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Solving log_a(y)^n

  1. Oct 12, 2013 #1
    1. The problem statement, all variables and given/known data
    I am wondering how you are meant to solve something like log_10(x)^2 = 0. What does multiplying and dividing do to the log function algebraically, I can easily tyoe into a calculator, but I don't know how to work it out on paper. My text book does not cover how to do these types of questions, but it does have those types of questions.

    2. Relevant equations
    The other log rules, maybe?

    3. The attempt at a solution
    It can be re-written as log_10(x)*log_10(x), but it really doesn't get me very far. I have tried a whole bunch of different methods, including rearranging some of the log rules, but to no avail.
  2. jcsd
  3. Oct 12, 2013 #2


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    If you want to solve [itex]n^2=0[/itex] then what's the only value that n can possibly be?
  4. Oct 12, 2013 #3
    Sorry, that was a bad example. An example that is actually what I'm looking to work out is:
    (log_10(x))^2 - 2log_10(x) - 3 = 0

    So what I have done is changed 2log_10(x) into log_10(x^2) and 3 into log_10(1000)

    It now looks like this: log_10(x)^2 - log_10(x^2) - log_10(1000) = 0

    So it would be: log_10(x)^2 - log_10(1000x^2) = 0

    But what do I do with the log_10(x)^2 part?
  5. Oct 12, 2013 #4

    Ray Vickson

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    Your method will not work, and it is unnecessary; just put y = log_10(x) and see what you get.
  6. Oct 12, 2013 #5
    What do you mean by that? Do you mean substitute y = log_10(x) into log_a(y) = x, or actually graph it using y = log_10(x) as the equation?
  7. Oct 12, 2013 #6


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    Right now we're doing algebra without the graphing, so stop thinking about graphs for just a moment :wink:

    You want to solve


    Now for the substitution [itex]y=\log_{10}{x}[/itex] that Ray Vickson mentioned, everywhere you see a [itex]\log_{10}{x}[/itex], replace it for y.

    For example, say we might want to solve


    and we can't quite figure out what to do. Well, if we substitute y=x+1 then we get


    And now we know how to solve for y, so we get y=0 as the solution, but we want our answer in x, so all we have to do is substitute back in now to get



    So try and see if you can apply the same technique to your problem.
  8. Oct 12, 2013 #7
    Oh, you substitute it like that. I always forget that as a potential way of solving things.

    So what I did was substitute log_10(x) for y which gave:
    y^2 - 3y +2 = 0
    (y - 1)(y-2) = 0
    y = 1,2

    Then I put that into the y = a^x formula, except the substituted y is actually x here:

    y = 10^1 or y = 10^2
    y = 10,100

    Which is the correct answer in the book. Thankyou very much for your help, and I'll try to remember to use substitution more often.
  9. Oct 12, 2013 #8


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    Good work :smile:

    The substitution isn't necessary by the way, but it does help to make things easier to spot. You could have always skip the substitution and done





    But clearly it's easier to work with the substitution, especially since you're more accustomed to dealing with quadratics in simple single variables like x.
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