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Homework Help: Need some help with solving a logarithmic equation

  1. Jan 26, 2015 #1
    1. The problem statement, all variables and given/known data
    The equation is as follows:

    2. Relevant equations
    What technique shall I use to solve the equation?

    3. The attempt at a solution
    Tried to solve it in any way that went through my mind, including making a substitution of terms but couldn't quite get the solution or find a path towards it.
  2. jcsd
  3. Jan 26, 2015 #2


    Staff: Mentor

    It looks like log base 3 might be a key.

    Show your work.
  4. Jan 26, 2015 #3


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    Gold Member

    Start by rewriting the RHS without a √ symbol but only indices.
  5. Jan 26, 2015 #4


    Staff: Mentor

    Yes, show your work. At this site we require homework posters to show what they have tried. It's debatable that what you have in your post counts as a valid attempt.
  6. Jan 27, 2015 #5
    Here's some of the work I made. The problem for me is that "x" is found on both the base and the power of the exponent, and I can't find a breakthrough to simplify it.
    Thanks for your help.

    Attached Files:

  7. Jan 27, 2015 #6
    [itex]A\ln {f(x)} = \ln {f(x)}^A[/itex] is something to keep in mind. I am using ln notation since I'm not used to logs other than base e, but regardless, the property holds. I noticed that you might not be aware of the property.
    Furthermore - when [itex]X^{f(x)} = X^{g(x)} [/itex]?
    What can you do to convert [itex]\sqrt 3[/itex] to a power of X?

    Another thing to point out - when you deal with logs, you should also define your domain in which you can search for solutions. The initial problem can only have solutions when x + 2 > 0 and you also have a square root of x9 - which only exists in real numbers if x is non-negative.

    The reason why this is important is if you do find a solution to the problem and you do not check if the solution fits into the domain, then you haven't solved the problem. A solution might not even exist, you always have to keep that possibility in mind.
    Last edited: Jan 27, 2015
  8. Jan 27, 2015 #7

    Ray Vickson

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    Often we solve such problems --- meaning problems whose solutions are not at all easy to see or get --- in two stages:

    (1) Plot both sides on graph paper to get a rough idea of where the two curves ##y = x^{3 \log_3 (x+2)}## and ##y = \sqrt{3 x^9}## cross (if ever).
    (2) Using the rough location obtained in (1), implement a numerical equation-solving routine to home in on a more accurate answer.

    In this case, step (1) leads to a somewhat tricky situation: it seems to produce a solution that is almost too easy and too good to be true. In fact, it IS too good to be true; you need to actually do step (2) to get a genuine solution.

    What about step (2)? You can use numerous methods manually, or you can nowadays use readily available computer tools. For example, you can use the EXCEL Solver tool if you have EXCEL on your computer. Or, you can use Maple or Mathematica if you have access to them (or other tools such as Maxima, etc.). If none of these are available, you can use the free on-line tool Wolfram Alpha (which is like Mathematica Lite) to deal with your problem. Alternatively, you can use one of the many free on-line equation-solving packages---just Google "solving equations" or something similar.
  9. Jan 27, 2015 #8


    Staff: Mentor

    Have you tried x=0,1,2...?
  10. Jan 27, 2015 #9
    What? If I had known the answer I wouldn't have asked it here. I do keep in mind all the rest of what you wrote, but that [itex]\sqrt 3[/itex] can simplify everything if I can represent it as a power of x in some form... but what form?

    I have tried x=0 and it does solve it, so it is one certain solution, but there are two other solutions, which shouldn't look pretty (actually one is x=4, which is) but what matters most is how can I reach them.

    Ray, you get it too complicated in this particular circumstances I'm afraid.. it should be too good to be true when solving it using algebraic technique as well.
  11. Jan 27, 2015 #10


    Staff: Mentor

    I was thinking that using log base 3 would help but it still doesn't fall apart cleanly when you use that technique. You'd take the log base 3 of both sides which reduces the sqrt of 3 to 1/2.
  12. Jan 27, 2015 #11


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    Staff: Mentor

    x=4 is not a solution. It is close to one, but it is not a solution.

    I fear there is no algebraic way to find the other two solutions in a reasonable form.
  13. Jan 27, 2015 #12

    Ray Vickson

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    You asked how to solve the equation, and I told you what you need to do. Drawing a graph is not complicated. Using a computerized solution package is not complicated. Before dismissing these steps you should at least give them a try.

    I think you should abandon the attempt to get an algebraic solution. If the problem is exactly as you stated it, I suspect there IS NO algebraic solution, and you MUST resort to numerical methods. There really is no choice.

    BTW: as has already been stated, x = 4 is not a solution (but it "looks like" one on a wide-ranging graph); that is what I meant when I said "too good to be true". If you draw a more detailed graph focussed on a region around x = 4 you will see the true situation.
    Last edited: Jan 27, 2015
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