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Exponential logarithm

  1. Aug 23, 2008 #1

    is this equation correct??
  2. jcsd
  3. Aug 23, 2008 #2
    You have y on the LHS, just solve for it. Why would you continue to take the logarithm?

    That is not correct. ln(a+b) =/= ln(a) + ln(b). I think you should take another look at what a logarithm means and not a rule sheet.
  4. Aug 23, 2008 #3
    ln126y=(6x6-11) ln8
    y=((6x6-11) ln8)/ln126

    Then is this correct??
  5. Aug 23, 2008 #4
    When multiply the log sign, can i just multiply one side??
  6. Aug 23, 2008 #5
    You seem to be over complicating things. From here just do some simple algebra to solve for y.
  7. Aug 23, 2008 #6
    9y + 14 = log86x6-log11
    = log86x6 /log11
    9y = log86x6 /log11 - 14
    y = (1/9)log86x6 /log11 - (14 / 9)

    Is this correct??
  8. Aug 23, 2008 #7


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    Staff Emeritus
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    No. lod a- log b is not log a/log b. It is log a/b

  9. Aug 23, 2008 #8
    9y + 14 = log86x6 -log11
    = log86x6 /11
    9y = (log86x6 /11) - 14
    y = ((1/9)log86x6 /11) - (14 / 9)

    Then, this would be correct, right??
  10. Aug 23, 2008 #9
    Usually, how does an equation have the condition of b(logbu) =u??
  11. Aug 23, 2008 #10
    Um, HallsofIvy explained why you CANNOT have that. Also please use parentheses and clear up the notation. You have no idea how frustrating it is to read logarithms without parentheses around the arguments, i.e. use the form ln(the expression inside of here).

    Start with the line that epkid08 quoted you on and simply solve for y. Don't take anymore logarithms!
  12. Aug 23, 2008 #11
    Ask yourself what logbu stands for. If your answer included the word exponent, then it's likely you'll immediately see that it holds by definition!
  13. Aug 24, 2008 #12
    The previous solution start off incorrectly, so i start off a new one, but there were all wrong, such as below:

    8log8(9y+14) =86x6-11
  14. Aug 24, 2008 #13


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    Yes, 8log_8(9y+ 14)= 9y+ 14 but what happened to the base 8 on the right side?

    [tex]9y+ 14= 8^{6x^6- 11}[/tex]
    That's easy to solve for y.

    Last edited: Aug 24, 2008
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