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Re: If x is an integer, is 4^x < 3^(x+1)? [#permalink]
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Bunuel wrote:
If x is an integer, is 4^x < 3^(x+1)?

(1) x is positive
(2) |x – 1| < 2


4^x < 3^(x+1)

(1) x is positive
4^1 < 3^2
4^4>3^5
INSUFFICIENT

(2) |x – 1| < 2
=> 0<= X <=2 because x is integer
=> 4^x < 3^(x+1) when x = 0, 1, 2
=> SUFFICIENT

Ans: B
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Re: If x is an integer, is 4^x < 3^(x+1)? [#permalink]
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If x is an integer, is 4^x < 3^(x+1)?

(1) x is positive
(2) |x – 1| < 2

(1) x is positive

x = 1
4 < 9 true then yes

and start increasing the value of x --- when x = 4

see the 4^x < 3^(x+1) is not valid ... hence not sufficient


(2) |x – 1| < 2

x -1 < 2
x < 3

when x is 2 or negative value then the inequality 4^x < 3^(x+1) holds true .

|x – 1| < 2
-x + 1 < 2
-x < 1
x > -1

for x equals 0 or more then then the inequality 4^x < 3^(x+1) holds true .

hence B sufficient

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Re: If x is an integer, is 4^x < 3^(x+1)? [#permalink]
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If x is an integer, is 4^x < 3^(x+1)?

(1) x is positive
(2) |x – 1| < 2

There is one variable in this question, and 2 equations from the conditions provided, meaning that the answer is likely to be (D). Looking at the conditions closely,
For condition 1, the answer is ‘yes’ for x=1, as 4<9, but ‘no’ for x=5, as 4^5=1,024>3^6=729 Hence, the condition is insufficient.
Looking at condition 2, the answer is ‘yes’ for all cases when x=0,1,2, so the condition is sufficient, and the answer becomes (B).

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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Re: If x is an integer, is 4^x < 3^(x+1)? [#permalink]
common logarithm helps a lot:

4^x<3^(x+1)

so lg4^x<lg3^(x+1)

so x<=4?

(1) x>0

insufficient

(2) |x-1|<2

-1<x<3

sufficient

B
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Re: If x is an integer, is 4^x < 3^(x+1)? [#permalink]
Bunuel wrote:
If x is an integer, is 4^x < 3^(x+1)?

(1) x is positive
(2) |x – 1| < 2


Ans: B

1) X>0, When X=4. 256<243; X=3 (64>27) - (Insufficient)
2) -1<x<3 Sufficient
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Re: If x is an integer, is 4^x < 3^(x+1)? [#permalink]
Bunuel wrote:
If x is an integer, is 4^x < 3^(x+1)?

(1) x is positive
(2) |x – 1| < 2



Method 1: Logical Solution


x=int
4^x < 3^x.3? => (4/3)^x < 3?

1) x=1 4/3 < 3 => Yes
x = 3 64/27 < 3 => Yes
x = 4 256/81 = 3.1 => No
Not Sufficient


2) |x - 1| < 2 => -1 < x < 3
x could be 0, 1, 2
x=0, 1 < 3 => Yes
x=1 4/3 < 3 => Yes
x=2 16/9 < 3 => Yes
Sufficient => ANSWER: B


Method 2: Plugging in Numbers


See the test cases for stmt 1 in Method 1
You’ll notice that the gap keeps decreasing.
Hence, we can infer that our answer will change from Yes to No.


This Q has been explained here: https://gmatclub.com/forum/inequalities ... l#p1582969
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Re: If x is an integer, is 4^x < 3^(x+1)? [#permalink]
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