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Log sums

  1. Jul 19, 2010 #1
    1. The problem statement, all variables and given/known data
    I have 2 problems ,which are:
    1) Solve :log 7 (x 3+27) + log 7 (x+3)
    2) Solve for x:log5 ( 5 1/x+ 125) log 5 6+1+1/2x


    2. Relevant equations



    3. The attempt at a solution
    The first one tried like this :
    log 7 { (x 3+27)/(x+3) }= 2
    or 7 2 = (x 3+27)/(x+3)
    But I don't know how to proceed beyond it please help!
    in second one
    log5{( 5 1/x+ 125) /6}=1+1/2x
    but I can't proceed here too.
    I need an urgent help!
    Thank You!
     
  2. jcsd
  3. Jul 19, 2010 #2

    eumyang

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    Homework Helper

    Please check the original problems as I see typos abound. For instance, in the first problem,

    log 7 (x 3+27) + log 7 (x+3)

    ...I see no right hand side. Also, if what you have above is correct, then you're next step...

    log 7 { (x 3+27)/(x+3) }= 2

    ...is already wrong. (Is is supposed to be divide?)

    Finally, if the last step above is supposed to have a divide, then review factoring special products -- specifically, the sum of two cubes.


    69
     
  4. Jul 19, 2010 #3

    Mentallic

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    For the first one, you could multiply through by x+3 to give

    [tex]49(x+3)=x^3+27[/tex]

    and then solve that cubic by simplifying and turning it into the form [tex]ax^3+bx^2+cx+d=0[/tex]
    but cubics are hard to solve, an easier way would be to realize that [tex]x^3+27[/tex] is a sum of two cubes, mainly [tex]x^3+3^3=(x+3)(x^2-3x+9)[/tex] which means you can cancel the x+3 from the numerator and denominator, giving you only a quadratic to solve, which is much simpler. Remember to look back at the original question to figure out what the domain of x is.

    For the second, do the same thing as you did above, turn the log into powers such that

    [tex]\frac{5^{\frac{1}{x}}+125}{6}=5^{1+\frac{1}{2x}}[/tex]

    Now simplify the right hand side. You should be able to factorize out 51/x and then solve.
     
  5. Jul 20, 2010 #4
    I agree with eumyang. Is the question to simply the above expressions? Also, with the second problem, are you missing any brackets/parenthesis at all?

    The Bob
     
  6. Jul 21, 2010 #5

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    At least in the second, 1/2" specifically says "solve for x". There are no equations to be solved!
     
  7. Jul 21, 2010 #6

    Mentallic

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    It's an honest mistake to mix up + and = considering they're the same key on the keyboard, but he did have lots of typos so it could have been re-interpreted as anything.

    And for the first equation, let's just assume he started to solve the correct problem, i.e. f(x)=2.
     
  8. Jul 23, 2010 #7
    Thanks for the help but I figured it out!
    Thanks everyone!
     
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