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Homework Help: Log5(x) = 16logx(5) solve for x

  1. Jan 12, 2010 #1
    log5(x) = 16logx(5)
    solve for x.

    With this one, I have no idea where to start. All I have even thought about doing is bringing up the 16 to make it 5^16 but that doesn't seem to help me.
     
  2. jcsd
  3. Jan 12, 2010 #2

    Mark44

    Staff: Mentor

    Re: Logarithms

    Convert one of your logs so that both logs are in the same base. Do you have a formula for converting from one log base to another?
     
  4. Jan 12, 2010 #3
    Re: Logarithms

    No i dont have a formula to do that
     
  5. Jan 12, 2010 #4
    Re: Logarithms

    the formula is:

    log5(x) = log (x) / log (5)
     
  6. Jan 12, 2010 #5

    Mark44

    Staff: Mentor

    Re: Logarithms

    Let y = logbx
    Then x = by
    So log x = log(by) = y log b
    And y = (log x)/(log b)

    Hence logbx = (log x)/(log b)

    In the third step above, you can use any log base you want. I used the common log (log10).
     
  7. Jan 12, 2010 #6
    Re: Logarithms

    ok thnx for the formulas, ok so i have the log base for one side which is (log10(x)) / (log10(5)) what do i do now?
     
  8. Jan 12, 2010 #7

    Mark44

    Staff: Mentor

    Re: Logarithms

    Replace log5(x) in your original equation.

    When you do that, what does your equation become?
     
  9. Jan 12, 2010 #8
    Re: Logarithms

    (log10(x)) / (log10(5)) = logx(5^16)
     
  10. Jan 12, 2010 #9

    Mark44

    Staff: Mentor

    Re: Logarithms

    Instead of changing log5 to log, why don't you change logx to log5? The goal is to be using the same log base on both sides of the equation.
     
  11. Jan 12, 2010 #10
    Re: Logarithms

    so are you saying change it so it is: log5(5^16) / log5(x) = log5(x) ????
     
  12. Jan 12, 2010 #11

    Mark44

    Staff: Mentor

    Re: Logarithms

    Yes. Now put it in the context of the original equation.

    log5x = logx516
    ==> log5x = [log5516]/log5x

    The numerator on the right can be simplified to just plain 16, and you can multiply both sides by log5x.
     
  13. Jan 12, 2010 #12
    Re: Logarithms

    omg, thankyou so much! i get it finally. I wish i could think like you
     
  14. Jan 13, 2010 #13

    Mark44

    Staff: Mentor

    Re: Logarithms

    Practice...
     
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