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)
Thanks!
What is value of x?
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This requires some factoring.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)
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
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]
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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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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Refer the following posts...
https://www.beatthegmat.com/2-x-2-x-2-3- ... tml#577790
https://www.beatthegmat.com/2-x-2-x-2-3- ... tml#577789
https://www.beatthegmat.com/2-x-2-x-2-3- ... tml#577790
https://www.beatthegmat.com/2-x-2-x-2-3- ... tml#577789
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Brent@GMATPrepNow wrote:Thanks Brent.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
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
Thanks - I follow the different appraoches to solving for x provided in the link.Anurag@Gurome wrote:Refer the following posts...
https://www.beatthegmat.com/2-x-2-x-2-3- ... tml#577790
https://www.beatthegmat.com/2-x-2-x-2-3- ... tml#577789
Neeg
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First of all, I made a slight error above. It should read: w^x + w^(x+5) = w^x(1 + w^5)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?
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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The problem occurs in blue.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 ?????
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