What is value of x?

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What is value of x?

by G- Unit30 » Tue Mar 05, 2013 4:18 pm
Hi,
I had this problem on GMATprep practice exam, so I don't have the full list of anwsers but the question is:

What is the value of x?
2^x - 2^(x-2) = 3(2^13)

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by Brent@GMATPrepNow » Tue Mar 05, 2013 4:24 pm
G- Unit30 wrote:Hi,
I had this problem on GMATprep practice exam, so I don't have the full list of anwsers but the question is:

What is the value of x?
2^x - 2^(x-2) = 3(2^13)
This requires some factoring.
2^x - 2^(x-2) = 3(2^13)
2^(x-2)[2^2 - 1] = 3(2^13)
2^(x-2)[3] = 3(2^13)
So, 2^(x-2)= 2^13
x-2 = 13
[spoiler]x = 15[/spoiler]

Cheers,
Brent
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by G- Unit30 » Tue Mar 05, 2013 4:35 pm
Ahh, thanks that helps. That tricky bit of factoring at the start was the part that was confusing me...I wasn't able to think of how to represent 2^x if I pulled out 2^(x-2) as a factor.

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by neeg » Tue Mar 05, 2013 5:17 pm
I'm unclear about the very fist step of factoring
2^(x-2)[2^2 - 1] = 3(2^13)....how is 2^(x-2)a common factor?

This requires some factoring.
2^x - 2^(x-2) = 3(2^13)
2^(x-2)[2^2 - 1] = 3(2^13)
2^(x-2)[3] = 3(2^13)
So, 2^(x-2)= 2^13
x-2 = 13
[spoiler]x = 15[/spoiler]

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by Brent@GMATPrepNow » Tue Mar 05, 2013 5:26 pm
neeg wrote:I'm unclear about the very fist step of factoring
2^(x-2)[2^2 - 1] = 3(2^13)....how is 2^(x-2)a common factor?

Good question.
To answer this, let's examine some analogous factoring cases:

k^5 - k^3 = k^3(k^2 - 1)
m^19 - m^15 = m^15(m^4 - 1)

Or how about:
w^x + x^(x+5) = w^x(1 + w^5)

Notice that, each time, the greatest common factor of both terms is the term with the smallest exponent.

So, in the expression 2^x - 2^(x-2), the term with the smallest exponent is 2^(x-2), so we'll factor out 2^(x-2)

Does that help?

Cheers,
Brent
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by Anurag@Gurome » Tue Mar 05, 2013 9:21 pm
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by neeg » Wed Mar 06, 2013 9:39 am
Brent@GMATPrepNow wrote:
neeg wrote:I'm unclear about the very fist step of factoring
2^(x-2)[2^2 - 1] = 3(2^13)....how is 2^(x-2)a common factor?

Good question.
To answer this, let's examine some analogous factoring cases:

k^5 - k^3 = k^3(k^2 - 1)
m^19 - m^15 = m^15(m^4 - 1)

Or how about:
w^x + x^(x+5) = w^x(1 + w^5)

Notice that, each time, the greatest common factor of both terms is the term with the smallest exponent.

So, in the expression 2^x - 2^(x-2), the term with the smallest exponent is 2^(x-2), so we'll factor out 2^(x-2)

Does that help?

Cheers,
Brent
Thanks Brent.

I understand for the below two as they have common base raised to the same powers and hence can be taken out as the common factor.
k^5 - k^3 = k^3(k^2 - 1)
m^19 - m^15 = m^15(m^4 - 1)

However, for w^x + x^(x+5) = w^x(1 + w^5)-----w^x does not appear in the second term at all so how does it become a common factor?

Similarly for the expression 2^x - 2^(x-2), I understand that smallest exponent is 2^(x-2)but don't see how this is a common factor in the first term which is 2^x?

Neeg

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by neeg » Wed Mar 06, 2013 9:41 am
Thanks - I follow the different appraoches to solving for x provided in the link.

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by Brent@GMATPrepNow » Wed Mar 06, 2013 10:00 am
neeg wrote: Thanks Brent.

I understand for the below two as they have common base raised to the same powers and hence can be taken out as the common factor.
k^5 - k^3 = k^3(k^2 - 1)
m^19 - m^15 = m^15(m^4 - 1)

However, for w^x + x^(x+5) = w^x(1 + w^5)-----w^x does not appear in the second term at all so how does it become a common factor?

Similarly for the expression 2^x - 2^(x-2), I understand that smallest exponent is 2^(x-2)but don't see how this is a common factor in the first term which is 2^x?
First of all, I made a slight error above. It should read: w^x + w^(x+5) = w^x(1 + w^5)

However, it is still true that 2^x - 2^(x-2) = 2^(x-2)[2^2 - 1]

Students typically have how issues with factoring when the exponents are integers. Things become more complicated when the exponents include variables. Nevertheless, it's important to understand how both types behave.

The basic principle is the same: (x^a)(x^b) = x^(a+b)

(w^3)(w^4) = w^(3+4) = w^7
(w^11)(w^8) = w^(11+8) = w^19
Similarly, (w^x)(w^5) = w^(x+5)
We can also say: w^x(1 + w^5) = w^x + w^(x+5)
Finally, we can say: 2^(x-2)[2^2 - 1] = 2^x - 2^(x-2)

Cheers,
Brent
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by J N » Wed Mar 20, 2013 4:21 pm
just an add on for info

does this make sense:

2^10 - 2^6 = whatever

pull out 2^10

2^10 (1 - 1^-4) ????

2^10 (1-1) ??????

=0 ?????


or more relevvant to exapmle

2^x - 2^(x-2) =

pull out 2^x

2^x (1-1^-2)

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by Brent@GMATPrepNow » Wed Mar 20, 2013 5:08 pm
J N wrote:
does this make sense:

2^10 - 2^6 = whatever

pull out 2^10

2^10 (1 - 1^-4) ????

2^10 (1-1) ??????

=0 ?????
The problem occurs in blue.

If you want to factor out 2^10, then 2^10 - 2^6 = 2^10[1 - 2^(-4)]
Notice that (2^10)(1) = 2^10, and (2^10)[2^(-4)] = 2^6
This is no different from saying x^10 - x^6 = x^10[1 - x^(-4)]

Of course, it makes more sense to factor out the 10^6 to get:
2^10 - 2^6 = 2^6(2^4 - 1)
This is the similar to x^10 - x^6 = x^6(x^4 - 1)

Cheers,
Brent
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