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lucenzo Group Title

I need to solve for x: log(base 5x + 9) (x^2 + 6x + 9) + log(base x+3) (5x^2 + 24x + 27) = 4

  • 2 years ago
  • 2 years ago

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  1. Dido525 Group Title
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    \[\log_{5x+9} (x^2+6x+9)+\log_{x+3} (5x^2+24x+27) = 4\] Is that the question?

    • 2 years ago
  2. Dido525 Group Title
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    @lucenzo

    • 2 years ago
  3. lucenzo Group Title
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    yeah, that's it

    • 2 years ago
  4. Dido525 Group Title
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    They don't have the same base?

    • 2 years ago
  5. Dido525 Group Title
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    Darn... This complicates things...

    • 2 years ago
  6. Dido525 Group Title
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    Okay I would factor.

    • 2 years ago
  7. Dido525 Group Title
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    \[\log_{5x+9}(x+3)^{2}+\log_{x+3} ((5x+9)(x+3))\]

    • 2 years ago
  8. Dido525 Group Title
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    =4

    • 2 years ago
  9. Dido525 Group Title
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    Can you see anyway to simplify this? :) .

    • 2 years ago
  10. lucenzo Group Title
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    no, I have no idea

    • 2 years ago
  11. Dido525 Group Title
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    You have two logs multiplied. What can you do? :) .

    • 2 years ago
  12. lucenzo Group Title
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    This is a type of a bonus, so we never learned this (I thought of factoring but idk how to xD). Is it possible to get both logs to the same base? By putting them both to the power of (5x+9)(x+3)

    • 2 years ago
  13. lucenzo Group Title
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    They don't have the same base so you can't do "loga + logb = log(a*b)"

    • 2 years ago
  14. Dido525 Group Title
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    I meant for the right log :) .

    • 2 years ago
  15. lucenzo Group Title
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    OH

    • 2 years ago
  16. Dido525 Group Title
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    Ohh you need help factoring? Okay: |dw:1351987084250:dw|

    • 2 years ago
  17. lucenzo Group Title
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    You do log (5x^2 + 24x + 27)/(x+3)

    • 2 years ago
  18. Dido525 Group Title
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    Wait I am still in the process of solving this :P .

    • 2 years ago
  19. lucenzo Group Title
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    and do that for both, lol okay. I'll try solving it here like that

    • 2 years ago
  20. Dido525 Group Title
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    Wow this is the hardest log I have done so far :P .

    • 2 years ago
  21. lucenzo Group Title
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    yeah, same. Never seen anything like it. But I think it'll work out nicely

    • 2 years ago
  22. asnaseer Group Title
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    @Dido525 your approach is correct. think of making use of the change of base for logs you'll be there.

    • 2 years ago
  23. Dido525 Group Title
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    That was my next plan :P .

    • 2 years ago
  24. asnaseer Group Title
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    :)

    • 2 years ago
  25. asnaseer Group Title
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    @lucenzo do you know about the change of base formula for logs?

    • 2 years ago
  26. lucenzo Group Title
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    no. I think I might have to solve this question w/o using the formula, since we never learned it

    • 2 years ago
  27. asnaseer Group Title
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    I can't see how to solve this without using that formula?

    • 2 years ago
  28. lucenzo Group Title
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    okay, nvm then. Can you tell me what the formula is?

    • 2 years ago
  29. Dido525 Group Title
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    But you have too O_o .

    • 2 years ago
  30. asnaseer Group Title
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    I would assume that if this is a /bonus/ question, then the teacher expected some of you to do some learning on your own to try and solve this.

    • 2 years ago
  31. Algebraic! Group Title
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    converting bases isn't advanced... iirc.

    • 2 years ago
  32. asnaseer Group Title
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    change of base is given by:\[\log_ax=\frac{\log_bx}{\log_ba}\]

    • 2 years ago
  33. Algebraic! Group Title
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    you usually learn it first...

    • 2 years ago
  34. asnaseer Group Title
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    that is what thought @Algebraic!

    • 2 years ago
  35. asnaseer Group Title
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    *what I thought

    • 2 years ago
  36. lucenzo Group Title
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    hm..

    • 2 years ago
  37. lucenzo Group Title
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    Oh, I've used that formula LOL. I actually was thinking that from the beginning. In this case I did \[\log_{a}n \] first

    • 2 years ago
  38. Dido525 Group Title
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    Well actually you don't have to use change of base :) .

    • 2 years ago
  39. Dido525 Group Title
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    Wait. Hold yon. You might :P .

    • 2 years ago
  40. Dido525 Group Title
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    Yeah you have too.\\

    • 2 years ago
  41. lucenzo Group Title
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    then I faactored both trinomials and it was log(base 10) so I was able to multiply the logs and now I've ended up w/ a strange trinomial: \[x^{2} + 6x - 9991 = 0\]

    • 2 years ago
  42. lucenzo Group Title
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    did you get that? @Dido525

    • 2 years ago
  43. Algebraic! Group Title
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    I got x =-3/2, -1 haven't checked it however...

    • 2 years ago
  44. lucenzo Group Title
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    did you get this trinomial tho: x^2+6x−9991=0

    • 2 years ago
  45. Dido525 Group Title
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    I got it!!!

    • 2 years ago
  46. Dido525 Group Title
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    No you don't get that...

    • 2 years ago
  47. lucenzo Group Title
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    I think I did it wrong

    • 2 years ago
  48. lucenzo Group Title
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    lol okay, yeah

    • 2 years ago
  49. Dido525 Group Title
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    @Algebraic! I got only -3/2 .

    • 2 years ago
  50. Dido525 Group Title
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    \[\log_{5x+9}(x+3)^{2}+\log_{x+3} ((5x+9)(x+3)) \] We were at that step.

    • 2 years ago
  51. asnaseer Group Title
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    I get 3 solutions

    • 2 years ago
  52. Dido525 Group Title
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    Use the properties of logs to expand that right log: @lucenzo

    • 2 years ago
  53. asnaseer Group Title
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    I think you have a typo there @Dido525 - they should all be base (x+3)

    • 2 years ago
  54. Dido525 Group Title
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    I do. I will fix.

    • 2 years ago
  55. asnaseer Group Title
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    \[\log_{x+3} ((5x+9)(x+3))=\log_{x+3} (5x+9)+\log_{x+3} (x+3)\]

    • 2 years ago
  56. Dido525 Group Title
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    Yep!

    • 2 years ago
  57. Dido525 Group Title
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    \[\log_{x+3} ((5x+9)(x+3))=\log_{x+3} (5x+9)+\log_{x+3} (x+3)=4\]

    • 2 years ago
  58. Dido525 Group Title
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    Are you there? @lucenzo

    • 2 years ago
  59. lucenzo Group Title
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    yeah. So we can expand all of the logs right?

    • 2 years ago
  60. Dido525 Group Title
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    Yep!

    • 2 years ago
  61. asnaseer Group Title
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    @Dido525 your last equation is wrong - you have forgotten the other log term

    • 2 years ago
  62. Dido525 Group Title
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    I am ignoring that one for now :P .

    • 2 years ago
  63. Dido525 Group Title
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    I didn't forget it.

    • 2 years ago
  64. asnaseer Group Title
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    then it shouldn't be set "= 4"

    • 2 years ago
  65. Dido525 Group Title
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    Yeah, my mistake.

    • 2 years ago
  66. lucenzo Group Title
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    it would be 3 then? Instead of 4

    • 2 years ago
  67. Dido525 Group Title
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    So expanding all that you should get: \[\log_{5x+9}(x+3)^{2}+\log_{x+3} (5x+9)+\log_{x+3} (x+3)=4\]

    • 2 years ago
  68. Dido525 Group Title
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    @lucenzo

    • 2 years ago
  69. Dido525 Group Title
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    Do you notice anything special about the last log term? @lucenzo

    • 2 years ago
  70. lucenzo Group Title
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    yeah, its the same thing over the same thing. So it is 1. You move that to the other side and the "4" becomes a "3"

    • 2 years ago
  71. Dido525 Group Title
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    Good!

    • 2 years ago
  72. Dido525 Group Title
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    So you should get : \[\log_{5x+9}(x+3)^{2}+\log_{x+3} (5x+9) =3\]

    • 2 years ago
  73. lucenzo Group Title
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    I got to that part no problem. The next part is what confuses me xD

    • 2 years ago
  74. Dido525 Group Title
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    Now what do you notice about the first log term?

    • 2 years ago
  75. Algebraic! Group Title
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    \[\log_{5x+9} (x+3)^2 = \frac{ 2}{\log_{x+3} (5x+9) }\] \[\log_{x+3} (5x+9)(x+3) = 1 +\log_{x+3} (5x+9)\] \[\log_{x+3} (5x+9)+\frac{ 2}{\log_{x+3} (5x+9) }=3\]

    • 2 years ago
  76. lucenzo Group Title
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    yeah, I expanded the first term

    • 2 years ago
  77. Dido525 Group Title
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    No no!!

    • 2 years ago
  78. Dido525 Group Title
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    Remember, in a log you can bring that power down :) .

    • 2 years ago
  79. Dido525 Group Title
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    |dw:1351988960498:dw|

    • 2 years ago
  80. lucenzo Group Title
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    powering every term in the equation by the same number?

    • 2 years ago
  81. Algebraic! Group Title
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    quadratic in \[\log_{x+3} (5x+9) \]

    • 2 years ago
  82. lucenzo Group Title
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    ah, that

    • 2 years ago
  83. lucenzo Group Title
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    I see now xD thanks you

    • 2 years ago
  84. Algebraic! Group Title
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    u^2 -3u+2=0

    • 2 years ago
  85. Dido525 Group Title
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    So you should get : \[2\log_{5x+9}(x+3)+\log_{x+3} (5x+9) =3\]

    • 2 years ago
  86. lucenzo Group Title
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    yes, I did

    • 2 years ago
  87. Dido525 Group Title
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    Now use change of base that you learned a little earlier. :) .

    • 2 years ago
  88. Dido525 Group Title
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    |dw:1351989182904:dw|

    • 2 years ago
  89. lucenzo Group Title
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    I know the normal way of using that rule where it would be: \[\log(x+3)/\log(5x+9) \]

    • 2 years ago
  90. Dido525 Group Title
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    Hint: Use it on the right log :) .

    • 2 years ago
  91. lucenzo Group Title
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    are we assuming that the "y" is just 10. so its log base of 10?

    • 2 years ago
  92. Dido525 Group Title
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    Ahh but in this case that rule is useless. Besides you can't use that unless they have the same base.

    • 2 years ago
  93. Dido525 Group Title
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    Well in this case it's base x+3 .

    • 2 years ago
  94. lucenzo Group Title
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    no, i meant for the first term. the term on the left

    • 2 years ago
  95. Dido525 Group Title
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    It's base 5x+9 .

    • 2 years ago
  96. lucenzo Group Title
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    just for the \[\log_{5x+9}(x+3) \]

    • 2 years ago
  97. Dido525 Group Title
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    The base is 5x+9 for that yes.

    • 2 years ago
  98. lucenzo Group Title
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    yeah, that's why i did \[\log(x+3)/\log(5x+9)\]

    • 2 years ago
  99. Dido525 Group Title
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    The trick for the right log is that you want to convert that right base to the base on the left log.

    • 2 years ago
  100. Dido525 Group Title
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    That's correct but you want base 5x+3 .

    • 2 years ago
  101. lucenzo Group Title
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    5x+9*

    • 2 years ago
  102. Dido525 Group Title
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    Yeah. My bad.

    • 2 years ago
  103. lucenzo Group Title
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    I got 1/\[1/\log_{5x+9}(x+3) \]

    • 2 years ago
  104. Dido525 Group Title
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    Great job!

    • 2 years ago
  105. lucenzo Group Title
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    whoops, just meant to put 1 "1/"

    • 2 years ago
  106. lucenzo Group Title
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    and now we put everything to the power of *5x+9* ?

    • 2 years ago
  107. Dido525 Group Title
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    So you should now have: \[2\log_{5x+9}(x+3)+1 =3\log_{5x+9} (x+3)\]

    • 2 years ago
  108. Dido525 Group Title
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    Move the common terms to one side.

    • 2 years ago
  109. lucenzo Group Title
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    nvm

    • 2 years ago
  110. lucenzo Group Title
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    yeah lol, I was about to say that

    • 2 years ago
  111. Dido525 Group Title
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    It's fine :) .

    • 2 years ago
  112. asnaseer Group Title
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    @Dido525 - are you sure of your last step?

    • 2 years ago
  113. Dido525 Group Title
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    Yes.

    • 2 years ago
  114. asnaseer Group Title
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    remember you started from:\[2\log_{5x+9}(x+3)+\frac{1}{\log_{5x+9} (x+3)} =3\]

    • 2 years ago
  115. Dido525 Group Title
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    Ohh!!!!

    • 2 years ago
  116. lucenzo Group Title
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    can we not put it all to ther power of 5x+9 tho? I'm still thinking it might be simpler

    • 2 years ago
  117. Dido525 Group Title
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    oops.

    • 2 years ago
  118. asnaseer Group Title
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    :)

    • 2 years ago
  119. asnaseer Group Title
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    from this point, it might be better to do what @Algebraic! did, i.e. let:\[u=log_{5x+9}(x+3)\]to get:\[2u+\frac{1}{u}=3\]and then solve this quadratic in u first.

    • 2 years ago
  120. Dido525 Group Title
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    Yep :) .

    • 2 years ago
  121. asnaseer Group Title
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    do you understand this step @lucenzo ?

    • 2 years ago
  122. lucenzo Group Title
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    Yeah, I'm gonna try it

    • 2 years ago
  123. Dido525 Group Title
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    I only get one solution >.> .

    • 2 years ago
  124. asnaseer Group Title
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    we got to:\[2u+\frac{1}{u}=3\]multiply both sides by u to get:\[2u^2+1=3u\]\[2u^2-3u+1=0\]this should have 2 solutions for u. then use the fact that:\[u=log_{5x+9}(x+3)\]to find all the x values.

    • 2 years ago
  125. Dido525 Group Title
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    I only get 1 answer :( .

    • 2 years ago
  126. asnaseer Group Title
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    there are 3 answers altogether - @lucenzo have you solved the quadratic in u yet?

    • 2 years ago
  127. lucenzo Group Title
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    yeah I got u = 1 or 1/2

    • 2 years ago
  128. lucenzo Group Title
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    so 1 or 0.5

    • 2 years ago
  129. lucenzo Group Title
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    what would be the third answer?

    • 2 years ago
  130. asnaseer Group Title
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    that is correct. so now you need to solve:\[log_{5x+9}(x+3)=1\]and:\[log_{5x+9}(x+3)=0.5\]

    • 2 years ago
  131. lucenzo Group Title
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    right

    • 2 years ago
  132. asnaseer Group Title
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    so, starting with:\[log_{5x+9}(x+3)=1\]what do you get for x?

    • 2 years ago
  133. Dido525 Group Title
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    Now I got it at last :D .

    • 2 years ago
  134. asnaseer Group Title
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    :)

    • 2 years ago
  135. asnaseer Group Title
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    remember that if:\[\log_ba=c\]then:\[a=b^c\]

    • 2 years ago
  136. asnaseer Group Title
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    are you stuck @lucenzo ?

    • 2 years ago
  137. lucenzo Group Title
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    yeah, but I just realized I had a 2 multiplied to the log term (I'm gonna try it without the 2)

    • 2 years ago
  138. lucenzo Group Title
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    I got 3/2 for the first one (if x = 1)

    • 2 years ago
  139. asnaseer Group Title
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    plz double check your work - I think you have made an error in the "sign" of the result. and it is for (u=1) not (x=1)

    • 2 years ago
  140. asnaseer Group Title
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    it might be better if you showed your steps here so that we can help you spot where you may have made a mistake

    • 2 years ago
  141. lucenzo Group Title
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    okay

    • 2 years ago
  142. Dido525 Group Title
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    I got -3/2 , 0 , 1.

    • 2 years ago
  143. lucenzo Group Title
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    I took the equation you had then I put both sides to the power of 5x+9 like so:

    • 2 years ago
  144. asnaseer Group Title
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    not quite right @Dido525

    • 2 years ago
  145. Dido525 Group Title
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    The 1 is incorrect right?

    • 2 years ago
  146. asnaseer Group Title
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    yes

    • 2 years ago
  147. lucenzo Group Title
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    \[5x+9^{\log_{5x+9}(x+3) } = 5x+9^1\]

    • 2 years ago
  148. Dido525 Group Title
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    erm...

    • 2 years ago
  149. asnaseer Group Title
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    @lucenzo you are over complicating this.

    • 2 years ago
  150. lucenzo Group Title
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    \[x+3=5x+9\]

    • 2 years ago
  151. Algebraic! Group Title
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    that's right, actually.

    • 2 years ago
  152. asnaseer Group Title
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    yes - that is correct - but you could have got there in one step. remember:\[\log_ba=c\implies a=b^c\]

    • 2 years ago
  153. lucenzo Group Title
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    \[x-5x=9-3\]

    • 2 years ago
  154. Dido525 Group Title
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    yeah it's correct but why would you do that? :( .

    • 2 years ago
  155. lucenzo Group Title
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    That's the way we learned to write it

    • 2 years ago
  156. Algebraic! Group Title
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    same thing...

    • 2 years ago
  157. lucenzo Group Title
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    Just a way of showing how to get there, it's just the same thing

    • 2 years ago
  158. asnaseer Group Title
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    ok - fair enough - if that is how you were taught then you should stick to that method.

    • 2 years ago
  159. lucenzo Group Title
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    I think I just showed the work

    • 2 years ago
  160. Dido525 Group Title
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    Yeah it's correct regardless.

    • 2 years ago
  161. Algebraic! Group Title
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    yep you did.

    • 2 years ago
  162. asnaseer Group Title
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    ok so what x value do you get from this?

    • 2 years ago
  163. lucenzo Group Title
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    3/2

    • 2 years ago
  164. lucenzo Group Title
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    6/4 which reduces to 3/2

    • 2 years ago
  165. Dido525 Group Title
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    I keep getting - 3/2 hmm...

    • 2 years ago
  166. asnaseer Group Title
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    this step was correct:\[x-5x=9-3\]which then leads to:\[-4x=6\]

    • 2 years ago
  167. lucenzo Group Title
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    uh-oh, I made a sign error xD. I wrote -6

    • 2 years ago
  168. lucenzo Group Title
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    you're right. It's -3/2

    • 2 years ago
  169. asnaseer Group Title
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    ok, so first solution is x=-1.5. next, you need to solve this for x:\[log_{5x+9}(x+3)=0.5\]

    • 2 years ago
  170. asnaseer Group Title
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    again here it would help if you listed your steps

    • 2 years ago
  171. lucenzo Group Title
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    okay

    • 2 years ago
  172. Dido525 Group Title
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    @asnaseer : I got so far: |dw:1351991548862:dw|

    • 2 years ago
  173. lucenzo Group Title
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    \[x+3=(5x+9)^{\frac{ 1 }{ 2 }}\]

    • 2 years ago
  174. Dido525 Group Title
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    |dw:1351991727301:dw|

    • 2 years ago
  175. lucenzo Group Title
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    LOL^

    • 2 years ago
  176. asnaseer Group Title
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    @Dido525 where did you derive that first equation from? @lucenzo that is correct - now square both sides

    • 2 years ago
  177. Dido525 Group Title
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    You get 0 and 1 if you solve for x.

    • 2 years ago
  178. Dido525 Group Title
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    If you solve x I keep getting 0 and 1.

    • 2 years ago
  179. Dido525 Group Title
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    This thread is so epic I am lagging >.< .

    • 2 years ago
  180. asnaseer Group Title
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    @Dido525 please let @lucenzo solve this

    • 2 years ago
  181. lucenzo Group Title
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    \[x^2 + x = 0\]

    • 2 years ago
  182. asnaseer Group Title
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    perfect!

    • 2 years ago
  183. asnaseer Group Title
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    now factorise and solve

    • 2 years ago
  184. Dido525 Group Title
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    Sorry about that anaseer.

    • 2 years ago
  185. asnaseer Group Title
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    np :)

    • 2 years ago
  186. Dido525 Group Title
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    I am kinda into the problem XD .

    • 2 years ago
  187. asnaseer Group Title
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    I know some problems are so intriguing that we feel compelled to try and solve them ourselves - it is only natural. :)

    • 2 years ago
  188. lucenzo Group Title
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    i did the quadratic formula and got x = -1 or 0

    • 2 years ago
  189. asnaseer Group Title
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    you have the right answer

    • 2 years ago
  190. asnaseer Group Title
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    but you did not need to use the quadratic formula

    • 2 years ago
  191. Dido525 Group Title
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    Yeah I got that too :) . I just had a silly algebra error.

    • 2 years ago
  192. Dido525 Group Title
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    You could have factored :P .

    • 2 years ago
  193. lucenzo Group Title
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    wha

    • 2 years ago
  194. asnaseer Group Title
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    \[x^2+x=0\]\[x(x+1)=0\]so\[x=0\]or:\[x=-1\]

    • 2 years ago
  195. Dido525 Group Title
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    ^

    • 2 years ago
  196. lucenzo Group Title
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    -.-

    • 2 years ago
  197. asnaseer Group Title
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    so final solution is: x = -1.5 or -1 or 0

    • 2 years ago
  198. lucenzo Group Title
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    yeah, and I think ALL of them work. so none are extraneous

    • 2 years ago
  199. Dido525 Group Title
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    or -3/2, -1, 0 .

    • 2 years ago
  200. Dido525 Group Title
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    Yeah. I checked. All work.

    • 2 years ago
  201. asnaseer Group Title
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    correct - and thanks for posting such an interesting problem @lucenzo :)

    • 2 years ago
  202. Dido525 Group Title
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    I agree! I am in uni and this was the toughest log I have done in ages :) .

    • 2 years ago
  203. lucenzo Group Title
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    lol. I'm the one to thank you guys. Thank you sooo much! You all really helped my very much. I'm in high school right now, and when I take calculus next semester and have to do a log question like this, I'll feel like I'm ahead of the others xD

    • 2 years ago
  204. asnaseer Group Title
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    yw :) and keep up the hunger for knowledge my friend! :)

    • 2 years ago
  205. Dido525 Group Title
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    Calculus is very fun! You will enjoy it a lot :) .

    • 2 years ago
  206. lucenzo Group Title
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    Thanks, I heard that it has less writing and that you learn 'tricks' for solving questions ALOT quicker (like never before)

    • 2 years ago
  207. Dido525 Group Title
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    Derivatives hehe...

    • 2 years ago
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