1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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;

    i put log5 on each side geting:


    then moving it about a bit i get:


    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

    User Avatar
    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



    that gives me that

    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

    User Avatar
    Homework Helper

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


    User Avatar
    Science Advisor

    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


    User Avatar
    Science Advisor

    -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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook