Simplification and solving of equation

[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.

 

[Mentallic]

Note: Hallsofivy was meant to say,

You can use the laws of exponents you cite to say that 2^n-2= 2ⁿ2^-2 so that 2ⁿ- 2^{n- 2}= 2ⁿ- 2ⁿ2^-2

I just want to clear that up to avoid more confusion.

chikis, you're misusing exponent laws, and you're struggling to apply simple factorizations. You need to understand these rules I'm going to give you. Memorize them. Never break them. If you're unsure with what the next step is in your problem and feel the need to break one of these rules to "solve" the problem, stop right there and slap yourself for thinking that.

a^b\cdot a^c=a^{b+c}(the dot means multiply). For example, 3^4\cdot 3^8=3^{4+8}=3^{12}3^2\cdot 3^{-4}=3^{2+(-4)}=3^{-2} Keep in mind that you cannot use this rule for anything other than multiplying values of the same base. The base is the value of "a". So, 3^2\cdot 9^5 cannot be simplified using this rule! What you can do however is transform your problem into something that does have the same base, and we do this by using this kind of rule:

(a^b)^c=a^{bc} For example, (4^{12})^3=4^{12\cdot 3}=4^{36}((-3)^3)^{-2}=(-3)^{3\cdot(-2)}=(-3)^{-6}=\frac{1}{(-3)^6}=\frac{1}{3^6}=3^{-6}

So now if we go back to our problem before that we couldn't simplify 3^2\cdot 9^5 we can realize that 9=3^2 so we can transform the problem into 3^2\cdot(3^2)^5=3^2\cdot 3^{10}=3^{12}

Now, what can't you do with exponents.
a^b+a^c\neq a^{b+c}
Remember this well, because you keep doing it, and it's wrong. What can we do with this though? We can factorize!

Some factorization rules:

a(b+c)=ab+ac
ab(cd+e)=abcd+abe
What this is telling us is that we can have as many factors as we want outside the brackets, as long as they are common to both terms, that is, ab is common to both abcd and abe. (note in this case ab means a times b).

So back to your question, 2^{n-2}-2^{n} and we want to simplify this. If you have understood the rules I have given you, you should be able to answer it.

Start with the rule a^b\cdot a^c=a^{b+c}thus 2^{n-2} can be simplified into what? And then use the factorization rule. What is the common factor in both terms?

 

[chikis]

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- 2^{n2-2. Can you factor 2n out of that?}

^{factoring out 2n from 2[sup ]-2(2n-2n*22
we have: 2n*2-2(1-22)
= 2n-2(1-22)
= 2n-2-2n-2*22
= 2n-2-2n-2+2
= 2n-2-2n
So what can we do next?}

 

[Mentallic]

factoring out 2ⁿ from 2[sup ]-2[/sup](2ⁿ-2ⁿ*2²
we have: 2ⁿ*2^-2(1-2²)
= 2^n-2(1-2²)
= 2^n-2-2^n-2*2²
= 2^n-2-2^n-2+2
= 2^n-2-2ⁿ
So what can we do next?

You just factorized and then expanded again.

That's like saying, ok, I know that by simplifying ab+ac I need to first factorize, so the steps you took were ab+ac = a(b+c) = ab+ac
You really went nowhere.

Look at the factorized expression: 2ⁿ*2^-2(1-2²) what is in the brackets? Simplify it.

 

[SammyS]

factoring out 2ⁿ from 2^-2)(2ⁿ-2ⁿ*2²
we have: 2ⁿ*2^-2(1-2²)

Another way to express this is:

2ⁿ*2^-2(1-2²)

\displaystyle <br /> =\frac{2^n}{2^2}(1-2^2)

\displaystyle <br /> =\frac{(-3)\left(2^n\right)}{4}

\displaystyle <br /> =-\left(\frac{3}{4}\right)2^n​

 

[chikis]

You just factorized and then expanded again.

That's like saying, ok, I know that by simplifying ab+ac I need to first factorize, so the steps you took were ab+ac = a(b+c) = ab+ac
You really went nowhere.

Look at the factorized expression: 2ⁿ*2^-2(1-2²) what is in the brackets? Simplify it.

Ok let's see:2^n-2-2ⁿ
= 1/4(2ⁿ-2ⁿ⁺²) if n=3.
You will see that each of the expression will be equal to -6 respectively, therefore they are equal and 2^n-2-2ⁿ is the final expression and answer.
If you work with 2ⁿ*2^-2(1-2²), you will see that 2ⁿ*2^-2(1-2²)
= 2ⁿ*2^-2(1+4)
= 2ⁿ*2^-2(5)
= 2^n-2(5)
and that is wrong because 2^n-2(5) is not = 1/4(2ⁿ-2ⁿ⁺²) if n=3. Can you notice that?

 

[gabbagabbahey]

Ok let's see:2^n-2-2ⁿ
= 1/4(2ⁿ-2ⁿ⁺²) if n=3.
You will see that each of the expression will be equal to -6 respectively, therefore they are equal and 2^n-2-2ⁿ is the final expression and answer.
If you work with 2ⁿ*2^-2(1-2²), you will see that 2ⁿ*2^-2(1-2²)
= 2ⁿ*2^-2(1+4)
= 2ⁿ*2^-2(5)
= 2^n-2(5)
and that is wrong because 2^n-2(5) is not = 1/4(2ⁿ-2ⁿ⁺²) if n=3. Can you notice that?

1-2²≠5. Can you notice that?

 

[chikis]

2[/SUP]≠5. Can you notice that?

Then who is really wrong here? Is either you or me. One person must be wrong.
1-2² = 5
because (-1)(-1)=1
and -2^-2
= (-2)(-2)=4
therefore (1-2^-2) = 1[(-2)(-2)] = 1+4 = 5. Can you notice that?

 

[Mentallic]

Then who is really wrong here? Is either you or me. One person must be wrong.
1-2² = 5
because (-1)(-1)=1
and -2^-2
= (-2)(-2)=4
therefore (1-2^-2) = 1[(-2)(-2)] = 1+4 = 5. Can you notice that?

-2² and (-2)² are two different things. The first says take the negative of 2² which is also take the negative of 4. The second says take the square of -2, which is also 4, but now you've lost the negative sign.

-2²= -4
(-2)²= 4

Oh and no offense but,

Then who is really wrong here? Is either you or me. One person must be wrong.

99 times out of a 100 it's going to be you.

 

[HallsofIvy]

Then who is really wrong here? Is either you or me. One person must be wrong.
1-2² = 5
because (-1)(-1)=1
and -2^-2
= (-2)(-2)=4
therefore (1-2^-2) = 1[(-2)(-2)] = 1+4 = 5. Can you notice that?

This is just bad arithmetic. You are completely wrong. 1- 2^2= -3. You need to go to your teacher and get help. You need a lot more than can be given here.

 

[chikis]

-2² and (-2)² are two different things. The first says take the negative of 2² which is also take the negative of 4. The second says take the square of -2, which is also 4, but now you've lost the negative sign.

-2²= -4
(-2)²= 4

Oh and no offense but,

99 times out of a 100 it's going to be you.

Yes, there is different between -2² and (-2)² and I have seen it. Bearing that in mind:
2ⁿ*2^-2(1-2²)
= 2ⁿ*2^-2(1-4)
= 2^n-2(-3)
that is the final simpilification.