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lucenzo

  • 2 years ago

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

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

  2. Dido525
    • 2 years ago
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    @lucenzo

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

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

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

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

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

  8. Dido525
    • 2 years ago
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    =4

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

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

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

  12. lucenzo
    • 2 years ago
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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)

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

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

  15. lucenzo
    • 2 years ago
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    OH

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

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

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

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

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

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

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

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

  24. asnaseer
    • 2 years ago
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    :)

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

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

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

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

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

  30. asnaseer
    • 2 years ago
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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.

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

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

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

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

  35. asnaseer
    • 2 years ago
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    *what I thought

  36. lucenzo
    • 2 years ago
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    hm..

  37. lucenzo
    • 2 years ago
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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

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

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

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

  41. lucenzo
    • 2 years ago
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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: \[x^{2} + 6x - 9991 = 0\]

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

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

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

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

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

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

  48. lucenzo
    • 2 years ago
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    lol okay, yeah

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

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

  51. asnaseer
    • 2 years ago
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    I get 3 solutions

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

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

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

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

  56. Dido525
    • 2 years ago
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    Yep!

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

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

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

  60. Dido525
    • 2 years ago
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    Yep!

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

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

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

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

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

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

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

  68. Dido525
    • 2 years ago
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    @lucenzo

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

  70. lucenzo
    • 2 years ago
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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"

  71. Dido525
    • 2 years ago
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    Good!

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

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

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

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

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

  77. Dido525
    • 2 years ago
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    No no!!

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

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

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

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

  82. lucenzo
    • 2 years ago
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    ah, that

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

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

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

  86. lucenzo
    • 2 years ago
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    yes, I did

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

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

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

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

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

  92. Dido525
    • 2 years ago
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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.

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

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

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

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

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

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

  99. Dido525
    • 2 years ago
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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.

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

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

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

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

  104. Dido525
    • 2 years ago
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    Great job!

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

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

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

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

  109. lucenzo
    • 2 years ago
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    nvm

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

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

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

  113. Dido525
    • 2 years ago
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    Yes.

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

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

  116. lucenzo
    • 2 years ago
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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

  117. Dido525
    • 2 years ago
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    oops.

  118. asnaseer
    • 2 years ago
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    :)

  119. asnaseer
    • 2 years ago
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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.

  120. Dido525
    • 2 years ago
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    Yep :) .

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

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

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

  124. asnaseer
    • 2 years ago
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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.

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

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

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

  128. lucenzo
    • 2 years ago
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    so 1 or 0.5

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

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

  131. lucenzo
    • 2 years ago
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    right

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

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

  134. asnaseer
    • 2 years ago
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    :)

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

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

  137. lucenzo
    • 2 years ago
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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)

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

  139. asnaseer
    • 2 years ago
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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)

  140. asnaseer
    • 2 years ago
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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

  141. lucenzo
    • 2 years ago
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    okay

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

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

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

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

  146. asnaseer
    • 2 years ago
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    yes

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

  148. Dido525
    • 2 years ago
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    erm...

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

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

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

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

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

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

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

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

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

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

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

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

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

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

  163. lucenzo
    • 2 years ago
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    3/2

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

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

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

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

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

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

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

  171. lucenzo
    • 2 years ago
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    okay

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

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

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

  175. lucenzo
    • 2 years ago
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    LOL^

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

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

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

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

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

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

  182. asnaseer
    • 2 years ago
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    perfect!

  183. asnaseer
    • 2 years ago
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    now factorise and solve

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

  185. asnaseer
    • 2 years ago
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    np :)

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

  187. asnaseer
    • 2 years ago
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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. :)

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

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

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

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

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

  193. lucenzo
    • 2 years ago
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    wha

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

  195. Dido525
    • 2 years ago
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    ^

  196. lucenzo
    • 2 years ago
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    -.-

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

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

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

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

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

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

  203. lucenzo
    • 2 years ago
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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

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

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

  206. lucenzo
    • 2 years ago
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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)

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

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