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GOODMAN

  • 4 years ago

Write in exponential form. (Will write in equation editor)

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  1. GOODMAN
    • 4 years ago
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    \[5^{\log5}^{25}\]

  2. GOODMAN
    • 4 years ago
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    wait..that came out funny..

  3. GOODMAN
    • 4 years ago
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    \[5^{\log5}\]

  4. GOODMAN
    • 4 years ago
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    There is suppoesd to be a 25 next to log 5

  5. UnkleRhaukus
    • 4 years ago
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    \[ 5^{log{5}^{25}}\] ?

  6. TuringTest
    • 4 years ago
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    \[\huge 5^{(\log5)^{25}}\]?

  7. UnkleRhaukus
    • 4 years ago
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    is 5 the base of the logarithm ?

  8. GOODMAN
    • 4 years ago
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    Turing Test wrote it correctly.

  9. GOODMAN
    • 4 years ago
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    I know how to do it, just am confused about the 5 in the front. what do i do with it?

  10. UnkleRhaukus
    • 4 years ago
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    the the exponent of the log can come out the front \[5^{log(5)^{25}}=5^{25(log(5)}\]

  11. GOODMAN
    • 4 years ago
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    Thats it? I am still confused.

  12. TuringTest
    • 4 years ago
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    \[5^{25\log5}\neq5^{25}\times5^{\log5}\]

  13. UnkleRhaukus
    • 4 years ago
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    I have made an error turing test?

  14. TuringTest
    • 4 years ago
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    \[\log_5(25)=2\]you are tired

  15. GOODMAN
    • 4 years ago
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    I am reading this off of my math book. 5 is the big dog, while \[\log _{5}25\] is above it.

  16. UnkleRhaukus
    • 4 years ago
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    oh yeah i see how wrong that is now

  17. TuringTest
    • 4 years ago
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    well now that we've got it sorted out the problem is trivial goodman please try to post more clearly, that is not what you originally wrote

  18. UnkleRhaukus
    • 4 years ago
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    errors ervery where, perhaps i should just watch and learn

  19. GOODMAN
    • 4 years ago
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    How did you get that? @TuringTest

  20. TuringTest
    • 4 years ago
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    \[\huge 5^{\log_5(25)}=5^2=25\]because\[\huge 5^2=25\]

  21. GOODMAN
    • 4 years ago
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    I appologize, srry, i did read it wrong. srry srry.

  22. GOODMAN
    • 4 years ago
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    I am new to log, so yea, srry.

  23. TuringTest
    • 4 years ago
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    but in general\[\huge a^{\log_a(x)}=x\]so we could have skipped that analaysis

  24. GOODMAN
    • 4 years ago
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    Oh yea, because it always is equal to the front number. Yea, okay, got it :D Thanx, srry for the mix up :/

  25. TuringTest
    • 4 years ago
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    it's ok, let me know if you have more questions

  26. GOODMAN
    • 4 years ago
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    You are familiar with log? Because dont understand em at all :P

  27. TuringTest
    • 4 years ago
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    yes, they can be a bit tricky at first think about small numbers first the definition as I like to give it:\[\huge\log_ax=b\text{ if }a^b=x\]in other words\[\large \log_ax\]asks "what power do we raise a to in order to get x?" for example...

  28. TuringTest
    • 4 years ago
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    \[\large\log_2(4)\]asks "2 raised to what power equals 4?" so what is the answer?

  29. GOODMAN
    • 4 years ago
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    2 rite?

  30. TuringTest
    • 4 years ago
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    right because 2^2=4 so what about\[\log_2(16)\]?

  31. GOODMAN
    • 4 years ago
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    4

  32. TuringTest
    • 4 years ago
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    right so now you should see why\[\log_5(25)=2\]yes?

  33. GOODMAN
    • 4 years ago
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    YES!! wow!! It is 5 because 5^5 is equal to 25

  34. TuringTest
    • 4 years ago
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    \[5^2\]

  35. GOODMAN
    • 4 years ago
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    Whoops..srry, so the exponent is the answer?

  36. TuringTest
    • 4 years ago
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    yeah\[\log_2(16)=4\]as you said, because\[2^4=16\]

  37. GOODMAN
    • 4 years ago
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    Is that always the case?

  38. TuringTest
    • 4 years ago
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    so\[\log_5(25)=2\]because\[5^2=25\]

  39. TuringTest
    • 4 years ago
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    in general we have\[\huge\log_a(x)=y\iff a^y=x\]so yes, that means always you could say that the question is on the left: "a to what power equals x?" the answer is on the right "a to the y equals x"

  40. TuringTest
    • 4 years ago
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    so do some more what is\[\log_4(16)\]?

  41. GOODMAN
    • 4 years ago
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    2

  42. TuringTest
    • 4 years ago
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    right\[\log_{10}(10000)\] ?

  43. GOODMAN
    • 4 years ago
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    umm..lol, 1000

  44. GOODMAN
    • 4 years ago
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    I think i got it wrong

  45. TuringTest
    • 4 years ago
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    the question is "10 to what power equals 10000" you are asserting that\[\large10^{1000}=10000\]which I dont think you believe what is\[10^2\] ?

  46. GOODMAN
    • 4 years ago
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    Wait..no, its 100

  47. TuringTest
    • 4 years ago
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    \[10^{100}\neq10000\]

  48. GOODMAN
    • 4 years ago
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    So what is it?

  49. GOODMAN
    • 4 years ago
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    I dont have a calculator rite now :P

  50. TuringTest
    • 4 years ago
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    \[10^2=10\cdot10=100\]\[10^3=10\cdot10\cdot10=1000\]the exponent on the ten is the number of zeros after the one so 10 to what power equals 10000 ?

  51. GOODMAN
    • 4 years ago
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    \[10^{4}?\]

  52. TuringTest
    • 4 years ago
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    exactly so\[\log_{10}(10000)=4\]because\[10^4=10000\]

  53. GOODMAN
    • 4 years ago
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    That makes its it so simple

  54. TuringTest
    • 4 years ago
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    yeah that's a handy rule, and is used in science to describe very large and small numbers last one: how about\[\log_3(81)\]

  55. GOODMAN
    • 4 years ago
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    4

  56. TuringTest
    • 4 years ago
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    right :D do you know the basic log rules\[\log(ab)=\log a+\log b\]\[\log(\frac ab)=\log a-\log b\]\[\log(a^b)=b\log a\]?

  57. GOODMAN
    • 4 years ago
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    No, we havent been taught those. We started log in class today.

  58. TuringTest
    • 4 years ago
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    when you start using that things will be easier \[\large \log(ab)=\log a+\log b\text{ because }x^a\cdot x^b=x^{a+b}\]for example it takes a bit to get used to those ideas though

  59. GOODMAN
    • 4 years ago
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    Yea, just by reading it, I understand the first function.

  60. GOODMAN
    • 4 years ago
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    I also understand the second one, the third one looks funny.

  61. TuringTest
    • 4 years ago
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    the third one can be illustrated by the problems we just did first I will ask you what is\[\log_22\]?

  62. GOODMAN
    • 4 years ago
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    1

  63. GOODMAN
    • 4 years ago
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    lol, i find myself saying "2 raised to what power is 2" thats a neat trick

  64. TuringTest
    • 4 years ago
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    I'm glad, it helps me! and in general\[\log_aa=1\]now look at\[\log_2(16)\]if we factor 16 we get\[\log_2(2^4)\]now apply the third rule\[4\log_2(2)=4(1)=4\]so we get the answer we already knew it also shows that\[\log_aa^x=x\]which is nice to know that's how I did your earlier problem) so by factoring and using these rules we can break down lots of numbers with logs

  65. TuringTest
    • 4 years ago
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    now here's the last grand example\[\log_2(24)=\log_2(2^3\cdot3)\]using the first rule\[\log_2(2^3)+\log_23\]and now the third\[3\log_2(2)+\log_23=3+\log_23\]and that's as far as it goes

  66. GOODMAN
    • 4 years ago
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    Wow, i never knew log could be so simple :D

  67. TuringTest
    • 4 years ago
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    I'm glad you feel that way :D

  68. GOODMAN
    • 4 years ago
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    Thanx a ton :D I am now cleared up on log, thank you thank you!!!

  69. TuringTest
    • 4 years ago
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    happy to help!

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