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anonymous

  • 4 years ago

log base 2 fifth root of 32 ?!?!?!?!?!??!?!?!?!!?!??! PLEASE HELP

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  1. Xishem
    • 4 years ago
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    \[\log_{2} \sqrt[5]{32}=\log_{2}2=1\]

  2. anonymous
    • 4 years ago
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    what?

  3. anonymous
    • 4 years ago
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    how did you get log base 2 then 2?

  4. anonymous
    • 4 years ago
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    the fifth root of 32 is 2, because \[2^5=32\]

  5. anonymous
    • 4 years ago
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    then shouldn't the answer be five ?

  6. anonymous
    • 4 years ago
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    so your job is to find \[\log_2(2)\] which asks the question, "what power do you raise the number "2" to to get an answer of 2?" and the answer is clearly 1

  7. anonymous
    • 4 years ago
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    \[\log_2(2^5)=5\] yes

  8. anonymous
    • 4 years ago
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    but you have \[\log_2(2)\]

  9. anonymous
    • 4 years ago
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    wait..what?!

  10. anonymous
    • 4 years ago
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    \[2^{y} = 2^{5} \]

  11. anonymous
    • 4 years ago
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    ok lets go slow

  12. anonymous
    • 4 years ago
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    if your question was \[\log_2(2^5)\] the answer would certainly be 5 correct?

  13. anonymous
    • 4 years ago
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    i.e. \[\log_2(32)=5\]

  14. anonymous
    • 4 years ago
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    wait one sec let me soak this in

  15. anonymous
    • 4 years ago
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    ok

  16. anonymous
    • 4 years ago
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    so you're telling me that 2^5 does not equal fifth root of 32?

  17. anonymous
    • 4 years ago
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    yes i am. 2^5 = 32 not the fifth root of 32

  18. anonymous
    • 4 years ago
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    so what is it? and how would u find what it is ?

  19. anonymous
    • 4 years ago
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    i want to know the fifth rote of 32

  20. anonymous
    • 4 years ago
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    but since 2^5 = 32 then we know that \[\sqrt[5]{32}=2\]

  21. anonymous
    • 4 years ago
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    it is 32^ 1/5

  22. anonymous
    • 4 years ago
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    you can think of it that way, yes

  23. anonymous
    • 4 years ago
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    im not understanding what you're telling me

  24. anonymous
    • 4 years ago
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    which means \[(32)^{\frac{1}{5}}=(2^5)^{\frac{1}{5}}=2^1=2\]

  25. anonymous
    • 4 years ago
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    in english: two the the fifth power is 32, therefore the fifth root of 32 is 2

  26. anonymous
    • 4 years ago
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    i think i understand that. thank you very much. can you help me with a few more?

  27. anonymous
    • 4 years ago
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    post i will reply

  28. anonymous
    • 4 years ago
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    \[3 \log _{7} \sqrt[6]{49}\]

  29. anonymous
    • 4 years ago
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    ok lets do this the easy way. first of all \[\log(x^n)=n\log(x)\]

  30. anonymous
    • 4 years ago
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    yep. i understand tha : - )

  31. anonymous
    • 4 years ago
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    and we can rewrite \[\sqrt[6]{49}\] as \[49^{\frac{1}{6}}\] yes?

  32. anonymous
    • 4 years ago
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    ummm im kind of confused on those

  33. anonymous
    • 4 years ago
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    is that just a general rule ?

  34. anonymous
    • 4 years ago
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    you need to understand exponential notation for logs, because logs are exponents.

  35. anonymous
    • 4 years ago
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    yes for example \[\sqrt{x}=x^{\frac{1}{2}}\] \[\sqrt[3]{x}=x^{\frac{1}{3}}\] etc

  36. anonymous
    • 4 years ago
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    and in general \[\sqrt[n]{x}=x^{\frac{1}{n}}\]

  37. anonymous
    • 4 years ago
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    okay awesome. just wrote that on my rules sheet XD

  38. anonymous
    • 4 years ago
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    so we start with \[3\log_7(49^{\frac{1}{6}})\] and then bring the 1/6 right out front as a multiplier (coefficient)

  39. anonymous
    • 4 years ago
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    \[\frac{1}{6}\times 3\log_7(49)\] \[\frac{1}{2}\log_7(49)\] and idea what to do next?

  40. anonymous
    • 4 years ago
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    i really have no idea :(

  41. anonymous
    • 4 years ago
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    how can you write 49 with an exponent?

  42. anonymous
    • 4 years ago
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    what would u do if you have a number in front of the log? like the 1/2

  43. anonymous
    • 4 years ago
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    hold on, lets just take care of the 49 first. we can do this problem a different way second

  44. anonymous
    • 4 years ago
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    we would write \[49=7^2\]

  45. anonymous
    • 4 years ago
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    and then you have \[\frac{1}{2}\log_77^2)\] which should be easy now

  46. anonymous
    • 4 years ago
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    how would we incorporate the 1/2?

  47. anonymous
    • 4 years ago
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    since \[\log_b(b^n)=n\]

  48. anonymous
    • 4 years ago
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    but its 1/2 log

  49. anonymous
    • 4 years ago
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    we leave it for a moment. the one half out front is just a number, let it sit there

  50. anonymous
    • 4 years ago
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    so the answer is 2

  51. anonymous
    • 4 years ago
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    we get \[\frac{1}{2}\log_7(7^2)=\frac{1}{2}\times 2\]

  52. anonymous
    • 4 years ago
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    the answer to \[\log_7(7^2)=2\] and then \[\frac{1}{2}\times 2=1\]

  53. anonymous
    • 4 years ago
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    oh okay!!!!! wow i get that!

  54. anonymous
    • 4 years ago
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    if you like we can do this problem a different way, but maybe one way is enough, you tell me

  55. anonymous
    • 4 years ago
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    answer is 1

  56. anonymous
    • 4 years ago
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    yes it is 1

  57. anonymous
    • 4 years ago
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    i like this way. do u mind helping me with more?

  58. anonymous
    • 4 years ago
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    go ahead

  59. anonymous
    • 4 years ago
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    \[50 \log 5 \sqrt{125} \]

  60. anonymous
    • 4 years ago
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    so we would have

  61. anonymous
    • 4 years ago
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    5 ^ x = square route 125 ^ 50?

  62. anonymous
    • 4 years ago
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    oh yikes leave the 50 out front!

  63. anonymous
    • 4 years ago
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    i mean you are right, but leave it there to make your life easier. think of how to write 125 as a number raised to a power

  64. anonymous
    • 4 years ago
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    okay. lol so i have 50 log base 5^x = 5^ 3

  65. anonymous
    • 4 years ago
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    i believe that is wrong

  66. anonymous
    • 4 years ago
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    close, you just forget the square root

  67. anonymous
    • 4 years ago
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    but i don't have the square route anymore i got it al out of the square rote

  68. anonymous
    • 4 years ago
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    \[125=5^3\] and so \[\sqrt{125}=5^{\frac{3}{2}}\]

  69. anonymous
    • 4 years ago
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    i dont understand that

  70. anonymous
    • 4 years ago
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    just a moment ago we found out that \[\sqrt{x}=x^{\frac{1}{2}}\] right?

  71. anonymous
    • 4 years ago
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    \[\sqrt{125}=125^{\frac{1}{2}}=(5^3)^{\frac{1}{2}}=5^{\frac{3}{2}}\]

  72. anonymous
    • 4 years ago
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    yes. oh okay i understand

  73. anonymous
    • 4 years ago
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    so then what would we have

  74. anonymous
    • 4 years ago
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    so now we have \[50\log_5(5^{\frac{3}{2}})\] \[50\times \frac{3}{2}\] etc

  75. anonymous
    • 4 years ago
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    as soon as you say \[\log_b(\text{whatever})\] you should think of ways to write "whatever" as a power of b

  76. anonymous
    • 4 years ago
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    *see

  77. anonymous
    • 4 years ago
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    oh okay!!! wow!

  78. anonymous
    • 4 years ago
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    can u give me an example to try?

  79. anonymous
    • 4 years ago
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    \[\log_3(\sqrt[5]{27})\]

  80. anonymous
    • 4 years ago
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    can u do one with a number in front of the log

  81. anonymous
    • 4 years ago
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    the number out front does not really make any difference, but if you like i can write \[10\log_3(\sqrt[5]{27})\]

  82. anonymous
    • 4 years ago
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    okay let me get my paper lol

  83. anonymous
    • 4 years ago
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    6?

  84. anonymous
    • 4 years ago
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    yup!

  85. anonymous
    • 4 years ago
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    OMG

  86. anonymous
    • 4 years ago
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    really? i got it right?

  87. anonymous
    • 4 years ago
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    don't be so shocked, you will get the hang of it

  88. anonymous
    • 4 years ago
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    can u help me with more? lol

  89. anonymous
    • 4 years ago
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    after a couple practice problems you will think "not much to this"

  90. anonymous
    • 4 years ago
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    ok we can do another before i go

  91. anonymous
    • 4 years ago
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    \[\log _{3} 4 - 1/2 \log _{3} (6x-5)\]

  92. anonymous
    • 4 years ago
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    holy moly what are the instruction, write as a single log?

  93. anonymous
    • 4 years ago
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    lol i have no idea what to do at all

  94. anonymous
    • 4 years ago
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    first you need to make sure what the instructions are. what does it say before the problem?

  95. anonymous
    • 4 years ago
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    all it says is : solve

  96. anonymous
    • 4 years ago
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    maybe u an help me with a different one instead?

  97. anonymous
    • 4 years ago
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    you cannot "solve' because there is not an equation here. maybe it says "write as one log"

  98. anonymous
    • 4 years ago
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    which we can do if you like

  99. anonymous
    • 4 years ago
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    sure lets write it as one log

  100. anonymous
    • 4 years ago
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    by what we know before \[n\log(x)=\log(x^n)\] so \[\frac{1}{2}\log(6x-5)=\log((6x-5)^{\frac{1}{2}})=\log(\sqrt{6x-5})\]

  101. anonymous
    • 4 years ago
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    also \[\log(a)-\log(b)=\log(\frac{a}{b})\] so you can write the whole thing as \[\log_3(\frac{4}{\sqrt{6x-5}})\]

  102. anonymous
    • 4 years ago
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    that is the answer?

  103. anonymous
    • 4 years ago
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    yes, although the text did not seem to provide a question

  104. anonymous
    • 4 years ago
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    okay. thank you so so so much for all your help! maybe i will pass math because of you! :D thanks again and have a wonderful day!!

  105. anonymous
    • 4 years ago
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    you too, good luck

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