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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

  • one year ago
  • one year 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?

    • one year ago
  2. Dido525 Group Title
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    @lucenzo

    • one year ago
  3. lucenzo Group Title
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    yeah, that's it

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

    • one year ago
  5. Dido525 Group Title
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    Darn... This complicates things...

    • one year ago
  6. Dido525 Group Title
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    Okay I would factor.

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

    • one year ago
  8. Dido525 Group Title
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    =4

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

    • one year ago
  10. lucenzo Group Title
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    no, I have no idea

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

    • one year 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)

    • one year 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)"

    • one year ago
  14. Dido525 Group Title
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    I meant for the right log :) .

    • one year ago
  15. lucenzo Group Title
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    OH

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

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

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

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

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

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

    • one year 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.

    • one year ago
  23. Dido525 Group Title
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    That was my next plan :P .

    • one year ago
  24. asnaseer Group Title
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    :)

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

    • one year 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

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

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

    • one year ago
  29. Dido525 Group Title
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    But you have too O_o .

    • one year 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.

    • one year ago
  31. Algebraic! Group Title
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    converting bases isn't advanced... iirc.

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

    • one year ago
  33. Algebraic! Group Title
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    you usually learn it first...

    • one year ago
  34. asnaseer Group Title
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    that is what thought @Algebraic!

    • one year ago
  35. asnaseer Group Title
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    *what I thought

    • one year ago
  36. lucenzo Group Title
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    hm..

    • one year 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

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

    • one year ago
  39. Dido525 Group Title
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    Wait. Hold yon. You might :P .

    • one year ago
  40. Dido525 Group Title
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    Yeah you have too.\\

    • one year 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: \[x^{2} + 6x - 9991 = 0\]

    • one year ago
  42. lucenzo Group Title
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    did you get that? @Dido525

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

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

    • one year ago
  45. Dido525 Group Title
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    I got it!!!

    • one year ago
  46. Dido525 Group Title
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    No you don't get that...

    • one year ago
  47. lucenzo Group Title
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    I think I did it wrong

    • one year ago
  48. lucenzo Group Title
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    lol okay, yeah

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

    • one year 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.

    • one year ago
  51. asnaseer Group Title
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    I get 3 solutions

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

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

    • one year ago
  54. Dido525 Group Title
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    I do. I will fix.

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

    • one year ago
  56. Dido525 Group Title
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    Yep!

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

    • one year ago
  58. Dido525 Group Title
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    Are you there? @lucenzo

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

    • one year ago
  60. Dido525 Group Title
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    Yep!

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

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

    • one year ago
  63. Dido525 Group Title
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    I didn't forget it.

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

    • one year ago
  65. Dido525 Group Title
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    Yeah, my mistake.

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

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

    • one year ago
  68. Dido525 Group Title
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    @lucenzo

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

    • one year 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"

    • one year ago
  71. Dido525 Group Title
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    Good!

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

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

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

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

    • one year ago
  76. lucenzo Group Title
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    yeah, I expanded the first term

    • one year ago
  77. Dido525 Group Title
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    No no!!

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

    • one year ago
  79. Dido525 Group Title
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    |dw:1351988960498:dw|

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

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

    • one year ago
  82. lucenzo Group Title
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    ah, that

    • one year ago
  83. lucenzo Group Title
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    I see now xD thanks you

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

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

    • one year ago
  86. lucenzo Group Title
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    yes, I did

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

    • one year ago
  88. Dido525 Group Title
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    |dw:1351989182904:dw|

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

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

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

    • one year 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.

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

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

    • one year ago
  95. Dido525 Group Title
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    It's base 5x+9 .

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

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

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

    • one year 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.

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

    • one year ago
  101. lucenzo Group Title
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    5x+9*

    • one year ago
  102. Dido525 Group Title
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    Yeah. My bad.

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

    • one year ago
  104. Dido525 Group Title
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    Great job!

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

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

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

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

    • one year ago
  109. lucenzo Group Title
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    nvm

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

    • one year ago
  111. Dido525 Group Title
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    It's fine :) .

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

    • one year ago
  113. Dido525 Group Title
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    Yes.

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

    • one year ago
  115. Dido525 Group Title
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    Ohh!!!!

    • one year 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

    • one year ago
  117. Dido525 Group Title
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    oops.

    • one year ago
  118. asnaseer Group Title
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    :)

    • one year 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.

    • one year ago
  120. Dido525 Group Title
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    Yep :) .

    • one year ago
  121. asnaseer Group Title
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    do you understand this step @lucenzo ?

    • one year ago
  122. lucenzo Group Title
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    Yeah, I'm gonna try it

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

    • one year 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.

    • one year ago
  125. Dido525 Group Title
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    I only get 1 answer :( .

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

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

    • one year ago
  128. lucenzo Group Title
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    so 1 or 0.5

    • one year ago
  129. lucenzo Group Title
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    what would be the third answer?

    • one year 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:\[log_{5x+9}(x+3)=0.5\]

    • one year ago
  131. lucenzo Group Title
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    right

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

    • one year ago
  133. Dido525 Group Title
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    Now I got it at last :D .

    • one year ago
  134. asnaseer Group Title
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    :)

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

    • one year ago
  136. asnaseer Group Title
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    are you stuck @lucenzo ?

    • one year 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)

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

    • one year 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)

    • one year 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

    • one year ago
  141. lucenzo Group Title
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    okay

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

    • one year 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:

    • one year ago
  144. asnaseer Group Title
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    not quite right @Dido525

    • one year ago
  145. Dido525 Group Title
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    The 1 is incorrect right?

    • one year ago
  146. asnaseer Group Title
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    yes

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

    • one year ago
  148. Dido525 Group Title
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    erm...

    • one year ago
  149. asnaseer Group Title
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    @lucenzo you are over complicating this.

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

    • one year ago
  151. Algebraic! Group Title
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    that's right, actually.

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

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

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

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

    • one year ago
  156. Algebraic! Group Title
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    same thing...

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

    • one year 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.

    • one year ago
  159. lucenzo Group Title
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    I think I just showed the work

    • one year ago
  160. Dido525 Group Title
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    Yeah it's correct regardless.

    • one year ago
  161. Algebraic! Group Title
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    yep you did.

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

    • one year ago
  163. lucenzo Group Title
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    3/2

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

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

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

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

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

    • one year 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:\[log_{5x+9}(x+3)=0.5\]

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

    • one year ago
  171. lucenzo Group Title
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    okay

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

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

    • one year ago
  174. Dido525 Group Title
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    |dw:1351991727301:dw|

    • one year ago
  175. lucenzo Group Title
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    LOL^

    • one year 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

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

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

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

    • one year ago
  180. asnaseer Group Title
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    @Dido525 please let @lucenzo solve this

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

    • one year ago
  182. asnaseer Group Title
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    perfect!

    • one year ago
  183. asnaseer Group Title
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    now factorise and solve

    • one year ago
  184. Dido525 Group Title
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    Sorry about that anaseer.

    • one year ago
  185. asnaseer Group Title
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    np :)

    • one year ago
  186. Dido525 Group Title
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    I am kinda into the problem XD .

    • one year 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. :)

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

    • one year ago
  189. asnaseer Group Title
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    you have the right answer

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

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

    • one year ago
  192. Dido525 Group Title
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    You could have factored :P .

    • one year ago
  193. lucenzo Group Title
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    wha

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

    • one year ago
  195. Dido525 Group Title
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    ^

    • one year ago
  196. lucenzo Group Title
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    -.-

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

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

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

    • one year ago
  200. Dido525 Group Title
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    Yeah. I checked. All work.

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

    • one year 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 :) .

    • one year 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

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

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

    • one year 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)

    • one year ago
  207. Dido525 Group Title
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    Derivatives hehe...

    • one year ago
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