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I used x=1/2,1,2 to plug in.. and I found 1 is sufficient.. but I didn't use 4 to plug in which yields no to the prompt..

How to determine what nos to plug in and when to stop plugging in?

Please help.

Regards, Sach

x is an integer. So firstly 1/2 need not be tested and deciding what numbers to plug in changes from problem to problem. After taking limiting factors into consideration, It is good to try out with a positive integer, a negative integer, a positive fraction, a negative fraction and 0.

In this case,

Is \(4^x < 3*3^x\)

x=1, 4 < 9 x=2, 16 < 27 x=3, 64 < 81

So we can see that the LHS is catching up to RHS with each increase in x. So for higher values of x, LHS > RHS. So the statement is not sufficient.

Kudos Please... If my post helped.
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ok.. so you mean to say that since lhs is catching up with rhs, it would eventually be greater than rhs..

Yes... We just have to recognize the pattern. Also in this question it would help to notice that with every increase in x, the LHS is increasing by a factor of 4 whereas, RHS is only increasing by a factor of 3. of course, that is only a restatement of my previous post.

Kudos Please... If my post helped.
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I used x=1/2,1,2 to plug in.. and I found 1 is sufficient.. but I didn't use 4 to plug in which yields no to the prompt..

How to determine what nos to plug in and when to stop plugging in?

Please help.

Regards, Sach

This will come through practice.

The good thing is - there are n number of ways to solve a question. the bad thing is - you need to find the fastest. There is no need to plug a 100 numbers in this case. You just need to find whether the situation holds true for every positive x or not. Your first step should always be to reduce the number of ' numbers you need to plug in' .

First, I'd suggest you put an '=' instead of the </> sign. so essentially 4^x = 3^x * 3.

This is one way of quickly finding what number to plug. Find the value of x that satisfies the equation. This is your critical point. Use a few values to the left of this and a few to the right.

so is 4^x<3^x*3 or (4/3)^x<3 (because 3^x is always positive you can bring it to the LHS) or (1.33)^x<3

for 1 it is true for 10000000 it is not. hence, the statement is insufficient.

I used x=1/2,1,2 to plug in.. and I found 1 is sufficient.. but I didn't use 4 to plug in which yields no to the prompt..

How to determine what nos to plug in and when to stop plugging in?

Please help.

Regards, Sach

There are two ways you can use:

1. Use logic/algebra instead of plugging in numbers. It's tough to prove something using numbers. It's easier to disprove.

Is 4^x < 3^(x+1)? Is (4/3)^x < 3?

4/3 is greater than 1. When you raise it to a high power, it will take a big enough value. There is no reason it should stay less than 3. When you raise 1 to a power, it stays 1. When you raise a number less than 1 to a positive integral power, the number becomes even lesser. These are some number properties you need to work through and be comfortable with.

2. Look for the transition point.

Find out where it will be equal. Then the behavior will be different on either side of the transition point (as done in the post above)

Remember, you cannot shut your mind and just plug in numbers and get your answer. If it were so easy, everyone would have scored Q51. The point is that you need to always keep your mind on alert and use logic even if you are using number plugging.
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