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anonymous

  • 5 years ago

Does anyone knows a really good book where I can learn about logarithms? I want a lot of theory and a million exercises.

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  1. anonymous
    • 5 years ago
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    logarithms arent fun

  2. anonymous
    • 5 years ago
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    I think the theory behind logarithms would lie in number theory

  3. anonymous
    • 5 years ago
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    also there is literally only like 4 things you need to any logarithm question

  4. anonymous
    • 5 years ago
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    you dont need "a million exercises"

  5. anonymous
    • 5 years ago
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    "logarithms arent fun" is that a book or is that what you think?

  6. anonymous
    • 5 years ago
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    yeh its a book :p

  7. anonymous
    • 5 years ago
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    Im trying to solve a lot of equations, inequalities, doing graphs, etc

  8. anonymous
    • 5 years ago
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    im out trying to promote my online tutoring....i tutor math and physics from low - hgh level....if u ever need it check me out http://mathphystut.tripod.com

  9. anonymous
    • 5 years ago
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    I know the theory behind logarithm in general, but i want something really challenging

  10. anonymous
    • 5 years ago
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    Solve \[2\ln(\sqrt{x}) - \ln(x) =1 \]

  11. anonymous
    • 5 years ago
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    an example

  12. anonymous
    • 5 years ago
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    but its all relatively standard

  13. anonymous
    • 5 years ago
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    You want challenging problems, or you want to clear your concept?

  14. anonymous
    • 5 years ago
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    want challenging problems

  15. anonymous
    • 5 years ago
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    yeh, well logarithms are that challenging to be honest

  16. anonymous
    • 5 years ago
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    arent*

  17. anonymous
    • 5 years ago
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    You can try one thing. You email me, and I will send you challenging problems

  18. anonymous
    • 5 years ago
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    As many as you want

  19. anonymous
    • 5 years ago
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    sure. can you give me your e-mail?

  20. anonymous
    • 5 years ago
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    About the problem above, elecengineer, I got 0=1. Is that right?

  21. anonymous
    • 5 years ago
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    Thanks "pi" I just sent you an email. Can you check it out

  22. anonymous
    • 5 years ago
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    It must take a little bit. Im from Ecuador and I suppose your in Europe lol

  23. anonymous
    • 5 years ago
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    I haven't yet received your email........... Still waiting

  24. anonymous
    • 5 years ago
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    Can you check it out again. Can take a little bit

  25. anonymous
    • 5 years ago
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    differentiate \[y= \ln [\frac{ (x-3)^4 \sqrt{x} }{x+1} ]\]

  26. anonymous
    • 5 years ago
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    thats prob the hardest question on logarithms every, but its still really easy

  27. anonymous
    • 5 years ago
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    Okay code is VILLAMAGUA CONZA LUIS MIGUEL VILLAMAGUA CONZA LUIS MIGUEL

  28. anonymous
    • 5 years ago
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    yean. I got 4LN(x-3)+1/2(LN(x))-LN(X+) cannot remember how to differenciate logs lol

  29. anonymous
    • 5 years ago
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    just joking

  30. anonymous
    • 5 years ago
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    yeah it is me PI, thanks for your help

  31. anonymous
    • 5 years ago
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    I have sent you a problem, but remember, you are not supposed to use a calculator

  32. anonymous
    • 5 years ago
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    lols, what was it

  33. anonymous
    • 5 years ago
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    By the way, are you a staff of UTPL ?

  34. anonymous
    • 5 years ago
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    what was the question!

  35. anonymous
    • 5 years ago
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    yeah. From UTPL

  36. anonymous
    • 5 years ago
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    You teach there?

  37. anonymous
    • 5 years ago
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    No teaching, but a student. Well finishing my major

  38. anonymous
    • 5 years ago
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    solve simultaneously \[5^{x+y} = \frac{1}{5}\ and \[5^{3x+2y} =1\]

  39. anonymous
    • 5 years ago
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    \[5^{x+y} = \frac{1}{5} , 5^{3x+2y} =1 \]

  40. anonymous
    • 5 years ago
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    I just replied you first question PI. Is that all right?

  41. anonymous
    • 5 years ago
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    there fairly standard lol all of logarithms are fairly standard. I have no idea why people find them hard , all it is is 4formulas to remember

  42. anonymous
    • 5 years ago
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    I dont know much about inequalities with logs.

  43. anonymous
    • 5 years ago
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    yeh , they come up a tiny bit in some financial maths topics

  44. anonymous
    • 5 years ago
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    inequalities and logs dont come up alot there is however one thing which you do need to look out for when solving inequalities with logs

  45. anonymous
    • 5 years ago
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    best seen by an example

  46. anonymous
    • 5 years ago
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    When I was in high school logarithms seems to be so hard. Now I tried them again and they look so easy. Im confused. I thought I would never be done with logs

  47. anonymous
    • 5 years ago
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    solve (1/3)^n > 0.5

  48. anonymous
    • 5 years ago
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    solve that, and see what happens

  49. anonymous
    • 5 years ago
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    ok

  50. anonymous
    • 5 years ago
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    I will put money on you being wrong :P

  51. anonymous
    • 5 years ago
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    it looks fairly simple, but there is a trick in it that would catch alot of people

  52. anonymous
    • 5 years ago
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    please dont go. Im a little bit slow. I like this :)

  53. anonymous
    • 5 years ago
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    well ... have you done the question yet inequalities with logs behave the exact same as equations

  54. amistre64
    • 5 years ago
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    the bigger n is.... the .......

  55. anonymous
    • 5 years ago
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    This is what i got

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  56. anonymous
    • 5 years ago
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    long way to go about things also you would need to use change of basic at the end if you actually wanted to get a number for it

  57. anonymous
    • 5 years ago
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    What is the other wat? Im excited lol

  58. anonymous
    • 5 years ago
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    what is the other way? sorry

  59. anonymous
    • 5 years ago
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    well the way I was taught was to take logarithms of both sides ( it doesnt matter what base, as long as you use the same on both sides ) so I will take ln of both sides so \[\ln [( \frac{1}{3})^n] = \ln(0.5) \]

  60. anonymous
    • 5 years ago
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    now \[\log_{a} x^r = r \log_{a} x \]

  61. anonymous
    • 5 years ago
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    yeah. I know where you go

  62. anonymous
    • 5 years ago
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    your bright man

  63. anonymous
    • 5 years ago
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    so then you get n ln(1/3) >0.5

  64. anonymous
    • 5 years ago
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    *so then you get n ln(1/3) > ln(0.5)

  65. anonymous
    • 5 years ago
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    \[n > \frac{\ln(0.5)}{\ln(\frac{1}{3} ) }\]

  66. anonymous
    • 5 years ago
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    there the problem people make!

  67. anonymous
    • 5 years ago
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    mm, I divide both sides by ln(1/3) , but am I allowed to do that? well , ln(1/3) is actually a negative number , so I must flip the inequality sign

  68. anonymous
    • 5 years ago
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    ln(a) is negative for all a in the interval 0<a<1

  69. anonymous
    • 5 years ago
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    Yeah I would have fallen there too. Can you email me some of this type of exercices.

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