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Solving for x log question

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  1. Nov 16, 2015 #1
    • Moved from a technical forum, so homework template missing.
    Hi folks,

    I'm revisiting logs for the first time in a long time through distance education and I was wondering if someone could have a look over a question I've answered and let me know if I've done it correctly or if I'm way off please

    Find x if Log3(10x – 1) – 2 = 2log3x

    I instantly divide by common log of log3 and work out from there

    (10x-1) - 2 / 2 = x
    (5x- 1/2) -1 = x
    5x - 1/2 = x + 1
    5x - x = 1 + 1/2
    4x = 3/2
    x = 3/8

    Any help would be greatly appreciated
     
  2. jcsd
  3. Nov 16, 2015 #2
    By 2Log3, does it mean that it's a log with basis of 2 or 2*Log3?
     
  4. Nov 16, 2015 #3
    its written as 2log3x, so I take it to mean 2 * log3X
     
  5. Nov 16, 2015 #4

    Samy_A

    User Avatar
    Science Advisor
    Homework Helper

    If it means
    Find ##x## if ##\log_3(10x-1) -2=2*\log_3x##,
    I don't understand what you did.

    You could take all the ##log_3## terms together, making use of:
    ##\log_3(a*b)=\log_3a+\log_3b##
    ##\log_3(a/b)=\log_3a-\log_3b##
    ##2*\log_3a=\log_3a^2##
    ##3^{log_3a}=a##

    and then solve for x.
     
  6. Nov 16, 2015 #5
    Since 2log3x = 2 (Log3 + Log x)
    which means there will 2 Log 3, so there would be still one Log 3 remaining.

    Or you can just actually change the 2Log3x to Log(3x)^2, and then manipulate the left side of equation to be Log, and equals them.
     
  7. Nov 16, 2015 #6
    Thanks everyone for all the help...its been far too long since I've looked at any of this sort of stuff....
    I now have it at
    (log310x - log31) - 2 = log3x2
    Then divide out the log3
    10x - 1 - 3 = x2
    0 = x2 -10x + 3

    I've obviously done something wrong again as this gives pretty silly results :(
     
  8. Nov 16, 2015 #7
    You've got to make (log310x - log31) - 2 into one Log3. You don't really divide out the log, but comparing the values inside the logs. They have to be equals, since the logs would result the same.
     
  9. Nov 16, 2015 #8

    Samy_A

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    How did you get the expression to the left?
    Note that ##\log_3(10x-1)\neq\log_3(10x)-\log_3(1)##

    Not sure what you mean by "divide out the log3"
     
  10. Nov 16, 2015 #9

    jedishrfu

    Staff: Mentor

    You start with finding what the value of 2 is in ##log_3##

    ##log_3(3)=1## and ## log_3(9)=log_3(3^2)=2##

    So the RHS becomes ##log_3(10x - 1) - log_3(9) = log_3( (10x - 1) * 1/9) ##

    and go from there.
     
  11. Nov 16, 2015 #10
    i'm just going round in circles here and getting nowhere I'm afraid
    ##\log_3(10x-1)=((log_3(10) + log_3(x)) / log_3(1))##

    I understood when i had a common log base, i could divide all by that log base
    hence ((10 + x) / 1) - 2 = x2
    0 = x2 - x - 8

    which doesn't give reasonable answers so i must have messed up somewhere again
     
  12. Nov 16, 2015 #11

    jedishrfu

    Staff: Mentor

    You're not dividing by log, you have to get that understanding straight first to solve this problem:

    ##log3(10x−1) =/= ((log3(10)+log3(x))/log3(1))##
     
  13. Nov 16, 2015 #12

    jedishrfu

    Staff: Mentor

  14. Nov 16, 2015 #13
  15. Nov 16, 2015 #14
    Right folks, ive gone away n come back with an improved effort i think...

    log3(10x-1) - 2 = log3(x2)

    =log3((10x-1)/x2) = 2

    = (10x-1)/x2 = 32 = 9

    9x2 -10x + 1 = 0

    =>x = 1, x = 1/9

    Does this look correct to you guys or am I missing something again?
     
  16. Nov 16, 2015 #15

    Samy_A

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    Looks correct.
     
  17. Nov 16, 2015 #16
    Thanks very much for your help Samy and the rest of you fine people, much appreciated
     
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