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taranicolee

  • 2 years ago

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

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

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

  3. taranicolee
    • 2 years ago
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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)

  4. JamesJ
    • 2 years ago
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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. :-)

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

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

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

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

  9. taranicolee
    • 2 years ago
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    yes

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

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

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

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

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

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

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

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

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

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

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

  22. taranicolee
    • 2 years ago
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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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

  37. taranicolee
    • 2 years ago
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    yes

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

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

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

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

  42. myininaya
    • 2 years ago
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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

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

  44. taranicolee
    • 2 years ago
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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?

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

  46. taranicolee
    • 2 years ago
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    no :/

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

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

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

  50. taranicolee
    • 2 years ago
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    *there

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

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

  53. taranicolee
    • 2 years ago
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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?

  54. myininaya
    • 2 years ago
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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

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

  56. myininaya
    • 2 years ago
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    yes! :)

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

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

  59. myininaya
    • 2 years ago
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    perfect! :)

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

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

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

  63. taranicolee
    • 2 years ago
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    ((:

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