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Homework Help: Isolating a variable in an exponent

  1. Feb 24, 2013 #1
    1. The problem statement, all variables and given/known data
    So I had to find out
    Where No=1.5,e is the elementary charge, u=-0.068 and x=0.07
    I came to the answer 1.221

    Now I need to find x if u =-0.036, N=1.221 and No remains 1.5

    2. Relevant equations

    3. The attempt at a solution
    I reduce the equation to the following


    The problem is that when I plug my numbers from the original problem into this I end up getting the following

    -3.02= x

    This is clearly incorrect since I know x should be 0.07.
    Any help would be greatly appreciated.
    Thank you.
    Last edited: Feb 25, 2013
  2. jcsd
  3. Feb 25, 2013 #2


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    I get something different.

    I'll assume u = -0.036 is a typo because it looks as though you want to use u = -0.068 again.

    Just check your answer for N again, and see if that changes anything.
  4. Feb 25, 2013 #3
    My order of operations was wrong then?

    So I get 1.273

    Now -0.036 was NOT a typo, however I am ignoring it, as I am using the original numbers to try to figure out how to isolate x. However, when I plug this new number in, I still end up with x = -2.413 instead of 0.07 which it should be.
    Clearly I am doing something very wrong and have no idea what.
  5. Feb 25, 2013 #4
    Let me illustrate step by step.


    Now I try to reverse it to figure out how to isolate x, this is where I do something very wrong I believe.


    But we know x=0.07. I'm lost.
  6. Feb 25, 2013 #5


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    Sorry, I mistook e as being the natural exponential 2.718...
    We're supposed to be using [itex]e=1.602\cdot10^{-19}[/itex], right?

    Using the figures:
    [tex]N_0 = 1.5[/tex]
    [tex]e=1.602\cdot 10^{-19}[/tex]

    You should get [itex]N\approx 1.22[/itex]
  7. Feb 25, 2013 #6


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    How did you go from the first to the second step?
  8. Feb 25, 2013 #7
    I got it.
  9. Feb 25, 2013 #8
    Haha Ok so I was correct in the first place when I posted 1.221. Then I was trying to figure out what I did wrong which would explain your confusion over my next process.
  10. Feb 25, 2013 #9


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    Yes, it's not correct.

    [tex](n\cdot a^b)^{cd} = n^{cd}\cdot a^{bcd}[/tex]

    But you've done something else.
    Of course, calculating them in this way is the long and inefficient way, so you'd instead enter it as so:


    EDIT: most of this is redundant, you already know it :smile:
    Last edited: Feb 25, 2013
  11. Feb 25, 2013 #10


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    Oh yeah sorry, that's my fault!

    As for the second part of the problem where you're trying to find x, you've done essentially what confused me as well.


    Remember that ln = loge is the natural logarithm, or in other words, it reverses the natural exponential operator.

    If [itex]a=e^{b}[/itex] then [itex]\ln(a)=b[/itex] but we don't have the natural exponential (e=2.718...) here, we have another base altogether, so what we'd need to do to reverse the exponent operation is to take the log with base [itex]1.6\cdot 10^{-19}[/itex]

    But your calculator probably doesn't have the option of choosing any base for the logarithm, so you instead use the formula


    So we have that

    [tex]\log_{1.6\cdot 10^{-19}}(1.221/1.5) = \frac{\ln (1.221/1.5)}{\ln(1.6\cdot 10^{-19})}[/tex]
  12. Feb 25, 2013 #11
    Let me try then.
  13. Feb 25, 2013 #12
    Are you saying that
  14. Feb 25, 2013 #13


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  15. Feb 25, 2013 #14
    Yeah, I got that. So using -0.036, I come up with 0.132. Excellent.
  16. Feb 25, 2013 #15
    Thank you very much.
  17. Feb 25, 2013 #16


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    You're welcome! And sorry about the confusion at the start.
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