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Homework Help: Solve log5(x-1) + log5(x-2) - log5(x+6) = 0

  1. Oct 3, 2013 #1
    1. The problem statement, all variables and given/known data


    log5(x-1) + log5(x-2) - log5(x+6) = 0

    2. Relevant equations

    3. The attempt at a solution

    I feel like this is really simple but I cannot get the answer.

    This is what I attempted

    Since they have the same base, I multiplied the top

    log5[(x-1)(x-2)] - log5(x+6) = 0

    This gives me

    log5(x2-3x+2) - log5(x+6) = 0

    At this point I did not now what to do, I tried 2 methods,

    Since they are the same base again, I divided the top this time

    log5(x2-3x+2) / (x+6) = 0

    but I don't know how to continue


    I moved - log5(x+6) to the other side

    log5(x2-3x+2) = log5(x+6)

    Since they have the same base I ignored them so

    x2-3x+2 = x+6

    Which gave me

    x2 - 4x - 4

    Which cannot be factored.

    Please help, thankyou :)
    Last edited: Oct 4, 2013
  2. jcsd
  3. Oct 4, 2013 #2


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    What you've done is incorrect because the rule of logarithms is

    [tex]\log{a}+\log{b} = \log{ab}[/tex]


    [tex]a^m\cdot a^n = a^{m+n}\neq a^{mn}[/tex]

    See if you can apply these algebraic formulae to correctly answer your question.
  4. Oct 4, 2013 #3
    I did use this logarithm law, since the base for all of them are common.


    I did not use at all throughout. I am not sure where I have gone wrong, can you point it out for me?
    Last edited: Oct 4, 2013
  5. Oct 4, 2013 #4


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    What is


    equivalent to?
  6. Oct 4, 2013 #5
    Oh, I think I found the problem... 5 is the base, not 10.

    So this is the correct equation:

    log5(x-1) + log5(x-2) - log5(x+6) = 0

    Terribly sorry, would my calculations be correct following the equation above?
  7. Oct 4, 2013 #6


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    Oh, in that case, you're on the right track in #2 in your first post. [itex]x^2-4x-4=0[/itex] cannot be factored in the manner you're thinking of, but it does have irrational roots. Use the quadratic formula to find them.

    Also, keep in mind that some (or even all) of your x solutions might not be valid. You need to plug them back into the original equation to see if it makes sense. If you end up with log(-1) for example, then that solution of x needs to be scrapped.
  8. Oct 4, 2013 #7


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    What is x2 - 4x - 4 equal to?

    It should be a quadratic equation, can you solve it? The roots need not be integer numbers!

  9. Oct 4, 2013 #8
    Oh I see, thank you!
  10. Oct 4, 2013 #9
    x2 - 4x - 4 = 0

    When you mean that they do not need to be intergers, how do I go about solving? As suggested by Mentallic, I would have to use the quadratic formula right?
  11. Oct 4, 2013 #10


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    The quadratic formula is the easiest method, but completing the square works too if you don't remember the formula.


    Now add 4 to both sides.



    This is how the quadratic formula is derived. Begin with


    and then complete the square to solve for x.
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