1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Natural Logs, adding variable exponents.

  1. Jun 1, 2012 #1
    1. The problem statement, all variables and given/known data

    e^2x * e^2

    2. Relevant equations



    3. The attempt at a solution

    Really, I'm just a little bit confused on how I would go about adding the exponent. None of the examples in the book cover that. Is it simply e^2x+2?
     
  2. jcsd
  3. Jun 1, 2012 #2

    Mark44

    Staff: Mentor

    You really need more parentheses if you write this as plain text. I'm assuming that you mean
    e^(2x) * e^2

    You can also write exponents (click Go Advanced to open the advanced menu, and then click the X2 button).

    Like this: e2x * e2
    But there are theorems that do, I'm sure.

    am * an = am + n

    What you wrote is e2x + 2. What I think you meant was e2x + 2, which is correct.

    If you write this as inline text, you HAVE TO USE PARENTHESES!
    I.e., e^(2x + 2)
     
  4. Jun 1, 2012 #3
    Lol, I think someone is getting missing parentheses fatigue.
     
  5. Jun 1, 2012 #4

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    It is really, really, really annoying :biggrin:
    I think you can easily distinguish a person who knows math from a person who doesn't know math by seeing how they handle parantheses.
     
  6. Jun 1, 2012 #5
    What would happen if I did something like this here: lne^2x * lne^2? Would I be butchering math?
     
  7. Jun 1, 2012 #6
    Not exactly butchering. But it is still unclear :tongue2: It could mean x*ln(e^2)*ln(e^2)...

    ln(e^(2x))*ln(e^2) is more appropriate, I think.

    since you have options like X2 in the advanced reply text boxes, and LaTeX, I don't see why you should write such math equations in the first place!
     
  8. Jun 1, 2012 #7
    I mean: lne^(2x) * lne^(2). That way I'd get 4x as opposed to e^(2x+2)

    Yeah, I'll check it out.
     
  9. Jun 1, 2012 #8
    See? These are the misunderstanding problems :wink:
     
  10. Jun 1, 2012 #9
    So,is lne^(2x) * lne^(2)= 4x acceptable just like e^(2x+2) is? Because, I am thinking I could do lne^(2x+2)=2x+2 and don't see how 2x+2 equals 4x.
     
  11. Jun 1, 2012 #10

    Mark44

    Staff: Mentor

    I'm not sure what you're asking here. The original problem was e2x * e2, which is equal to e2x + 2, and that's not much more you can do with it. You can't just come in an take the log of each factor in a product. Since 2x + 2 is not identically equal to 4x, it should be clear that ln(e^(2x)) * ln(e^2) ≠ ln(e^(2x + 2)).

    Am I missing what you're asking?
     
  12. Jun 2, 2012 #11
    I think I have gone through something like this before. Once I had to simplify:

    (x^2-y^2)^(1/2)* (x-y)^(3/2)* (x+y)^(-1/2).

    What I did was:

    (x^2-y^2)^(1/2*2)* (x-y)^(3/2*2)* (x+y)^(-1/2*2)

    I just couldn't understand why my "simplification" didn't produce the right answer till I re-framed what I did in this manner:

    [(x^2-y^2)^(1/2)* (x-y)^(3/2)* (x+y)^(-1/2)]^2

    Only then I could see I was just squaring the whole expression. It was like squaring x^(2) and not quiet getting why x^(2) doesn't equal x^(4).

    Does what I wrote above sound something like me coming and taking the log of each factor in a product in e^(2x)* e^(2) ?
     
  13. Jun 2, 2012 #12

    Mark44

    Staff: Mentor

    Yes, it does. Here's where things break down.

    e2x * e2 = e2x + 2

    You can take the ln of both sides to get:
    ln(e2x * e2) = ln(e2x + 2)

    You cannot rewrite the left side as ln(e2x) * ln(e2). There is no property of logs that says that log(AB) = logA * logB

    There is a property that says that log(AB) = logA + logB, so we can rewrite the equation above as
    ln(e2x) + ln(e2) = 2x + 2

    Simplifying the left side gives
    2x + 2 = 2x + 2, which is identically true.
     
  14. Jun 3, 2012 #13
    Nice breakdown.

    My brain has a tendency to see equations where there aren't any. It's learning, though.

    Thanks.
     
  15. Jun 3, 2012 #14

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Note: [tex] X \log(B) = \log(B^X) \text{ if } B > 0.[/tex] Apply this to [itex] X = \log(A)[/itex] to get
    [tex] \log(A) \cdot \log(B) = \log(B^{\log(A)}).[/tex] This also equals
    [tex] \log(A^{\log(B)}),[/tex]
    so, apparently we have
    [tex] A^{\log(B)} = B^{\log(A)}[/tex] for all A, B > 0. Hmmm...I've never seen that before, but it does check out.

    RGV
     
  16. Jun 4, 2012 #15
    Looks very interesting. Thanks.
     
  17. Jun 4, 2012 #16
    Sorry guys. I'm really not used to writing equations in text format. If I wanted to show that the 2x+2 was a variable on paper I would just make it higher and small. ;P
     
  18. Jun 5, 2012 #17

    Mark44

    Staff: Mentor

    I assume that by "variable" you really mean "exponent."
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Natural Logs, adding variable exponents.
  1. Variable in Exponents (Replies: 2)

  2. Natural log (Replies: 1)

  3. Natural logs (Replies: 5)

  4. Adding exponents (Replies: 4)

Loading...