1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Log 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

    Office_Shredder

    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

    Mark44

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Log problem
  1. Log problems (Replies: 3)

  2. Log problem (Replies: 5)

  3. Log problem (Replies: 1)

  4. Log and > problem (Replies: 2)

  5. Log problem (Replies: 6)

Loading...