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Pre Calculus problem in need of help

  1. Apr 13, 2014 #1
    1. The problem statement, all variables and given/known data

    Solve for X in terms of Natural Logs;
    Both my friend and I (top scores in the class) are struggling with this problem.

    2. Relevant equations
    e^ax=c*2^bx

    3. The attempt at a solution
    We both decided to divide both sides by 2^bx which gave us
    e^ax / 2^bx = C

    Then I flipped the 2^bx up to the top by making the ^-bx
    Then I multiplied the 2 and the e and added the ^ax and ^-bx

    thus resulting in
    2e^ax-bx=c

    Then divide both sides by two and natural log both sides
    getting

    ax-bx= ln(c/2)

    Factor out a x and divide both sides by a-b

    x= ln(c/2) / a-b

    We both suspect our answers are wrong due to the multiplying of 2 and the e, other than that we have no idea how to get the right answer.
     
  2. jcsd
  3. Apr 13, 2014 #2

    SammyS

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    You're right regarding the error of multiplying 2 and e .

    The general way to solve an exponential equation is to take some logarithm of both sides. In this case using the natural log, ln, should work fine.
     
  4. Apr 13, 2014 #3
    Trying what I believe you just said still confuses me and I am still not getting an answer. Would you possibly explain further?
     
  5. Apr 13, 2014 #4

    SammyS

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    Show what you get when you take ln of both sides of
    eax=c*2bx

    (Assuming that's the equation you started with. You really should have used parentheses to set off the exponents.)
     
  6. Apr 13, 2014 #5
    Alright, so my results are now looking like this....

    x= (ln(2c))/(a+(1/b))

    Is this now correct?
    (I do not have an answer key for this)
     
  7. Apr 13, 2014 #6

    SammyS

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    No.

    Show some of your steps.

    2 and c shouldn't combine like that at all.
     
  8. Apr 13, 2014 #7
    I took the ln of both sides, then the ln(e^(ax)) went to ax=ln(c*2^(bx)) From here I put the bx in front of the ln(c*2) but now I suspect that is wrong as well. I am not sure what to do with the ln(c*2^(bx))
     
  9. Apr 13, 2014 #8

    micromass

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    [tex]\log(ab) = \log(a) + \log(b)[/tex]
     
  10. Apr 13, 2014 #9
    Right, I feel pretty stupid for that now.

    Ok well, now...
    ax= ln(c) + ln(2^(bx))
    Bring the bx to the front and subtract to the other side.
    ax-bxln(2)=ln(c)

    Factor out x

    x(a-bln(2))= ln(c)

    divide by (a-bln(2))
    x= ln(c)/(a-bln(2))

    Alright.....now I feel like I am getting closer to the answer, if not finally hit it.
     
  11. Apr 14, 2014 #10

    Mark44

    Staff: Mentor

    That's it.
     
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