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(1) \(x^{x+3}=(2x)^{x-1}\) \(x^x * x^3 = 2^{x-1} * x^{-1} * x^x\) Assuming x is not 0, we can cancel x^x out \(x^4 = 2^{x-1}\) The only solution to this is x=1. 1 has only one factor Sufficient

(2) x>=7/3 then 3x-7=2x+2, x=9 x<7/3 then 3x-7=-2x-2 , x=1 x can have 1 factor or 3 factors Insufficient

how can the answer be A wont x = 0 satify equation no 1 ?...expert plz reply !!

If \(x=0\) in (1) then the left hand side of the equation becomes \(0^{-1}=\frac{1}{0}=undefined\), (while the right hand side becomes \(0^3=0\)). Remember you cannot raise zero to a negative power.

(2) \(|3x-7| = 2x+2\) --> \(x=1\) or \(x=9\). Not sufficient.

Answer: A.

\(x^4=2^{(x-1)}\) --> \(x=1\)---------How do we arrive at this. How to solve these????? just by testing values vaguely or is there any particular mathematics we can do with this?????

(2) \(|3x-7| = 2x+2\) --> \(x=1\) or \(x=9\). Not sufficient.

Answer: A.

\(x^4=2^{(x-1)}\) --> \(x=1\)---------How do we arrive at this. How to solve these????? just by testing values vaguely or is there any particular mathematics we can do with this?????

Number plugging is the best way to solve \(x^4=2^{(x-1)}\):

Now, if x>1 (2, 3, ...), then the left hand side of the equation is always greater than the right hand side of the equation: 2^4=16>2^1=2, 3^4=81>2^2=4, ... (as you can see LHS increases "faster" than RHS).

Hence, x=1 is the only positive integer solution of \(x^4=2^{(x-1)}\).

\(x^4=2^{(x-1)}\) --> \(x=1\)---------How do we arrive at this. How to solve these????? just by testing values vaguely or is there any particular mathematics we can do with this?????

I believe Bunuel has already explained this. Just another way to look at it is this:

We end up with: \(x^4=2^{(x-1)} \to x = 2^{\frac{(x-1)}{4}}\) .

We know that if x is an integer, the power of 2,i.e. \(\frac{x-1}{4}\) has to be a non-negative integer.

I.\(\frac{x-1}{4}>0\) As x equals some positive power of 2, x is even.Also, the only way x can be an integer is if the exponent of 2 is raised to an integer\(\to\) x is a multiple of 4.

If (x-1) is a multiple of 4, then x has to be odd and this contradicts out previous statement, that x is even.

II. \(\frac{x-1}{4}=0\)\(\to x=1\) and this satisfies the equality.
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Re: How many factors does the integer X have? [#permalink]

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10 Jul 2013, 11:37

How many factors does the integer x have?

(1) x^(x+3) = (2x)^(x-1) x^(x+3) = (2x)^(x-1) x^(x+3) = x^(x-1) 2^(x-1) (x^(x-1)) / x^(x+3) = x^(x-1) 2^(x-1) / (x^(x-1)) x^(x+3)-(x-1) = 2^(x-1) x^(4) = 2^(x-1) (This cannot be simplified any more. Thus, look for any value of x that will satisfy the equation. x=1 is the only value that will make x^(4) = 2^(x-1) valid) SUFFICIENT

(2) |3x-7| = 2x+2

x≥7/3, x<7/3

x≥7/3 |3x-7| = 2x+2 (3x-7) = 2x+2 x=9 Valid

x<7/3 |3x-7| = 2x+2 -(3x-7) = 2x+2 -3x+7 = 2x+2 5=5x x=1 Valid We have two valid x values. INSUFFICIENT

(A)

I'm not entirely sure why b) is insufficient. The stem doesn't tell us how many solutions we are looking for, it just asks for the number of factors. Isn't saying "number 2 has two factors" just as valid as saying "number one has one factor"?

(1) x^(x+3) = (2x)^(x-1) x^(x+3) = (2x)^(x-1) x^(x+3) = x^(x-1) 2^(x-1) (x^(x-1)) / x^(x+3) = x^(x-1) 2^(x-1) / (x^(x-1)) x^(x+3)-(x-1) = 2^(x-1) x^(4) = 2^(x-1) (This cannot be simplified any more. Thus, look for any value of x that will satisfy the equation. x=1 is the only value that will make x^(4) = 2^(x-1) valid) SUFFICIENT

(2) |3x-7| = 2x+2

x≥7/3, x<7/3

x≥7/3 |3x-7| = 2x+2 (3x-7) = 2x+2 x=9 Valid

x<7/3 |3x-7| = 2x+2 -(3x-7) = 2x+2 -3x+7 = 2x+2 5=5x x=1 Valid We have two valid x values. INSUFFICIENT

(A)

I'm not entirely sure why b) is insufficient. The stem doesn't tell us how many solutions we are looking for, it just asks for the number of factors. Isn't saying "number 2 has two factors" just as valid as saying "number one has one factor"?

When a DS question asks about the value, then the statement is sufficient ONLY if you can get the single numerical value.

From (2) we have that x=1 or x=9. If x=1, then the answer is 1 (1 has 1 factor) but if 9, then the answer is 3 (9 has 3 factors), so we have TWO different answers to the question. Therefore this statement is NOT sufficient.

Re: How many factors does the integer X have? [#permalink]

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30 Sep 2014, 11:10

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Re: How many factors does the integer X have? [#permalink]

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14 Jul 2016, 11:56

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