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Solving exponential equations using logarithms

  1. Oct 27, 2003 #1
    Hi, I have a couple of problems that involve solving exponential equations using logarithms. One of them I got an answer but I'm not positive whether I did it right, and one of them I have no idea...

    3^(4logx)= 5
    (4logx)log3=log5
    logx=log5/4log3
    logx=.698/1.91
    logx=.365
    10^.365=x
    x=2.32
    did I do it right?
    and then...

    5(1.044)^t=t+10
    I've gotten this, again, not sure if I'm on the right track...
    t-logt=15.89
    and even if that is right, where do I go from there?
    Oh man...
    :frown:
    -Kate
     
  2. jcsd
  3. Oct 27, 2003 #2

    HallsofIvy

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    Why is this in the calculus section? It doesn't have anything to do with calculus.

    "log" means logarithm base 10, I take it? It would have been a good idea to say so at the start.

    Yes, if 3^(4log(x))= 5, then 4log(x) log(3)= log(5) so
    log(x)= log(5)/(4log(3)). I get 0.366 to three decimal places. If you are using a calculator, try not to round off until the end.
    Yes, x= 2.32.

    There is no elementary way to solve an equation that has the unknown both as an exponent and not. You might try "Lambert's W function" which is defined as the inverse of the function xe<sup>x</sup> but I suspect that is more advance than you want to use. If you really must solve such an equation, try a numerical method such as Newton's method.
     
  4. Oct 29, 2003 #3
    "Why is this in the calculus section? It doesn't have anything to do with calculus."
    I didn't realize that wasn't calculus. What should it have been under?

    "log" means logarithm base 10, I take it? It would have been a good idea to say so at the start."
    And I'm sorry you were confused about the log base10 issue. My math analysis teacher said that when you wrote "log" it was assumed that it was base 10.
    Anyway thank you very much.
    -Kate
     
  5. Oct 29, 2003 #4
    PS

    Oh and also, I forgot to post this part,
    we just learned how to do that second equation in class, as you said there is no simple way to solve it algebraically so you have to graph the two equations and find their intersection.
     
  6. Oct 29, 2003 #5

    HallsofIvy

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    "Calculus" is limits, derivatives, integrals and such. This problem might have come up in a Calculus class as an introduction to something later.

    I would like to have a discussion with your "math analysis" teacher! Yes, in elementary mathematics, log commonly means the "common" logarithm (base 10) but in mathematics at the calculus or higher level, log almost always means "natural" logarithm.
    I may have mistook the level of the course from the fact that you put it in the calculus section.
     
  7. Oct 29, 2003 #6

    HallsofIvy

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    Oh, and I meant to say, it probably should have been under "General Math"- or, even better, the "Homework" section.
     
  8. Oct 29, 2003 #7
    i think general math section? but yeah, that gu7y is right, in calculus they use "e" and ln or natural log more often, however, the derivative of log is seen quite often...
     
  9. Oct 30, 2003 #8
    Ok, thanks
    :smile:
     
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