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Log Law Problems

  1. Oct 17, 2013 #1
    1. The problem statement, all variables and given/known data


    Write as a single logarithm:


    2. Relevant equations

    Logarithm Laws:

    [itex]log_a(xy) = log_a(x) + log_a(y)[/itex]

    [itex]log_a(\frac{x}{y}) = log_a(x) - log_a(y)[/itex]
    ___________

    Problem Set:

    [itex]log_{10}A + log_{10}B - log_{10}C[/itex]

    [itex]\frac{1}{2}logX - 2log4[/itex]

    [itex]2logN + 3logX[/itex]

    3. The attempt at a solution

    I simplified the first question to [itex]log_{10}(\frac{AB}{C})[/itex] Am I correct?

    I wasn't sure about how to approach the second question. I multiplied [itex]\frac{1}{2}[/itex] by [itex]X[/itex] and [itex]2[/itex] by [itex]4[/itex] and simplified as follows:

    [itex]log_{10}{\frac{1}{2}X} - log_{10}8 [/itex]

    to get [itex]log_{10}(\frac{0.5x}{8})[/itex]

    I'm not sure if this is correct though.

    If it is wrong, how would I solve it correctly?

    For the third problem, I solved it to:

    [itex]log_{10}[ (2n)(3x) ][/itex]

    Thanks,
     
  2. jcsd
  3. Oct 17, 2013 #2

    jedishrfu

    Staff: Mentor

    Your AB/C is correct.

    For the 1/2 log X you haven't listed the loglaw for it which is:

    C * log (x) = log (x^C)

    given that rework your answer.
     
  4. Oct 17, 2013 #3
    Yes..Thats right.

    That is not the correct way .

    Use the following property of logarithms : logb(xn) = n logbx.
     
  5. Oct 17, 2013 #4
    I solved the second one to:

    [itex]log_{10}\frac{X^{0.5}}{16}[/itex]

    Is this correct?

    Thanks
     
  6. Oct 17, 2013 #5
    Correct
     
  7. Oct 17, 2013 #6
    And would the second one be

    [itex]log_{10}(N^2X^3)[/itex]?

    Thanks
     
  8. Oct 17, 2013 #7
    :thumbs:
     
  9. Oct 17, 2013 #8
    Thank you very much!
     
  10. Oct 17, 2013 #9

    jedishrfu

    Staff: Mentor

    dont forget to use the Thanks button to thank everyone.
     
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