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anonymous

  • one year ago

Check this.? Im not very good with logarithms and i watched a video and it said to do it like this but im not sure Log\/2(3)+log\/2(x)=3 then cancel out the logs as they are the same base so, 3+x=3 -3 -3 x=0???

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  1. Nnesha
    • one year ago
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    what's the original question ?

  2. anonymous
    • one year ago
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    Log\/2(3)+log\/2(x)=3

  3. Nnesha
    • one year ago
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    i mean equation \[\log_2(3)+\log_2 x=3\] like this OR

  4. Nnesha
    • one year ago
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    \[\frac{ \log_2 3 }{ \log_2 x} =3\]

  5. Nnesha
    • one year ago
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    maybe or maybe division hmm which one is correct ? graysongraddyLOL

  6. anonymous
    • one year ago
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    the first one is the original

  7. Nnesha
    • one year ago
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    oh okay familiar with the log properties ??

  8. anonymous
    • one year ago
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    Not exactly.

  9. Nnesha
    • one year ago
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    quotient rule\[\large\rm log_b x - \log_b y = \log_b \frac{ x }{ y}\] to condense you can change subtraction to division product rule \[\large\rm log_b x + \log_b y = \log_b( x \times y )\] addition ----> multiplication power rule \[\large\rm log_b x^y = y \log_b x\]

  10. Nnesha
    • one year ago
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    look at that and tell me which one you should apply ?

  11. anonymous
    • one year ago
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    The last one.?

  12. anonymous
    • one year ago
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    Math just absolutely dumbfounds me.

  13. Nnesha
    • one year ago
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    hmm alright there is PLUS sign between log now look at the log identities

  14. anonymous
    • one year ago
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    the subexponent of 2 is the identity correct.?

  15. Nnesha
    • one year ago
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    2 is base

  16. Nnesha
    • one year ago
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    quotient rule\[\large\rm\color{reD}{ log_b x - \log_b y = \log_b \frac{ x }{ y}}\] to condense you can change subtraction to division product rule \[\large\rm \color{reD}{log_b x + \log_b y = \log_b( x \times y )}\] addition ----> multiplication power rule \[\large\rm \color{Red}{log_b x^y = y \log_b x}\] you don't need ^power rule for this e quation we should apply it when there is a number at front of log

  17. anonymous
    • one year ago
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    so with the equation that i am using, How do i get a Log on the other side of the equal sign.?

  18. Nnesha
    • one year ago
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    alright we need to apply product rule which is \[\large\rm \color{reD}{log_b x + \log_b y = \log_b( x \times y )}\] log_b same base so take it out and multiply x y \[\log_b x+ \log_b y\] \[\log_b (x \times y)\]

  19. anonymous
    • one year ago
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    So multiply 3 and x.?

  20. Nnesha
    • one year ago
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    \[\huge\rm log_2(3)+\log_2 x=3\] like this OR yes right!!

  21. Nnesha
    • one year ago
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    that's how we should condense log equations \[\huge\rm log_2 (3 \times x)=3\] now we should convert log to exponential form

  22. Nnesha
    • one year ago
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    |dw:1439575720373:dw|

  23. anonymous
    • one year ago
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    That just absolutely confused me.

  24. Nnesha
    • one year ago
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    well you will see how easy it's just have to move that variables around

  25. Nnesha
    • one year ago
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    right side term becomes exponent of base and move the left side term to the right side |dw:1439575981638:dw|

  26. anonymous
    • one year ago
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    so log2 (3) = 3x.?

  27. Nnesha
    • one year ago
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    we need to convert to exponential there shouldn't be an log b=2 how would write 2 with 3 exponent ?

  28. Nnesha
    • one year ago
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    2 to the 3 ppower

  29. anonymous
    • one year ago
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    so 2^3=3?

  30. Nnesha
    • one year ago
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    2^3 = 3x not just x

  31. Nnesha
    • one year ago
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    now solve for x

  32. anonymous
    • one year ago
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    \[x=\frac{ 8 }{ 3 }\]

  33. Nnesha
    • one year ago
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    :)

  34. anonymous
    • one year ago
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    Okay cool :)

  35. anonymous
    • one year ago
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    Can you give me a example logarithm so i can see if i got the hang of it.?

  36. Nnesha
    • one year ago
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    why not!!

  37. anonymous
    • one year ago
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    Thank you so much :D

  38. Nnesha
    • one year ago
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    \[\huge\rm log_3 (4) - \log_3 (y) = 2\] look at the log properties

  39. anonymous
    • one year ago
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    So id use the first one right.?

  40. Nnesha
    • one year ago
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    yea quotient property

  41. anonymous
    • one year ago
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    then it would go to \[\log_{3}(4 \div x)=3\]

  42. Nnesha
    • one year ago
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    yep right

  43. Nnesha
    • one year ago
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    it's y not x but it's okay :P

  44. Nnesha
    • one year ago
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    i'm not gonna take off points like teachers do lol

  45. anonymous
    • one year ago
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    Oh whoops lol

  46. anonymous
    • one year ago
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    So it would go to \[3^{4}=3x\]

  47. Nnesha
    • one year ago
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    hmm no

  48. anonymous
    • one year ago
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    Where did i go wrong.?

  49. Nnesha
    • one year ago
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    the number which is opposite of the log supposed to be the exponent of base \[\log_3 ( \frac{ 4 }{ x }) =2\] and btw it's equal to 2 i would take off points for this mistake

  50. Nnesha
    • one year ago
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    try again!

  51. anonymous
    • one year ago
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    okay so it would go to \[\frac{ 4 }{ x } = 2x \]

  52. Nnesha
    • one year ago
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    how did you get two x's ?

  53. anonymous
    • one year ago
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    wait it would just be 2 on the right side

  54. anonymous
    • one year ago
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    then i multiply 4 from the left side to get 8 on the right side so x=8.?

  55. Nnesha
    • one year ago
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    |dw:1439577454761:dw|

  56. Nnesha
    • one year ago
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    here is the example to convert log to exponential form \[\huge\rm log_\color{ReD}{b} x = \color{blue}{y}\] b is base \[\huge\rm \color{ReD}{b}^\color{blue}{y} =x\]

  57. anonymous
    • one year ago
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    \[3^{2} =4x.?\]

  58. Nnesha
    • one year ago
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    not really hmm it's 4/x so that would stay same at right side \[3^2=\frac{ 4 }{ x }\]now solve for x

  59. anonymous
    • one year ago
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    \[x=\frac{ 4 }{ 9 }\]

  60. Nnesha
    • one year ago
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    yep!

  61. anonymous
    • one year ago
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    Thank you so much!

  62. Nnesha
    • one year ago
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    my pleasure!

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