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Log values problem

  1. Nov 7, 2009 #1
    1. The problem statement, all variables and given/known data
    A professor is doing a problem on the black board and ends up with the expression

    [tex]\frac{Log A}{Log B}[/tex] = [tex]\frac{2}{3}[/tex]

    He absentmindedly cancels the “log”, making the left-hand side A/B. (A very wrong thing to do!)
    Luckily, he ends up with the correct values for A and B. What are these values?

    3. The attempt at a solution

    I wasn't sure what to do here. I figured 2/3 is .667

    from the question I have Log A- Log B= .667
    I set A to be equal to 10 which gives me a value of 1, then Log B would equal .333 because 1-.667 =.333

    raising both sides to be powers of ten, I get B= 10.333
    which is 2.15

    This works except if the proffessor "cancelled the Log) Then 10/2.15 obviously isn't 2/3.
    What should I do here?
  2. jcsd
  3. Nov 7, 2009 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You can't just set A to be equal to 10. You have two equations

    [tex] \frac{logA}{logB}=\frac{2}{3}[/tex] is one of them. Also, [tex] \frac{A}{B}=\frac{2}{3}[/tex] is another. You have to use both of them to solve for A and B
  4. Nov 7, 2009 #3


    Staff: Mentor

    You have an error in your work: (log A)/(log B) != log A - log B. What is true is the log(A/B) = log A - log B.

    There are two equations to work with:
    A/B = 2/3, and
    (log A)/(log B) = 2/3

    Solve for one variable in the first equation, and substitute for it in the second equation.
  5. Nov 8, 2009 #4
    Okay. Here is what I have..

    [tex]\frac{log A}{log B}[/tex] = [tex]\frac{2}{3}[/tex]

    3 log A = 2 log B

    log A3 =log B2

    log A3- log B2 = 0

    log [tex]\frac{A^3}{B^2}[/tex] = 0

    [tex]\frac{A^3}{B^2}[/tex] = 1

    A3 = B2

    from the question, we know that [tex]\frac{A}{B}[/tex] = [tex]\frac{2}{3}[/tex]

    B = [tex]\frac{3}{2}[/tex] A

    ([tex]\frac{3}{2}[/tex]A)2 =A3

    [tex]\frac{9}{4}[/tex] A2 =A3

    A= [tex]\frac{9}{4}[/tex]

    B= [tex]\frac{3}{2}[/tex] x [tex]\frac{9}{4}[/tex] = [tex]\frac{27}{8}[/tex]

    Thanks for your input.
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