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exponent problem- correct rule for exp?

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exponent problem- correct rule for exp?

by kapitalist » Sat Mar 14, 2009 3:44 pm
For all real numbers V, the operation V* is defined by the equation

V* = V - (V/3)

If (V*)* = 8, then V = ?

Is this correct solution?

V** = V^2* or (V*)^2

(2v/3)^2 = 4V/9

4v/9=8, v=18

(lower and capital letter are the same, just typed fast!)

but if you square a fractioned variable...do you square the variable too??

is (2v/3)^2 equal to (4v^2/9) or is it just (4v/9)????
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by Feep » Sat Mar 14, 2009 4:03 pm
If there is a variable within the parenthesis, you would square it, yes. But you're operating on incorrect assumptions.

When you do an operation twice, it is not accurate to say that you "do the operation, squared." If V* = V+2, (V*)* would not equal (V+2)^2. You simply do the operation twice.

As you correctly reasoned, what V* really means to to get (2/3)*V, or in other words, multiply V by (2/3). You do this twice, resulting in V * (2/3) * (2/3) = 8, which you did obtain (via shady methods), and thus the answer is indeed V = 18.
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by kapitalist » Sat Mar 14, 2009 4:38 pm
but what I dont understand is if v* means v(2/3) and your doing it twice, how come its not v(2/3) x v(2/3) which would be v^2(4/9) again?
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by Feep » Sat Mar 14, 2009 11:27 pm
Because the second * operation is not on V. It's on V*, which is (2/3*V).

If 2v means 2*v (and, uh, it does), what is 2(2v)? This is the exact same thing. You do not introduce a new variable. You merely take the necessary actions on what is inside the parenthesis.
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Correct Answer

by gmat740 » Sun Mar 15, 2009 12:15 am
@kapitalist
V** = V^2* or (V*)^2

This is just an Operation denoted by *, which happens to be in eponential power in this case.

Take a look into this:

7% 5 =35

Here the % is an operator.

So in this case its not doing its usual function,rather it is multiplying 7 and 5

So if a question ask you like this:

4% 100=?

You are no going to say its 4 percent of 100
Rather your answer will have to be 400.

Hope this makes sense now.

V* = V - (V/3)

If (V*)* = 8, then V = ?

Here V*= V-(V/3)

so When (V*)* = [V-(V/3)]*

==> now V-(v)/3 assumes the value of V

==> [V-(V/3)]*= [(V-V/3)-V/3]

==> V-(2/3V)
==> V/3

and this value is given to be 8

This V/3 =8
So, v=24

@Feep

Please check your answer
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by avenus » Sun Mar 15, 2009 3:09 am
so When (V*)* = [V-(V/3)]*

==> now V-(v)/3 assumes the value of V

==> [V-(V/3)]*= [(V-V/3)-V/3]
wrong...

if V-(V/3) assumes the value of V as you say (it would be easier to simplifly to 2V/3 in advance by the way) then:

[V-(V/3)]* = [(V-(V/3)) - (V-(V/3))/3] = 2V/3 - 2V/9 = 4V/9

and hence V = 18

so feep is right
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Yes you are correct

by gmat740 » Sun Mar 15, 2009 3:23 am
Thanks avenus for the rectification,
this is the problem with me.I always end up committing a calculation mistake
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