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How to x in this eqaution

  1. Jul 7, 2011 #1
    1. The problem statement, all variables and given/known data

    4(3^2x+1) + 17(3^x) - 7 = 0


    2. Relevant equations



    3. The attempt at a solution

    4 + 17 (3^2x+1+x) - 7 = 0
     
  2. jcsd
  3. Jul 7, 2011 #2

    rock.freak667

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    I am assuming your problem reads:

    4(32x+1) + 17(3x) - 7 = 0

    In which case remember that am+n=aman and amn = (am)n.

    So after you expand out the first term you can use a substitution to solve for x such as y=3x.
     
  4. Jul 7, 2011 #3
    and how will i expand it...becaues i dont know how to
     
  5. Jul 7, 2011 #4

    rock.freak667

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    Take the first term 4(32x+1) using the first rule I posted with am+n=aman, how can you rewrite 32x+1 as?
     
  6. Jul 7, 2011 #5
    so is it (3^2x)(3^1)
     
  7. Jul 7, 2011 #6
    Yes. So following up on what rock.freak said, you will have

    4(3)(32x)+17(3x)-7=0

    So now, you will substitute y=3x. After that, you should be able to solve it.
     
  8. Jul 7, 2011 #7
    i have no idea what to do next
     
  9. Jul 7, 2011 #8

    SammyS

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    32x = (3x)2

    & 4(3) = 12

    Now, does Ivan's suggestion make more sense to you?
     
  10. Jul 7, 2011 #9
    4(3)(3^2x)+17(3x)-7=0

    32x = (3x)2

    & 4(3) = 12
    so after that would it be something like this:

    12 + (3^x)^2 + 17(3x)-7=0
     
  11. Jul 7, 2011 #10
    You are not to add 12 but multiply it to the first term. I also noticed that your notation is a bit wrong. Your equation shouldould look like this:

    12(3x)2+17(3x)-7=0

    Then do what rocky said earlier, which is y=3x. Then you should be able to see what you are to do next.
     
  12. Jul 7, 2011 #11
    then
    12*17(3x)-7=0
     
  13. Jul 7, 2011 #12
    No, you would multiply 12 to the (3x)2.

    Where ever you see 3x you substitute in y. When you do that, you should form a quadratic equation. Then do what is necessary to solve it.

    12(3x)2+17(3x)-7=0

    Here is the equation again.
     
  14. Jul 7, 2011 #13
    so should it be this
    (36x)2 + (51x)-7=0


    which would be
    36x^2 + 51x - 7=0
     
  15. Jul 7, 2011 #14
    You just did a common algebraic error:

    a(bx) [itex]\neq[/itex](ab)x

    Just simply put in y for where ever you see 3x. Then do what I told you before.
     
  16. Jul 7, 2011 #15
    12(3x)2+17(3x)-7=0

    so are you saying i should do this:

    12(y)2+17(y)-7=0
     
  17. Jul 7, 2011 #16

    SammyS

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    See this equation as corrected by Ivan:

    [tex]12(3^x)^2+17(3^x)-7=0[/tex]

    Everyplace you see [itex]3^x[/itex] in the equation, replace it with [itex]y\,.[/itex]

    You should obtain a quadratic equation, but the variable is y instead of the usual x which you are more accustomed to.

    The expression on the left hand side of your new equation can be factored, which should lead to a solution. Alternatively, you should be able to solve it using the quadratic formula.

    Added in Edit:

    I see you got the quadratic equation while I was entering my post. I'm a slow typist.
     
  18. Jul 7, 2011 #17
    ok understand now, so i would be

    12y^2+17y-7=0
     
  19. Jul 7, 2011 #18
    Yep. So now all you would have to do now is solve for y. Then when you solve for y, replace y with 3x then solve for x.
     
  20. Jul 8, 2011 #19
    i solve foe y but how should i replace it with 3x

    for eg. y= 1.75
    would it then be
    3x=1.75
     
  21. Jul 8, 2011 #20
    Yes, exactly and then use logs to solve it.
     
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