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

VERITAS PREP OFFICIAL SOLUTION:

Our moral duty is to first give you the logical solution since we would like you to think in those terms as far as possible. (Though in this question, plugging in numbers might seem easier.) We will discuss how to get the answer by plugging in appropriate numbers in this case.

Method 1: Logical Solution

The question stem only tells us that x can take any integral value. We need to find whether \(4^x\) is less than \(3^{(x+1)}\).

Consider statement I: x is positive. Given that x is positive, is \(4^x\) less than \(3^{(x+1)}\)? If we put x = 1, it is easy to see that 4^x is less than 3^(x+1). So the inequality holds in this case. The point is how do we prove that this will be true for all positive values of x? It’s tough to prove that something holds for a lot of numbers. It’s easier to show that it doesn’t hold for at least one number since we need only one suitable value in that case.

Question Stem: Is \(4^x < 3^{(x+1)}\)?

Reframe it as: Is \(4^x < 3^x * 3\)?

Or Is \((\frac{4}{3})^x < 3\)?

Note that 4/3 (= 1.33) is greater than 1. When you raise it to a very high power, it will take a very large value. There is no reason it should stay less than 3. Hence the inequality will not hold for large values of x. Hence this statement alone is not sufficient.

Some properties to note:
* When you raise a positive number greater than 1 to a large positive integer power, it takes a large value.
* When you raise 1 to a large positive integer power, it stays 1.
* When you raise a positive number less than 1 to a large positive integer power, the number becomes even smaller than the original value.

These are some number properties you need to work through and be comfortable with.

Consider statement 2: \(|x – 1| < 2\)

Hopefully, you understand modulus well now. We can say that this inequality implies that x is a point at a distance less than 2 from the point 1 on the number line i.e. -1 < x < 3.

Is \((\frac{4}{3})^x < 3\)?

For small values of x e.g. x = 0, we know the inequality holds. Let’s check for only the largest value x can take i.e. 2 since x must be an integer. Even if x were 2, \((\frac{4}{3})^x = \frac{16}{9}\) i.e. less than 2. \((\frac{4}{3})^x\) would still be less than 3. Hence the inequality will hold in this case. This statement alone is sufficient to answer the question.

Note that we did use some basic number plugging here too but that number plugging helps us get a clear picture and makes us ask the right questions. There is nothing wrong with plugging in some numbers here and there to understand the logic. If you know why you are plugging in a particular number, it means you are on the right track. Blindly plugging in is the problem.

2. Putting in Numbers

Let’s look at this method too now.

Is \(4^x < 3^{(x+1)}\)?

Here it’s not easy to find the transition point. We would have to plug in numbers again to find where the two sides of the inequality are equal! So let’s ignore the transition points and directly start plugging in numbers.

Statement 1: x is positive.

Put x = 1, you get 4 < 9 (Holds)

Put x = 2, you get 16 < 27 (Holds)

Put x = 3, you get 64 < 81 (Holds)

What do we do now? How long are we supposed to keep putting in numbers? We cannot do it for all positive integers. How do we decide when to stop? Note that the relative difference between the left hand side and the right hand side is reducing. 9 is more than twice of 4. 27 is more than 16 but not quite twice. It is more than 1.5 times of 16. 81 is somewhat more than 64 but not more than 1.5 times of 64. What this means is that soon enough, the difference will go to 0 and left hand side will become more than the right hand side. If you want to check and you are comfortable with higher powers of numbers,

Put x = 4, you get 256 < 243 (Does not hold)

What you need to do in case the transition point is not apparent is focus on the pattern of the numbers. Is the difference between them narrowing or widening?

This statement alone is not sufficient to answer the question.

Statement II: |x – 1| < 2

As above, we get -1 < x < 3 so this is simply a matter of putting in x = 0, 1 and 2 to see that the inequality holds in each case. Sufficient.

Answer (B)
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Ans is B.

Question stem asks whether 4^x < 3^(x+1)

First analyse under what circumstance will this be true. It will be certainy true for all values (Integers) of x <= 0 and for initial three positive integers (1,2 & 3 only) as illustrated below.

Case 1) If x = -1, 1 / 4 < 3^0....................................True.
Case 2) If x = -4, 1 / (4^4) < 1 / 3^3. ........................True.
Case 3) If x = 0, 4^0 < 3^1......................................True.
Case 4) If x = 2, 4^2 < 3^3......................................True.
Case 5) If x = 4, 4^4 < 3^5 , Is 256< 243, ..No therefore False.

Simillarly it will be false for all positive integer x > 3.

Statement 1. x is positive ----> Refer case 4 & 5, contradictory solutions , Hence not suffucient.

Statement 2. On solving we come to know the range of x. i.e -1 < x < 3.

Accordingly x can be 0,1 or 2. In all three cases the stem holds true. Hence Sufficiant.

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

Ans
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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

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.

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

Ans: B

1) X>0, When X=4. 256<243; X=3 (64>27) - (Insufficient)
2) -1<x<3 Sufficient

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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