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taranicolee

solve log5=log4+log(-4x+5)

  • 2 years ago
  • 2 years ago

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  1. JamesJ
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    first simplify the RHS (right hand side), what do you get?

    • 2 years ago
  2. Mr.crazzy
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    x=1.225

    • 2 years ago
  3. taranicolee
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    well it will equal 5=4+(-4x+5) and you would solve from there, but OH WAIT. haha for some reason when i was trying to solve it i wrote 4 times (-4x+5)

    • 2 years ago
  4. JamesJ
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    @crazzy: it's actually almost never helpful just to give an answer without explaining how you got the answer. But now that you've given the answer, you can now explain step by step how you got there. :-)

    • 2 years ago
  5. taranicolee
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    could you help me with another problem? \[9^{x-5} = 81^{3}\]

    • 2 years ago
  6. taranicolee
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    waiiit.. james j :/ hahah

    • 2 years ago
  7. myininaya
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    \[\log(ab)=\log(a)+\log(b)\]

    • 2 years ago
  8. myininaya
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    Do you recall this?

    • 2 years ago
  9. taranicolee
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    yes

    • 2 years ago
  10. myininaya
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    \[\log(5)=\log(4(-4x+5))\]

    • 2 years ago
  11. myininaya
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    so now we can set the insides equal

    • 2 years ago
  12. myininaya
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    \[5=4(-4x+5)\]

    • 2 years ago
  13. taranicolee
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    THAT'S WHY I HAD IT MULTIPLYING. haha sweet (:

    • 2 years ago
  14. sasogeek
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    • 2 years ago
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  15. myininaya
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    \[5=-16x+20\] I distributed the 4 here

    • 2 years ago
  16. myininaya
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    Now I'm going to subtract 20 on both sides \[5-20=-16x\]

    • 2 years ago
  17. myininaya
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    \[-15=-16x => -16x=-15 \]

    • 2 years ago
  18. myininaya
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    Now we divide both sides by -16 \[x=\frac{-15}{-16}\]

    • 2 years ago
  19. myininaya
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    But a negative divided by a negative is positive

    • 2 years ago
  20. myininaya
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    \[x=\frac{15}{16}\]

    • 2 years ago
  21. sasogeek
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    what did i do wrong? my answer doesn't concur with yours myininaya... :|

    • 2 years ago
  22. taranicolee
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    i think that's right though because when i tried to solve it on my own i got one, and 15/16 .9375

    • 2 years ago
  23. sasogeek
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    i know my working is wrong, i don't see where i went wrong though...

    • 2 years ago
  24. taranicolee
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    myininaya could you help me with the other equation i put up there please?

    • 2 years ago
  25. myininaya
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    @saso I can't read yours that well

    • 2 years ago
  26. taranicolee
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    sasogeek, you didn't just cancle out the log, you cancled the five as well, right?

    • 2 years ago
  27. Mr.crazzy
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    thanks for ur advice james

    • 2 years ago
  28. sasogeek
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    log5-log5=0

    • 2 years ago
  29. myininaya
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    Taranicolee you can post if you wish

    • 2 years ago
  30. taranicolee
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    sasogeek you have to distribute the four before you subtract from in the parenthesis

    • 2 years ago
  31. taranicolee
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    9 x−5 =81 3

    • 2 years ago
  32. taranicolee
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    oh wait that's not it

    • 2 years ago
  33. myininaya
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    by the way \[\log(4x+5) \neq \log(4x)+\log(5) \]

    • 2 years ago
  34. taranicolee
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    \[9^{x -5}=81^{3}\]

    • 2 years ago
  35. sasogeek
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    i'm just noticing that, thanks for pointing it out. i forgot the log(ab)=log(a)+log(b) rule :)

    • 2 years ago
  36. myininaya
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    \[81=9^2 \] agree?

    • 2 years ago
  37. taranicolee
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    yes

    • 2 years ago
  38. myininaya
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    \[9^{x-5}=(9^2)^3\]

    • 2 years ago
  39. myininaya
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    so i just replaced 81 with 9 squared

    • 2 years ago
  40. taranicolee
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    ohh to get the bases the same so you can cancle out right?

    • 2 years ago
  41. myininaya
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    now law of exponents tells us \[(x^a)^b=x^{a \cdot b}\]

    • 2 years ago
  42. myininaya
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    if you get the same base on both sides then the exponents have to be equal in order for the equation to hold

    • 2 years ago
  43. myininaya
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    \[9^{x-5}=9^6 =>x-5=6\]

    • 2 years ago
  44. taranicolee
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    you multiplied the exponents on the right side of the equation right ? then cancled out the bases and now to solve you just add five to 6?

    • 2 years ago
  45. myininaya
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    do you recall the law of exponent i mentioned?

    • 2 years ago
  46. taranicolee
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    no :/

    • 2 years ago
  47. myininaya
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    \[(x^a)^b=x^{a \cdot b} \]?

    • 2 years ago
  48. taranicolee
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    nope. i wasn't here for the teaching of this

    • 2 years ago
  49. myininaya
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    yes if is it in the form you just multiply the exponents to simplify

    • 2 years ago
  50. taranicolee
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    *there

    • 2 years ago
  51. myininaya
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    so for example if we have \[(3^2)^3=3^{2 \cdot 3} =3^6 \]

    • 2 years ago
  52. myininaya
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    or if we have \[(43^4)^3=43^{4 \cdot 3} =43^{12}\]

    • 2 years ago
  53. taranicolee
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    oh sweet ! you're awesome. wanna move to italy and be my tutor?? hahah. if the equation were \[16^{x}=8\] not in that form, how would you solve?

    • 2 years ago
  54. myininaya
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    I would love to go to Italy. write both sides with same base both 16 and 8 can be written as 2 to the power of something

    • 2 years ago
  55. taranicolee
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    2^4 and 2^3

    • 2 years ago
  56. myininaya
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    yes! :)

    • 2 years ago
  57. myininaya
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    \[(2^4)^x=2^3\] so this is our equation

    • 2 years ago
  58. taranicolee
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    cancle bases, 4x=3 divide by 4 on both sides?

    • 2 years ago
  59. myininaya
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    perfect! :)

    • 2 years ago
  60. taranicolee
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    YAYYYY. thank you !!!!

    • 2 years ago
  61. myininaya
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    You're welcome.

    • 2 years ago
  62. myininaya
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    Thank you for being eager to learn this stuff.

    • 2 years ago
  63. taranicolee
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    ((:

    • 2 years ago
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