# algebraic problem

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

1. ### BodybldgDoc Guest

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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

3. ### DDSDakoo

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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

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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 Ultra Senior Member Dentist

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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 Guest

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If A=(2B)^2 then wouldnt it be A=4B^2?

7. ### BodybldgDoc Guest

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but the answer says a multiplied by 81. How do they get a? The correct answer is e though

8. ### DDSDakoo

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yeah you are right. clarity at last!

9. ### DDSDakoo

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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 Guest

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yeah that makes sense. thanks dakoo and street

11. ### Streetwolf Ultra Senior Member Dentist

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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.

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