Need an idea how to start a problem

[PetarR3]

Hello! Quite new here, actually a couple of minutes, just thought to make a first post and try to get myself started
Edit; Just how do I get the same base here, after that it should be rather simple

 

[Deleted member 4993]

Hello! Quite new here, actually a couple of minutes, just thought to make a first post and try to get myself started
Edit; Just how do I get the same base here, after that it should be rather simple

If I were to do this problem, I would start with:

2^(2x-1) + 4^x = [2^(2x)]/2 + [2^(2x)] = (3/2) * 2^(2x)

continue.....

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem

 

[Harry_the_cat]

What is the whole question here? Are you asked to solve this inequality or are you asked to prove it for specific x?

 

[Steven G]

You mention base case. Does that mean you want to prove this inequality? By induction? Don't you think that you should state this--new or not new on this forum. No one here is a mind reader--ok, one helper here is a mind reader. If you do not state the problem clearly it is hard to help you.

If you want to prove this by induction and need a base case you need to tell us when this inequality is true for! Is it true for x>5, or x>2 or maybe x>=0? It is not fair to require us to figure out what you mean.

3^(2x+1)/2 -3^(2x-3)/2 <= (2^(2x-1) + 4^x

x=0: 3^(2*0+1)/2 -3^(2*0-3)/2 <= (2^(2*0-1) + 4⁰
3^(1)/2 -3^(-3)/2 <= 2^(-1) + 1
3^1/2-1/3^3/2 <= 1/2 + 1 Is this statement true? If yes, then maybe this is your base case

 

[Dr.Peterson]

You mention base case. Does that mean you want to prove this inequality? By induction? Don't you think that you should state this--new or not new on this forum. No one here is a mind reader--ok, one helper here is a mind reader. If you do not state the problem clearly it is hard to help you.

If you want to prove this by induction and need a base case you need to tell us when this inequality is true for! Is it true for x>5, or x>2 or maybe x>=0? It is not fair to require us to figure out what you mean.

3^(2x+1)/2 -3^(2x-3)/2 <= (2^(2x-1) + 4^x

x=0: 3^(2*0+1)/2 -3^(2*0-3)/2 <= (2^(2*0-1) + 4⁰
3^(1)/2 -3^(-3)/2 <= 2^(-1) + 1
3^1/2-1/3^3/2 <= 1/2 + 1 Is this statement true? If yes, then maybe this is your base case

Nothing was said about a base case -- just about a base (as in exponentials).

It's a nice problem. Just expanding each exponential lets you turn the LHS into a multiple of 3^x and the right side into a multiple of 2^x or 4^x, and then it works out nicely. (But not everything will have the same base -- that's a wrong expectation.)

We just need to see some work.

 

[Deleted member 4993]

You mention base case. Does that mean you want to prove this inequality? By induction? Don't you think that you should state this--new or not new on this forum. No one here is a mind reader--ok, one helper here is a mind reader. If you do not state the problem clearly it is hard to help you.

If you want to prove this by induction and need a base case you need to tell us when this inequality is true for! Is it true for x>5, or x>2 or maybe x>=0? It is not fair to require us to figure out what you mean.

3^(2x+1)/2 -3^(2x-3)/2 <= (2^(2x-1) + 4^x

x=0: 3^(2*0+1)/2 -3^(2*0-3)/2 <= (2^(2*0-1) + 4⁰
3^(1)/2 -3^(-3)/2 <= 2^(-1) + 1
3^1/2-1/3^3/2 <= 1/2 + 1 Is this statement true? If yes, then maybe this is your base case

..ok, one helper here is a mind reader

Some of our tutors (helper) cannot even read the original post correctly. Need to go back to middle school - revive "reading comprehension".....