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Help a newbie!

  1. Dec 13, 2007 #1
    [SOLVED] Help a newbie!

    Hey everyone. This will be my first post. A little background info: I have a bit of math anxiety due to doing very poorly with it in school despite a bit of effort. I had finished college in May with a degree that required only Statistics. My last year, I had enrolled in Calculus (I got to Calculus in Highschool, but didn't understand it, but made very little effort) but dropped out because of panic attacks. I saved my text book so I could learn on my own, as I am very disappointed in my horrible record and sentiment regarding mathematics. Anyway, I've been reading through, trying to teach myself, and i'm on page 114 (not yet Calculus). Everything has been smooth sailing, until I got stuck on some log questions.


    [​IMG]

    I don't understand how C and D was done, as there are no steps to follow. I've been staring at a few couple pages for quite a while and am very frustrated. On a side note, are there any textbooks that are fairly easy to comprehend for Calculus beginners? I'm not too fond of this book, but it's all I have. Thanks for the help in advance, I'm glad a forum like this exists.
     
  2. jcsd
  3. Dec 13, 2007 #2
    I realize this is too blurry. Let me transcribe it in text.

    C) e^xlog(base_e)b = e^log(base_e)b^x = b^x

    D) log(base_e)x/log(base_e)b = log(base_e)(b^log(base_b)x)/log(base_e)b = (log(base_b)x)(log(base_e)b)/log(base_e)b = log(base_b)x
     
  4. Dec 13, 2007 #3

    cristo

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    In (c) they use the rule that alogb=logb^a

    In (d) there is a trick used. They rewrite x as [tex]b^{\log_bx}[/tex]. They then use the same rule as used in (c).

    Welcome to the forums!
     
  5. Dec 13, 2007 #4
    In C, one of the fundamental property of logarithm is used. It is, x log b = log b^x.

    Now, you have e^log b^x, which IS equal to b^x. Let e^(log b^x) be some number m. You'll get an equation in exponential form. Convert it into logarithmic form. What do you get?

    Regards,
    Sleek.
     
  6. Dec 13, 2007 #5
    Man, i'm not too good at this.. I really appreciate the quick responses, though. I understand the fundmental rule of moving the x to the exponent. However, I'm wondering why b^x suddenly equals e^log(base_e)b^x..
     
  7. Dec 13, 2007 #6

    cristo

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    Ok, well we have [tex]e^{x\log_eb}[/tex]. So, perform your usual identity on the exponent to give [tex]e^{\log_eb^x}[/tex] and letting b^x=y we have [tex]e^{\log_e y}[/tex]. But now recall that e^z and log_e(z) are inverse functions, and a function composed with its inverse is equal to the identity function, so our expression is equal to y=b^x.
     
  8. Dec 13, 2007 #7
    Oh! Haha, thank you so much. It makes sense now. Thanks a ton guys.
     
  9. Dec 13, 2007 #8

    cristo

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    You're welcome! :smile:
     
  10. Dec 13, 2007 #9
    Gotcha. I'm just so slow with this stuff. I feel like, because I can't get it intuitively, I'm going to be for a rude awakening when I get deeper into the book. For example, I can read that "this is just another way to write this" but I have trouble comprehending the mathematical logic behind it. Ah well, no worries.
     
  11. Dec 13, 2007 #10

    Gib Z

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    In case you haven't already, I would recommend you try prove [tex] x \log_b a = \log_b a^x[/tex]. It will really help you understand why some of these steps can be done and if you know how to prove it, theres almost a zero chance you'll forget that rule, or confuse it with something that looks similar.

    If you have any problems, we'll be right here =]
     
  12. Dec 13, 2007 #11

    cristo

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    Don't worry about it. Anything can seem a little tricky to begin with, especially if you don't spot what's happening. But now, since you know, next time it crops up you'll be able to spot it. There's nothing wrong with not getting something the first time, so long as you learn, and let every experience with a question build up your repertoire. The homework forums here are a very useful place to pick up such tricks. I hope you'll stick around!
     
  13. Dec 13, 2007 #12
    Gib z: I assume because the inverse of exponents is multiplication.

    Cristo: Thanks for the vote of confidence. I'm very used to math being a frightening topic, but I actually feel pretty good sitting down and getting answers correctly for once! Haha. I'm definitely going to stick around - lots to learn here.
     
  14. Dec 13, 2007 #13

    Gib Z

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    The inverse of exponents are logarithms, not multiplication :( Multiplication is the inverse of division, (well, except for 0 but don't worry about that).

    As cristo said in post 6, if a function is composed with its inverse, which basically just means you apply the function, then apply the inverse, you get the identity function, which means just the same number you did the thing to. For example, my original function could be f(x) = 3x, which is just multiplication by 3. The inverse would then be division by 3, or g(x)=x/3.

    So if we apply one, and then the other - f( g(x) ) = 3 (x/3) = x , just as expected.
     
  15. Dec 13, 2007 #14
    Ack, of course. That was dumb of me. Oh, okay, I think I understand now. The question itself is composed of it's inverse. I think I'm gunna re-review this just to get a better grasp on how logs transcribe to exponents.
     
  16. Dec 13, 2007 #15
    Can anyone please tell me the term for the slope that runs parallel with an axis x or y yet never touches the line. Thanks
     
  17. Dec 13, 2007 #16

    Gib Z

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    Can you express your question more clearly? Are you looking for the equation of the line, or the gradient of the line?
     
  18. Dec 13, 2007 #17

    Dick

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    A-S-Y-M-P-T-O-T-E. Don't post your own question on other peoples threads.
     
  19. Dec 13, 2007 #18

    Gib Z

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    damn my Brain must be deflated for me not to have got that :(
     
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