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algebraic problem

Discussion in 'DAT Discussions' started by BodybldgDoc, Apr 17, 2007.

  1. BodybldgDoc

    BodybldgDoc Guest

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    Jan 13, 2007
    In your closet
    How do we solve this?

    The nonzero numbers a and b are such that a=b^4. If the value of b is tripled, the new value is which of the following.
    a. a raised to the 4th power
    b. a multiplied by 3
    c. a multiplied by 4
    d. a multiplied by 12
    e. a multiplied by 81
     
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  3. DDSDakoo

    DDSDakoo 7+ Year Member

    56
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    Feb 27, 2007
    Texas
    im gonna take a staaab it at this.

    i think the answer is E.

    reasoning:
    if a=b^4

    then if you plug in 1 for a and b. you get:
    1=1^4
    then you triple B (1 * 3) right?
    a=3^4
    a=81

    i hope that makes sense! i hope more that its right! haha
     
  4. DDSDakoo

    DDSDakoo 7+ Year Member

    56
    0
    Feb 27, 2007
    Texas
    ok i just had a friend explain this to me, the end of it sort of didnt make sense in my head

    if A=B^2,
    then you double B, what is the new value:

    A=2B^2

    A=4B

    the answer would be: A multiplied by 4

    since it is an equality equation. what you do to one side, you have to do to the other.

    so correction:
    if A= (3^4)b
    then A would be multiplied by a factor of 81. since the equality had changed when you multiplied by a factor of 3 on the side of b^4.
     
  5. Streetwolf

    Streetwolf Ultra Senior Member Dentist 7+ Year Member

    1,801
    5
    Oct 25, 2006
    NJ
    You just sub in 3b for b and see what happens, since 3b is b tripled:

    Instead of b^4 you have (3b)^4 which equals 3^4*b^4 = 81b^4.
     
  6. BodybldgDoc

    BodybldgDoc Guest

    712
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    Jan 13, 2007
    In your closet
    If A=(2B)^2 then wouldnt it be A=4B^2?
     
  7. BodybldgDoc

    BodybldgDoc Guest

    712
    0
    Jan 13, 2007
    In your closet
    but the answer says a multiplied by 81. How do they get a? The correct answer is e though
     
  8. DDSDakoo

    DDSDakoo 7+ Year Member

    56
    0
    Feb 27, 2007
    Texas
    yeah you are right. clarity at last!
     
  9. DDSDakoo

    DDSDakoo 7+ Year Member

    56
    0
    Feb 27, 2007
    Texas
    ok i "think" i got it.

    so if a= (3b)^4

    then you change the right side of the equation like this when you increase the factor of 3:
    what you do on the right side, you must do on the left.
    a = (3^4)b^4 ; (3^4) being the part u introduced into the equation.
    (3^4)a= 81b^4

    81*a=81b^4
     
  10. BodybldgDoc

    BodybldgDoc Guest

    712
    0
    Jan 13, 2007
    In your closet
    yeah that makes sense. thanks dakoo and street
     
  11. Streetwolf

    Streetwolf Ultra Senior Member Dentist 7+ Year Member

    1,801
    5
    Oct 25, 2006
    NJ
    Take 3b as your tripled value of b. Plug in:

    a_original = b^2 (original)
    a_new = (3b)^2 (b is tripled)
    a_new = 81b^2.

    But we know a_original = b^2 so sub that in:

    a_new = 81*a_original.

    The new value of a is 81 times the original value of a.
     
  12. DDSDakoo

    DDSDakoo 7+ Year Member

    56
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    Feb 27, 2007
    Texas

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