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Solving a log proplem!

  1. Feb 22, 2009 #1
    1. The problem statement, all variables and given/known data
    http://img205.imageshack.us/img205/2327/nummer1.jpg [Broken]


    2. Relevant equations
    If you use ti-89 titanium, you write log with base(b) (c=number) as log(c,b) fx; the log of 3x in base 5 is writen log(3x,5) (log5(3x))took me forever to figure out
    3. The attempt at a solution
    I filled out about 5 handwriten papers with attempts, all wich failled. dont force me to write all my attempts since i find this to be very tedius.
    simplified, the equation is;
    10*5x=3*5-x+1

    i put log5 on each side geting:

    log5(5x+3)-x=x+log5(10)

    then moving it about a bit i get:

    2x-log5(5x+3)=log5(10)

    But here i get stuck...
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 22, 2009 #2
    Try to Substitute: 5^x = y (with some restrictions on y)

    You'll end up solving a quadratic equation in y!

    Good luck
     
  4. Feb 23, 2009 #3
    uh? sorry i couldt not get that to work, one of those 5x has a negative sign... (i am not sure i did it right. where in the the proces shuld i put it in?)... sorry, i still need help.
     
  5. Feb 23, 2009 #4
    Is there any way that i can get 3 out the log (or in a seperate) in of 2x-log5(5x+3)
     
  6. Feb 23, 2009 #5

    Gib Z

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    Homework Helper

    You should try phymatast's suggestion again. Multiply both sides by 5^x. Then see if you can pull out any factors and make the equation look like

    [tex] a 5^{2x} + b 5^x + c[/tex].

    Since 5^(2x) = (5^x)^2, its just a quadratic equation, like phymatast said.
     
  7. Feb 23, 2009 #6
    i see, then i get 10*(5x)2-5x-3=0

    then:

    5x=-(1[tex]\pm\sqrt{}[/tex]-12-4*10*(-3))/2*10

    that gives me that
    5x=1/2
    5x=-3/5

    with 5x=-3/5 my calculator goes haywire so gues thats an imposiple answer (because of the -)

    then i take log5(5x)=log(1/2)
    wich gives me x=-log5(2)
    wich gives me an asnwer that dosent fulfil that the anser shuld be given in a+log5b (exept if i say 0-log52 i quess)

    Is that the correct answer tough? need a litlle bit more help here!
     
  8. Feb 23, 2009 #7

    Gib Z

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    Check your working in the quadratic formula, it should be -1, not 1 !
     
  9. Feb 23, 2009 #8

    HallsofIvy

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    You shouldn't need the quadratic formula to solve [itex]10y^2- y- 3= (2y+ 1)(5y- 3)[/itex]: which does NOT give y= 1/2, y= -3/5.
     
  10. Feb 23, 2009 #9
    it's 3/5 and not -3/5 and this will give you your answer using basic properties of logarithmic functions.

    hint: after you get 5^x = 3/5

    take log base 5 on both sides
     
  11. Feb 24, 2009 #10
    aah, it seem i have made a mistake; the two possible anser is 3/5 and -1/2.
    So, -1/2 is imposible (could someone explain why?). that leaves me with 3/5 using log5 on both side i get x =log5(3/5) using the rule that log(a/b)=log(a)+log(b) i get x=log5(5)+log5(3) wich is the same as x=1+log5(3)

    Heruka, i finnaly got it! thanks guys, you are all great!:biggrin:
     
  12. Feb 24, 2009 #11

    HallsofIvy

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    -1/2 is impossible because the original equation would then involve log5(-1/2) and logarithm is defined for negative values of x.
     
  13. Feb 24, 2009 #12
    Rigth stupid me; its because log5(-1/2) Reads
    "what power must 5 be raised to, to get -1/2" And it is imposible because you can't get a negative number by raising a positive number.
     
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