Logarithms

Discussion in 'Homework Help' started by MohtasaUnique, Jan 23, 2011.

Logarithms
  1. Unread #1 - Jan 23, 2011 at 8:52 PM
  2. MohtasaUnique
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    Logarithms

    How do you solve these: log(2x-1) = 2+log(x-2)
     
  3. Unread #2 - Jan 23, 2011 at 10:35 PM
  4. Koot
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    Logarithms

    Subtract the logs to one side and using the log rules, you can combine them.

    log A - log B = log A/B

    The you can change the logs to 10^ and solve.
     
  5. Unread #3 - Feb 6, 2011 at 4:17 AM
  6. panzerjager
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    Logarithms


    Yea so : log(2x-1) = 2+log(x-2)
    becomes log(2x-1)-log(x-2)= 2
    which is: log[(2x-1)/(x-2)] = 2
    Turns into: 10^2 = (2x-1)/(x-2) = 100

    i dont have a calculator on me but its gonna be a relatively small #
     
  7. Unread #4 - Feb 10, 2011 at 3:22 AM
  8. shelnaz2
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    Logarithms

    dude is it natural log or log base 10?
     
  9. Unread #5 - Feb 13, 2011 at 3:25 AM
  10. valvemasher21
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    Logarithms

    Subtract the log(x-2) to the other side, use the log identity to combine them(log A-log B=log(A/B), and raise 10 to the power of the other side and set it equal to (2x-1)/(x+2).
     
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