1. Not finding help here? Sign up for a free 30min 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!

Simply Logarithms problem

  1. Jul 8, 2006 #1
    1) ((logx)/(log4))^2=((logx^5)/(log4))-4

    how do you these problmes...i mean..is there anyway of simplfing the expression ...i am stuck..
    regards
    vijay
     
  2. jcsd
  3. Jul 8, 2006 #2
    Use the fact [itex]\log {a^b} = b\log{a}[/itex] and simplify.
     
  4. Jul 8, 2006 #3
    yea....i used it....but this sum is different..seriously..try it
     
  5. Jul 8, 2006 #4
    Can you show your work?
     
  6. Jul 8, 2006 #5
    mr neutrino...
    i can show my work but it is too long....
    i ll jus ask you a simple question...
    what is (logx/log4)^2...i mean...how do you simplyfy it in terms of logx^5/log4?
     
  7. Jul 8, 2006 #6

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    First, write your equation as:
    [tex](\frac{\log(x)}{\log(4)})^{2}=(\frac{\log(4)}{\log(x^{5})})^{4}[/tex]
    Then, collect the terms on the LHS and rewrite your equation as:
    [tex](\frac{\log(x)}{\log(4)}-(\frac{\log(4)}{\log(x^{5})})^{2})*(\frac{\log(x)}{\log(4)}+(\frac{\log(4)}{\log(x^{5})})^{2})=0[/tex]
    Now, you should be able to continue a bit!
     
  8. Jul 8, 2006 #7
    Actually you've got to simplify the second term (logx^5/log4) using the rule I mentioned earlier, and then see if the resulting equation resembles something you've come across before. Remember that log4 is just a constant; don't worry too much about that part.
     
    Last edited: Jul 8, 2006
  9. Jul 8, 2006 #8
    I believe it was a -4 at the end, and not a power, unless it was a typo.
     
  10. Jul 8, 2006 #9

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    That minus sign was removed by flipping the fraction. :smile:

    EDIT:
    Oh dear, it was MINUS 4 not that the exponent was minus 4 as I thought..:cry:
     
  11. Jul 8, 2006 #10
    Exactly. :)
     
  12. Jul 8, 2006 #11
    Can this be solved analytically?
     
  13. Jul 8, 2006 #12

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Of course.
    It is a linear equation in the variable log(x).
     
  14. Jul 8, 2006 #13
    Not quadratic?
     
  15. Jul 8, 2006 #14

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    It might be..:uhh:
     
  16. Jul 8, 2006 #15

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Assuming this is
    [tex]\left(\frac{log x}{log 4}\right)^2= \frac{log x^5}{log 4}- 4[/tex]
    and not
    [tex]\left(\frac{log x}{log 4}\right)= \left(\frac{log x^5}{log 4}\right)^{- 4}[/tex]
    as I also initially interpreted it, let y= log x. since log x5= 5 log x, this can be written as
    [tex]\frac{1}{(log 4)^2}y^2- \frac{5}{log 4} y+ 4= 0[/tex]
    a quadratic equation for y.
    ("log 4" is, of course, simply a constant.)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Simply Logarithms problem
  1. Logarithm problem (Replies: 16)

  2. Logarithmic problem (Replies: 18)

  3. Logarithmic Problem (Replies: 4)

Loading...