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!

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
    N=Noe-ux
    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

    ln(N/No)=-ux

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

    ln(1.221/1.5)=-(-0.068)x
    -0.2058=0.068x
    -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

    Mentallic

    User Avatar
    Homework Helper

    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.

    N=Noe-ux
    N=1.5*1.6x10-19(0.068)(0.07)
    N=1.5*0.0960.07
    N=1.5*0.849
    N=1.274

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

    N=Noe-ux
    1.274=1.5*1.6x10-19(0.068)(x)
    0.849=1.6x10-19(0.068)(x)
    ln0.849=0.068x
    -0.164=0.068x
    -2.41=x

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

    Mentallic

    User Avatar
    Homework Helper

    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]
    [tex]u=-0.068[/tex]
    [tex]x=0.07[/tex]

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

    Mentallic

    User Avatar
    Homework Helper

    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

    Mentallic

    User Avatar
    Homework Helper

    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:

    (1.6E-19)^(0.068*0.07)

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

    Mentallic

    User Avatar
    Homework Helper

    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.

    ln(N/No)=-ux

    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

    [tex]\log_ab=\frac{\log_cb}{\log_ca}[/tex]

    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
    [itex]\frac{ln(1.221/1.5)}{ln(1.6E-19)}[/itex]=-ux?
     
  14. Feb 25, 2013 #13

    Mentallic

    User Avatar
    Homework Helper

  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

    Mentallic

    User Avatar
    Homework Helper

    You're welcome! And sorry about the confusion at the start.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Isolating a variable in an exponent
  1. Variable in Exponents (Replies: 2)

  2. Isolating A Variable (Replies: 7)

Loading...