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Solve the simultaneous equations

  1. Oct 23, 2004 #1
    Sovle the simultaneous equations
    [tex]log_2 (x-14y) = 3[/tex]
    [tex]lgx - lg(y+1) = 1[/tex]
    How do i start?
     
  2. jcsd
  3. Oct 23, 2004 #2
    for the first equation, make it

    [tex] 2^3= x-14y [/tex]

    then carry on from there......

    :smile:
     
  4. Oct 23, 2004 #3

    HallsofIvy

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    Of course, without knowing what "lg" means we can't help you with the second equation!
     
  5. Oct 23, 2004 #4
    I suppose lg means [tex]log_{10} ?[/tex] I got the question from my book.
     
    Last edited: Oct 23, 2004
  6. Oct 23, 2004 #5

    Hurkyl

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    Weird. I'm most familiar with lg denoting log base two, from my computer science courses.
     
  7. Oct 23, 2004 #6
    Well u might be correct considering that u are probably smarter than me. But i was thaught that lg is [tex]log_{10}[/tex] if the base isn't stated.
     
  8. Oct 23, 2004 #7

    arildno

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    lg is used, for example in certain fluid mechanics formulae, as the Briggsian logarithm, that is, [tex]log_{10}[/tex]
     
  9. Oct 23, 2004 #8
    The answer for the question is x=15, y=1/2
     
  10. Oct 23, 2004 #9

    arildno

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    Since you've learnt to use "lg" as log10, stick with that!
    The second equation can then be rewritten as:
    [tex]\frac{x}{y+1}=10[/tex]
    Do you agree with that reasoning?
     
  11. Oct 23, 2004 #10
    Yes. Thank you
     
  12. Oct 23, 2004 #11
    How about this one. Solve 1 + 2 lg (x+1) = lg (2x+1) + lg (5x+8)
    Sorry but i still haven't got the hang of log yet
     
  13. Oct 23, 2004 #12

    arildno

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    Look first at your RIGHT-HAND side:
    Can you write the sum of two logs as a single log?
     
  14. Oct 23, 2004 #13
    ok so i got [tex]lg(10x^2 + 21x + 8)[/tex]
    Then i will get [tex] 10+(x+1)^2 = (10x^2 + 21x + 8)[/tex]right?
     
  15. Oct 23, 2004 #14

    arildno

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    Right, so you can use that expression as your right-hand side instead (agreed?).
    Now, consider the 2lg(x+1)-term on your original left-hand side.
    Can you rewrite that into log(something..)
     
  16. Oct 23, 2004 #15

    arildno

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    No, your suggestion at exponentiating the equation is wrong, even though you made a correct rewriting of your right-hand side
     
    Last edited: Oct 23, 2004
  17. Oct 23, 2004 #16
    Isn't the left hand side in a log form already? Except the 1.
     
  18. Oct 23, 2004 #17

    arildno

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    It is completely wrong:
    We have:
    [tex]1+lg((x+1)^{2})=lg(....)[/tex]
    We must move the log term on the left-hand side over and get:
    [tex]1=lg(\frac{(...)}{(x+1)^{2}})[/tex]
    Or :
    [tex]10=\frac{(...)}{(x+1)^{2}}[/tex]

    Do you see the difference?

    I've used (...) to denote what stood on the right-hand side.
     
  19. Oct 23, 2004 #18
    Ahhh... Finally i get it. I can solve it from here.
     
    Last edited: Oct 23, 2004
  20. Oct 23, 2004 #19
    What notation you use for logarythms in base "e" and in base "10"? In Spain we use "ln" for the first, and "log" for second. I think that your notation is not the same...
     
  21. Oct 23, 2004 #20
    Same for me in my country
     
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