Simplification and solving of equation

[chikis]

Homework Statement

The problem are two:
(1) Simplify: 1/4(2ⁿ- 2ⁿ⁺²)

(2) Solve the equation 2^(2x+1)- 9(2^x) + 4 = 0

Homework Equations

The Attempt at a Solution

for the number one question, simplify:
1/4(2ⁿ- 2ⁿ⁺²)

1/4(2ⁿ- 2ⁿ⁺²)
= 4^-1(2ⁿ-2ⁿ⁺²)
= 2^(-2)(2ⁿ-2ⁿ⁺²)
= 2^(-2)(2ⁿ-2ⁿ*2²
I can't go further please I need asistance.

For the second question, solve the equation:
2^(2x+1)- 9(2^x) + 4 = 0

I don't just know how to start dealing with the question. It is really a troubsome question.

 

[Villyer]

I'm confused as to your original equations.

$$(1)\frac{1}{4}2^n-2^{n+2}$$
$$(2)\ 2^{2x+1} - 9(2^x + 4)$$

Are those correct?

 

[Ray Vickson]

Homework Statement

The problem are two:
(1) Simplify: 1/4(2n[sup/]- 2n+2[sup/])

(2) Solve the equation 2(2x+1)- 9(2x[sup/] + 4 = 0

Homework Equations

The Attempt at a Solution

for the number one question, simplify:
1/4(2<sup>n[sup/]- 2n+2[sup/])

1/4(2n[sup/]- 2n+2[sup/])
= 4-1[sup/](2n[sup/]-2n+2[sup/])
= 2(-2)[sup/](2n[sup/]-2n+2[sup/])
= 2(-2)[sup/](2n[sup/]-2n[sup/]*22[sup/]
I can't go further please I need asistance.

For the second question, solve the equation:
2(2x+1)- 9(2x[sup/] + 4 = 0
I don't just know how to start dealing with the question. It is really a troubsome question.</sup>

<sup>Use [S U P] and [/S U P] (but with no spaces), not [sup ] and [ /sup]. In fact, just use the "X2" button on the palette at the top in the input panel.

RGV</sup>

 

[chikis]

Use [S U P] and [/S U P] (but with no spaces), not [sup ] and [ /sup]. In fact, just use the "X²" button on the palette at the top in the input panel.

RGV

I have edited the problem so that it can be seen clearly.

 

[chikis]

I'm confused as to your original equations.

$$(1)\frac{1}{4}2^n-2^{n+2}$$
$$(2)\ 2^{2x+1} - 9(2^x + 4)$$

Are those correct?

I have edited my opening post so that it can be seen clearly. Maybe you can look at the problem again.

 

[Villyer]

For (1), you are close. Combine like terms.

For (2), it is actually a quadratic. Can you see what the variable is?

 

[chikis]

For (1), you are close. Combine like terms.

If like terms is collected, the expression will look like this:
2^(-2)*2²(2ⁿ-2ⁿ)
= 2(2ⁿ-2ⁿ)
= 2ⁿ⁺¹-2ⁿ

For (2), it is actually a quadratic. Can you see what the variable is?

No! I didn't see the variable.

 

[Mentallic]

If like terms is collected, the expression will look like this:
2^(-2)*2²(2ⁿ-2ⁿ)

This does not follow from the last line you had written in your opening post:

= 2^(-2)(2ⁿ-2ⁿ⁺²)
= 2^(-2)(2ⁿ-2ⁿ*2²

What we have so far is

2^{-2}(2^n-2^n\cdot 2^2)

Can you expand this?

No! I didn't see the variable.

There is a basic rule of exponentiation that says that if you exponentiate an exponent, you multiply the exponents together. What this is saying is that
(a^m)^n=a^{mn}

So then what does (2^x)^2 simplify to?

 

[eumyang]

If like terms is collected, the expression will look like this:
2^(-2)*2²(2ⁿ-2ⁿ)
= 2(2ⁿ-2ⁿ)
= 2ⁿ⁺¹-2ⁿ

Nope. Look at the last step in your original post:

1/4(2ⁿ- 2ⁿ⁺²)

1/4(2ⁿ- 2ⁿ⁺²)
= 4^-1(2ⁿ-2ⁿ⁺²)
= 2^(-2)(2ⁿ-2ⁿ⁺²)
= 2^(-2)(2ⁿ-2ⁿ*2²)

Answer this: what is x - 4x?
Now answer this: what is 2ⁿ - 4*2ⁿ?
That's what Villyer meant by combining terms.

No! I didn't see the variable.

2^{2x+1} - 9\cdot 2^x + 4
Rewrite the 1st term so it looks like (something) times 2^2x.EDIT: Beaten to it.

 

[chikis]

I don't understand what you guys are writing, write using the same format I used in showing my raise to power in my opening post. I will understand it better and therefore be able to respond to the numerous replies you guys are posting.

 

[Mentallic]

I don't understand what you guys are writing, write using the same format I used in showing my raise to power in my opening post. I will understand it better and therefore be able to respond to the numerous replies you guys are posting.

You can't see Latex? I think you should try find out and fix what the problem is on your end, because Latex is commonly used on this forum and not just in this thread.

 

[chikis]

You can't see Latex? I think you should try find out and fix what the problem is on your end, because Latex is commonly used on this forum and not just in this thread.

Please help me, I believe I would be able to understand and write using latex with time. Please help me, I really need this help please!

 

[Mentallic]

Please help me, I believe I would be able to understand and write using latex with time. Please help me, I really need this help please!

I'm not asking you to learn to use latex so suddenly, but rather I'm confused as to why you can't read what we've written. Correct me if I'm wrong, but I think that Latex is software that's inbuilt into this forum, so basically everyone can read it (unless they have some horribly outdated internet browser).

Anyway, I'll edit what emuyang and I wrote in the format you've requested.

This does not follow from the last line you had written in your opening post:What we have so far is

2^-2*(2ⁿ - 2ⁿ*2²)

Can you expand this?There is a basic rule of exponentiation that says that if you exponentiate an exponent, you multiply the exponents together. What this is saying is that
(a^m)ⁿ=a^mn

So then what does (2^x)² simplify to?

Nope. Look at the last step in your original post:

Answer this: what is x - 4x?
Now answer this: what is 2ⁿ - 4*2ⁿ?
That's what Villyer meant by combining terms.2^2x+1 - 9*2^x + 4
Rewrite the 1st term so it looks like (something) times 2^2x.EDIT: Beaten to it.

EDIT: Changed the ^ to [sup ] tags

 

[Ray Vickson]

Please help me, I believe I would be able to understand and write using latex with time. Please help me, I really need this help please!

You have been given lots of help---any more help and we would be doing your homework for you. If you are having trouble understanding what people write you need to sit down with a pencil and several sheets of paper and start writing things down in whatever notation you like.

RGV

 

[chikis]

You have been given lots of help---any more help and we would be doing your homework for you. If you are having trouble understanding what people write you need to sit down with a pencil and several sheets of paper and start writing things down in whatever notation you like.

RGV

I did not ask you to do my home work. What I ask is this: give me directions that will help me do the homework. That's all that I ask, anything far from that then you are doing my homework. I know that you are giving me some asistance, but how can I work with the language I don't understand. I know that apart from latex, there are other formats that can be used and I will be able to understand all the replies you are posting and be able to respond accordingly.

 

[Ray Vickson]

I did not ask you to do my home work. What I ask is this: give me directions that will help me do the homework. That's all that I ask, anything far from that then you are doing my homework. I know that you are giving me some asistance, but how can I work with the language I don't understand. I know that apart from latex, there are other formats that can be used and I will be able to understand all the replies you are posting and be able to respond accordingly.

Here is what one responder said to you:

"Answer this: what is x - 4x?
Now answer this: what is 2ⁿ - 4*2ⁿ? "

So, what are your answers? Do you see why the answers are important?

RGV

 

[chikis]

Here is what one responder said to you:

"Answer this: what is x - 4x?

Is just an expression an algebraic expression. If factored it will look like this: x(1-4)
= x(1-2)(2+1)

Now answer this: what is 2ⁿ - 4*2ⁿ? "

So, what are your answers? Do you see why the answers are important?

RGV

That is 2 raise power n minus 4 times 2 raise to power n. It can be simplified further to appear like this: 2ⁿ - 2² * 2ⁿ
= 2ⁿ/2²⁺ⁿ
= 2^n-(2+n)
= 2^n-2-n

 

[Mark44]

Is just an expression an algebraic expression. If factored it will look like this: x(1-4)
= x(1-2)(2+1)

But neither of these is the simplest way to write it. It's not very useful to factor 1 - 4.

That is 2 raise power n minus 4 times 2 raise to power n. It can be simplified further to appear like this: 2ⁿ - 2² * 2ⁿ

The above is not wrong, but it's not helpful to write 4 as 2².
Your work below here is incorrect.

= 2ⁿ/2²⁺ⁿ
= 2^n-(2+n)
= 2^n-2-n

These two problems are similar, and you are making them much more complicated than they deserve.

 

[chikis]

But neither of these is the simplest way to write it. It's not very useful to factor 1 - 4.
The above is not wrong, but it's not helpful to write 4 as 2².
Your work below here is incorrect.


These two problems are similar, and you are making them much more complicated than they deserve.

What else can I do then?

 

[SammyS]

What else can I do then?

1 - 4 = -3, Correct ?

 

[chikis]

1 - 4 = -3, Correct ?

Oh no! You just reminded me that x-4x is also -3x. Anyway thanks for that. The question is what else can I do?

 

[Mark44]

So now what is 2ⁿ - 4*2ⁿ in its simplest form?

 

[chikis]

So now what is 2ⁿ - 4*2ⁿ in its simplest form?

It can be expressed as: 2ⁿ-2²*2ⁿ
= 2ⁿ-2²*2ⁿ
= 2ⁿ/2²⁺ⁿ
= 2^n-2-n
= 2²
= 4

 

[Mark44]

It can be expressed as: 2ⁿ-2²*2ⁿ
= 2ⁿ-2²*2ⁿ
= 2ⁿ/2²⁺ⁿ

No, there's no property of exponents that let's you do this (above).

= 2^n-2-n
= 2²
= 4

Start with this: 2ⁿ-2²*2ⁿ, and factor out 2ⁿ.

 

[chikis]

No, there's no property of exponents that let's you do this (above).


Start with this: 2ⁿ-2²*2ⁿ, and factor out 2ⁿ.

what I can make out from: 2ⁿ-2²*2ⁿ
Is that 2 is common here, but how to factorise 2ⁿ out, I don't know because am not use to that kind of factorisation where the whole terms is raised to a power, different powers for that matter.

 

[SammyS]

what I can make out from: 2ⁿ-2²*2ⁿ
Is that 2 is common here, but how to factorise 2ⁿ out, I don't know because am not use to that kind of factorisation where the whole terms is raised to a power, different powers for that matter.

2ⁿ is in common .

Let x = 2ⁿ. then you have x - (2²)*x .

What does that simplify to?

 

[chikis]

2ⁿ is in common .

Let x = 2ⁿ. then you have x - (2²)*x .

What does that simplify to?

How can 2ⁿ be common, when the whole three terms are raised different powers? What about the other 2 that is raised to power two?
If x = 2ⁿ
----> x - (2²)*x .

= x - 4x
= -3x

 

[SammyS]

...

If x = 2ⁿ
----> x - (2²)*x .

= x - 4x
= -3x

Right.

And if x = 2ⁿ , that gives you

-3x → (-3) (2ⁿ).​

 

[gabbagabbahey]

How can 2ⁿ be common, when the whole three terms are raised different powers? What about the other 2 that is raised to power two?

This is a simple application of a basic property of multiplication, called the distributive property. Surely you must be familiar with this property, right?

A(B+C)=A*B+A*C

Applying this property with A=2ⁿ B=1 and C=-2², and recognizing that 2ⁿ=2ⁿ*1, gives you 2ⁿ-2²*2ⁿ=2ⁿ(1-2²). This is what is meant by "factoring out" 2ⁿ.

 

[Ray Vickson]

How can 2ⁿ be common, when the whole three terms are raised different powers? What about the other 2 that is raised to power two?
If x = 2ⁿ
----> x - (2²)*x .

= x - 4x
= -3x

Congratulations. You managed too have others complete your homework for you. Well done.

RGV

 

[Mentallic]

Congratulations. You managed too have others complete your homework for you. Well done.

RGV

Give the guy a break, he's been struggling with applying a simple factorization. Dragging him along for days to answer such a brief question is probably not the best approach here.
He's also taken multiple attempts at the answer thus far.

 

[Ray Vickson]

Give the guy a break, he's been struggling with applying a simple factorization. Dragging him along for days to answer such a brief question is probably not the best approach here.
He's also taken multiple attempts at the answer thus far.

I genuinely believe that the best way for him to learn is to get a mark of zero on that problem and to then study the posted solutions after getting his paper back. That way he might actually, finally, absorb and *understand* the material.

RGV

 

[chikis]

Congratulations. You managed too have others complete your homework for you. Well done.

RGV

Nothing has been completed yet because I cannot work with what I don't understand. So the thread is still going on.

 

[chikis]

Give the guy a break, he's been struggling with applying a simple factorization. Dragging him along for days to answer such a brief question is probably not the best approach here.
He's also taken multiple attempts at the answer thus far.

Nothing has been completed yet because I cannot work with what I don't understand. So the thread is still going on.

 

[chikis]

I genuinely believe that the best way for him to learn is to get a mark of zero on that problem and to then study the posted solutions after getting his paper back. That way he might actually, finally, absorb and *understand* the material.

RGV

You want me to fail? Then you are not helping matters.

 

[Mark44]

We are 36 posts into this thread, most of which has been spent on a fairly simple problem: simplifying (1/4)(2ⁿ - 4*2ⁿ).

What part of the explanation did you not understand?

 

[chikis]

This is a simple application of a basic property of multiplication, called the distributive property. Surely you must be familiar with this property, right?

A(B+C)=A*B+A*C

Applying this property with A=2ⁿ B=1 and C=-2², and recognizing that 2ⁿ=2ⁿ*1, gives you 2ⁿ-2²*2ⁿ=2ⁿ(1-2²). This is what is meant by "factoring out" 2ⁿ.

Yes I know that A(B +C) =A*B+A*C
A=2ⁿ B=1 and C=-2²
How can those things work when you give me 2ⁿ - 2² *2ⁿ
when I can't even figure out how 2ⁿ is a common factor. I need proof it, proove it.

 

[Mark44]

Yes I know that A(B +C) =A*B+A*C
A=2ⁿ B=1 and C=-2²
How can those things work when you give me 2ⁿ - 2² *2ⁿ
when I can't even figure out how 2ⁿ is a common factor. I need proof it, proove it.

In this expression -- 2ⁿ - 2² *2ⁿ -- there are two[/color] terms.

What are the two terms?

Each term has a factor of 2ⁿ. This makes 2ⁿ a common factor (both terms have this factor in common).

 

[chikis]

We are 36 posts into this thread, most of which has been spent on a fairly simple problem: simplifying (1/4)(2ⁿ - 4*2ⁿ).

What part of the explanation did you not understand?

I think you missed it. Is not (1/4)(2ⁿ - 4*2ⁿ). It is 1/4(2^n-2^(n+2)). The problem maybe simple to you but harder to me. Is not the number of post that matters. What matters is this, "do I understand what you are trying to point out?"

 

[chikis]

In this expression -- 2ⁿ - 2² *2ⁿ -- there are two[/color] terms.

What are the two terms?

Each term has a factor of 2ⁿ. This makes 2ⁿ a common factor (both terms have this factor in common).

Yes, now I understand. There are two terms, and 2ⁿ is common to the two terms. The mistake I made at first, that made it difficult for me to see that 2ⁿ is common, is that I was seeing the expression as three terms, failing to recognise that the multiplication sighn between - 2² and 2ⁿ has joined the two terms and made them a single term. Anyway, thanks for bringing me back from the wilderness where am lost. Shall we procced?

 

[chikis]

Lets move on... 2ⁿ is a common factor for the expression: 2^-2(2ⁿ-2ⁿ*2²)
Factoring we have:
2ⁿ(2^-2-2²)
= 2ⁿ(2^-2/2²)
= 2ⁿ(2^-2-2)
= 2^n-4
I need asistance please.

 

[gabbagabbahey]

I think you missed it. Is not (1/4)(2ⁿ - 4*2ⁿ). It is 1/4(2^n-2^(n+2)).

These two expressions are mathematically equivalent though. To see this, you need to know only 4 basic concepts (and these are very important concepts for you to learn, practice, and remember):

(1) The property of exponentiation that says a^b+c = a^b*a^c. This concept/property allows you to conclude that 2ⁿ⁺²=2ⁿ*2². Do you understand this part?

(2) The fact that 2²=4. This allows you to conclude that 2ⁿ*2²= 2ⁿ*4. Do you understand this part?

(3) The basic property of multiplication called commutativity, which says a*b=b*a. This allows you to conclude that 2ⁿ*4=4*2ⁿ. Do you understand this part?

(4) The concept of algebraic substitution that tells you if a=b and a=c, then b=c. This allows you to conclude that

(1/4)(2ⁿ-2ⁿ⁺²)
= (1/4)(2ⁿ-2ⁿ*2²)
= (1/4)(2ⁿ-2ⁿ*4)
= (1/4)(2ⁿ-4*2ⁿ)

Do you understand this part?

The problem maybe simple to you but harder to me. Is not the number of post that matters. What matters is this, "do I understand what you are trying to point out?"

I agree. I think we are all trying to help increase your understanding of this subject matter.

Yes I know that A(B +C) =A*B+A*C
A=2ⁿ B=1 and C=-2²
How can those things work when you give me 2ⁿ - 2² *2ⁿ
when I can't even figure out how 2ⁿ is a common factor. I need proof it, proove it.

Normally, we do not give out proofs/solutions to homework problems, but seeing as you've made multiple efforts and still have some misunderstanding which I have yet to identify, I will show you how to prove this part of your problem in the hopes of identifying exactly where your misunderstanding(s) are.

(1) Recognize that the expression 2ⁿ - 2² *2ⁿ has two terms; 2ⁿ and -2² *2ⁿ

Do you understand this part?

(2) Apply the identity element property of multiplication (see here for a list of multiplication properties) to the first term (2ⁿ) to get

2ⁿ - 2² *2ⁿ = 2ⁿ*1 - 2² *2ⁿ

Do you understand this part?

(3) Apply the commutativity property to the second term to conclude that 2ⁿ - 2² *2ⁿ = 2ⁿ*1 - 2ⁿ *2².

Do you understand this part?

(4) Recognize that the distributive property tells you that 2ⁿ(1-2²)=2ⁿ*1 - 2ⁿ *2² and use the algebraic substitution property to conclude that 2ⁿ - 2² *2ⁿ=2ⁿ(1-2²).

Do you understand this part?

 

[gabbagabbahey]

Lets move on... 2ⁿ is a common factor for the expression: 2^-2(2ⁿ-2ⁿ*2²)
Factoring we have:
2ⁿ(2^-2-2²)
= 2ⁿ(2^-2/2²)
= 2ⁿ(2^-2-2)
= 2^n-4
I need asistance please.

You've made mistakes in both your 1st and 2nd steps here.

2^-2(2ⁿ-2ⁿ*2²)=2^-2*2ⁿ(1-2²)≠2ⁿ(2^-2-2²)

2^-2-2²≠2^-2/2²

 

[Ray Vickson]

You want me to fail? Then you are not helping matters.

No, I want you to pass, but I think you are going about it in the wrong way. If you are having so much trouble with a simple factorization problem you are going to need more help than you can get here. You need to go visit your instructor, or tutor, and explain your difficulties to him/her, and try to get personal help, face-to-face with a real, live person. If you can afford it you should consider hiring a tutor to help you understand the needed concepts and to try to help you over your apparent "blocks" in basic understanding.

RGV

 

[Mark44]

I concur with Ray. Also, regarding the number of posts in this thread, which is something that I mentioned, if you are still struggling with factoring, and exponents, this might indicate that you don't understand material that is prerequisite to what you're working on. Unlike many other subjects, math builds on preceding topics. If you don't have a good understanding of the topics that came before, you will really have tough time on later topics that use those concepts.

 

[Villyer]

I think it will help you to stop writing 4 as 2² and simplify it the 2ⁿ terms, and then go back and deal with the 4.

 

[chikis]

You've made mistakes in both your 1st and 2nd steps here.

2^-2(2ⁿ-2ⁿ*2²)=2^-2*2ⁿ(1-2²)≠2ⁿ(2^-2-2²)

2^-2-2²≠2^-2/2²

Yes, I made a mistake. What that implies is that, is not only 2ⁿ is common. Both 2ⁿ and 2^-2 are common to the two terms:
Working with that in mind we have:
2ⁿ*2^-2-2ⁿ*2^-2*2²
= 2^n-2-2ⁿ
= 2^n-2/2ⁿ
= 2^n-2-n+1
= 2^-1
= 1/2

 

[Mentallic]

= 2^n-2-2ⁿ
= 2^n-2/2ⁿ

No. 2^n-2-2ⁿ \neq 2^n-2/2ⁿ

What you need to do is factor out 2ⁿ from both terms because 2ⁿ is a common factor.

= 2^n-2/2ⁿ
= 2^n-2-n+1

Even though what you have up until this point is wrong, these two lines do not follow either. Where did the +1 come from?

Also, I'll give you a tip with what you can do to check if you're doing your algebra correctly. Let's say you're asked to factorize x²-x, and you come up with the answer x(x-1) but you're unsure about it. Since both these expressions should be equal, it means that for every x they should be equal. So why not try a few values of x out?

We want to test if
x²-x = x(x-1)

For x=0 we get
0²-0 = 0(0-1)
0=0
Correct.

For x=1 we get
1²-1 = 1(1-1)
0=0
Correct.

For x=-278 we get
(-278)²-(-278) = -278(-278-1)
If you use a calculator you'll find that you get
77,562 = 77,562
Again, correct.

You can then be quite sure that you factorized correctly. It doesn't prove that you factorized it correctly, but it's a very strong indicator.

The reason it doesn't prove it is because say you instead factorized x²-x into x(x+1) ---- which is wrong

You test for x=0 and find that 0=0. Does this mean that you factorized correctly then? No. You haven't tested for all values of x. If you try it for x=1 you get 0=2 which is incorrect.
This is why you need to test for a few values, and once you get more comfortable with testing values, you'll intuitively begin to understand how many values you should test for before you can be fairly certain your answer is correct.

Now, how can you apply this value testing to your question?

You wrote that 2^n-2-2ⁿ = 2^n-2/2ⁿ
This means it should work out for all values of n! If we try n=0, we have
2^0-2-2⁰ = 2^0-2/2⁰
2^-2-1 = 2^-2/1
1/4-1=1/4
-3/4 = 1/4
Which is clearly wrong, so that means those two expressions are not equal and you need to go back and try again.

 

[chikis]

No. 2^n-2-2ⁿ \neq 2^n-2/2ⁿ

What you need to do is factor out 2ⁿ from both terms because 2ⁿ is a common factor.


Even though what you have up until this point is wrong, these two lines do not follow either.

Where did the +1 come from?

Oh sorry, I mean to write 2^n-2-2ⁿ
= 2^n-2-n
= 2²
= 4
because in indices, when base A divides base A, then their power will subtract and we will take a single base and raise it to the power of what we got when we subtracted the powers. That's why is like that.

Also, I'll give you a tip with what you can do to check if you're doing your algebra correctly. Let's say you're asked to factorize x²-x, and you come up with the answer x(x-1) but you're unsure about it. Since both these expressions should be equal, it means that for every x they should be equal. So why not try a few values of x out?

We want to test if
x²-x = x(x-1)

For x=0 we get
0²-0 = 0(0-1)
0=0
Correct.

For x=1 we get
1²-1 = 1(1-1)
0=0
Correct.

For x=-278 we get
(-278)²-(-278) = -278(-278-1)
If you use a calculator you'll find that you get
77,562 = 77,562
Again, correct.

You can then be quite sure that you factorized correctly. It doesn't prove that you factorized it correctly, but it's a very strong indicator.

The reason it doesn't prove it is because say you instead factorized x²-x into x(x+1) ---- which is wrong

You test for x=0 and find that 0=0. Does this mean that you factorized correctly then? No. You haven't tested for all values of x. If you try it for x=1 you get 0=2 which is incorrect.
This is why you need to test for a few values, and once you get more comfortable with testing values, you'll intuitively begin to understand how many values you should test for before you can be fairly certain your answer is correct.

Now, how can you apply this value testing to your question?

You wrote that 2^n-2-2ⁿ = 2^n-2/2ⁿ
This means it should work out for all values of n! If we try n=0, we have
2^0-2-2⁰ = 2^0-2/2⁰
2^-2-1 = 2^-2/1
1/4-1=1/4
-3/4 = 1/4
Which is clearly wrong, so that means those two expressions are not equal and you need to go back and try again.

Yes, I have tested and seen that, if n is taken as a any value say 3 and is subtituded both into the original expression and my final answer, the expressions we not be equal and that clearly indicatates that either my working, final answer or both are wrong. But I must confess, that's is where I can stop. I can't go further than this. If I keep solving at this point, then I will be making no sense mathmatically, what that means is that you have to start from where I stopped: 2^n-2-2ⁿ , then at the end, I will ask you questions as to how you got your final answer such that when any value of n is subtituted into the orignal expression, 1/4(2ⁿ-2ⁿ⁺²) will be equal to what you got as your final expression, when the value of n is subtituted into your final expression.

 

[HallsofIvy]

Oh sorry, I mean to write 2^n-2-2ⁿ
= 2^n-2-n
= 2²
= 4
because in indices, when base A divides base A, then their power will subtract and we will take a single base and raise it to the power of what we got when we subtracted the powers. That's why is like that.

But you have been told repeatedly that this is not division. It is true that A^m/Aⁿ= A^m-n but that does not apply here! You have 2^n-2- 2ⁿ. This is subtracton, not division. You have been told in post 2 to factor 2ⁿ out but have not done that and this is post 50!
You can use the laws of exponents you cite to say that 2^n-4= 2ⁿ2^-2 so that 2ⁿ- 2^{n- 2}= v2ⁿ- 2n2^-2. Can you factor 2ⁿ out of that?

Yes, I have tested and seen that, if n is taken as a any value say 3 and is subtituded both into the original expression and my final answer, the expressions we not be equal and that clearly indicatates that either my working, final answer or both are wrong. But I must confess, that's is where I can stop. I can't go further than this. If I keep solving at this point, then I will be making no sense mathmatically, what that means is that you have to start from where I stopped: 2^n-2-2ⁿ , then at the end, I will ask you questions as to how you got your final answer such that when any value of n is subtituted into the orignal expression, 1/4(2ⁿ-2ⁿ⁺²) will be equal to what you got as your final expression, when the value of n is subtituted into your final expression.