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I can't remember how to solve equations with logarithms/exponents!

  1. Mar 13, 2014 #1
    This is frustrating me so much. I've been out of college for 1 year and I already forgot how to solve logarithm problems (though, to my surprise, I've encountered one I need to solve in real life.)

    t2/t1 = (d2/d1)1.06

    and I need to solve for d2.

    I know the equation is the equivalent of

    logd2/d1(t2/t1) = 1.06,

    but I still can't figure out how to isolate the d2.
  2. jcsd
  3. Mar 13, 2014 #2


    Staff: Mentor

    Start by taking the log of both sides.
  4. Mar 13, 2014 #3
    log1.06 of both sides?
  5. Mar 13, 2014 #4


    Staff: Mentor

    log10 or ln, either one.
  6. Mar 13, 2014 #5
    Raise both sides to the 1/1.06 power.

  7. Mar 13, 2014 #6
    Wow, I feel like an idiot now. Was Mark44 trying to lead me down a rabbit hole?
  8. Mar 13, 2014 #7
    No. Mark is a serious guy. He had a different approach in mind, probably motivated by your questions about logarithms.

  9. Mar 13, 2014 #8
    When I originally tried doing the "log to both sides, ..." approach, I started going in circles.
  10. Mar 13, 2014 #9
  11. Mar 15, 2014 #10
    Starting with logs seems like the hard way to me. Do what you usually do when you want to change the subject of a formula: get d2 on its own on one side.

    First use the exponent rule: (a/b)^n = a^n/b^n, then re-arrange to get

    d_2^1.06 = ...

    Then continue.

    EDIT: 'log rule' changed to 'exponent rule'
    Last edited: Mar 15, 2014
  12. Mar 15, 2014 #11


    Staff: Mentor

    That would be an exponent rule.
    I agree that this approach is simpler, but taking logs of both sides isn't that much longer. After you take the natural log of both sides of the original equation, you have
    $$ln(\frac{t_2}{t_1}) = 1.06 ln(\frac{d_2}{d_1})$$
    Now divide both sides by 1.06 and exponentiate to get d2/d1 by itself. One more step and you're done.

    The approach I suggested was just the first one to come to mind.
  13. Mar 15, 2014 #12
    Slip of the tongue, so to speak ...:p
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