
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    I dont thnk it's extraneous? am i right?

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