Simplification and solving of equation

[Millennial]

At last, correct.

 

[Mark44]

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​

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.

chikis, do you see that your final expression is identically equal to what Sammy shows? I am not confident that you do or that you would be able to show that they are the same.

 

[chikis]

chikis, do you see that your final expression is identically equal to what Sammy shows? I am not confident that you do or that you would be able to show that they are the same.

Show that what are the same?

 

[Mark44]

Look at post #62. I showed both expressions.

 

[chikis]

Look at post #62. I showed both expressions.

There is no number assigned to each post. I have no strenght to start counting from opening post uptilll post #62. I would apreciate it if you will go to the post, then copy it then repost, that way I can see what you are talking about.

 

[Mark44]

The post number appears in the upper right corner of each post, at least on my browser. This is post #66. Count back four posts to #62, which is a post from me. I copied what Sammy wrote and what you wrote. Can you see that the two results are identically the same.

 

[chikis]

The post number appears in the upper right corner of each post, at least on my browser. This is post #66. Count back four posts to #62, which is a post from me. I copied what Sammy wrote and what you wrote. Can you see that the two results are identically the same.

I have told you repeatedly, that I don't understand latex format of writing. I would still prefer it better, if you can go to that post and copy the particular something you want me to show.

 

[chikis]

Now let's consider this possibility, suppose the original problem that I brought is changed to something like this:
1/4(2ⁿ 2ⁿ⁺²) and am ask to simplify and I start:

1/4(2ⁿ 2ⁿ⁺²)
= 2^-2(2ⁿ 2ⁿ⁺²)
= 2^-2*2ⁿ*2ⁿ⁺²
= 2^-2+n+n+2
= 2²ⁿ
Is my simplification correct?

 

[chikis]

Is everybody tired? Yet the problem is not yet finished. I brought two problem here, one has been dealt with, what about the other one? Is that the other is beyond what you can handle? Don't make me think like that please!

 

[chikis]

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.

 

[azizlwl]

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.

2^(2x+1)=2^2x×2¹
2^2x=(2^x)²

 

[chikis]

2^(2x+1)=2^2x×2¹
2^2x=(2^x)²

What about the - 9(2^x) + 4? What can I do with that one?

 

[azizlwl]

2^(2x+1)=2^2x×2¹
2^2x=(2^x)²

2(2^x)²+9(2^x)+4=0

(2^x)²+(9/2)(2^x)=-2

(2^x)²+(9/2)(2^x) +(9/4)²=-2+(9/4)²

If 2^x confuse you, , replace it with y. Remember to replace it to original value in the final calculation.

 

[Mentallic]

Is everybody tired? Yet the problem is not yet finished. I brought two problem here, one has been dealt with, what about the other one? Is that the other is beyond what you can handle? Don't make me think like that please!

Excuse me? You need to quit with the attitude mate.

I've also told you that you need to fix your browser or whatever the problem is so that you can read latex, because we aren't going to be bending over backwards for your every command.
It's quite amazing that you expect all the helpers in this thread to toss out their primary math display tool and be forced to work with a more crude system just for you, especially when it seems like you haven't appreciated any of the help given thus far.

 

[Ray Vickson]

Excuse me? You need to quit with the attitude mate.

I've also told you that you need to fix your browser or whatever the problem is so that you can read latex, because we aren't going to be bending over backwards for your every command.
It's quite amazing that you expect all the helpers in this thread to toss out their primary math display tool and be forced to work with a more crude system just for you, especially when it seems like you haven't appreciated any of the help given thus far.

I agree. As far as I am concerned the poster does not deserve to be helped. He wants us to do all the work, at his convenience, and to keep working away until we have finally presented him with a solution he can turn in and receive credit for (meanwhile, learning nothing). He is arrogant and stubborn, and just about the most impolite poster I have seen on this Forum.

RGV

 

[DeIdeal]

Is everybody tired? Yet the problem is not yet finished. I brought two problem here, one has been dealt with, what about the other one? Is that the other is beyond what you can handle? Don't make me think like that please!

Take a look at the Advanced Physics and Calculus & Beyond sections. Do you honestly think that there aren't people here who could solve that in a minute?

I can't believe someone has the guts to be this impolite to people who are trying to help him. You've dragged this thread on for five pages. Don't you think it's time to follow the advice people have given to you and ask your teacher about this. I'm just repeating what others have said, but you clearly need more help in maths than you can get here.

 

[chikis]

Take a look at the Advanced Physics and Calculus & Beyond sections. Do you honestly think that there aren't people here who could solve that in a minute?

I can't believe someone has the guts to be this impolite to people who are trying to help him. You've dragged this thread on for five pages. Don't you think it's time to follow the advice people have given to you and ask your teacher about this. I'm just repeating what others have said, but you clearly need more help in maths than you can get here.

Sorry please,I don't mean to be this rude. Is just that the problem is still remaining one.

 

[chikis]

I agree. As far as I am concerned the poster does not deserve to be helped. He wants us to do all the work, at his convenience, and to keep working away until we have finally presented him with a solution he can turn in and receive credit for (meanwhile, learning nothing). He is arrogant and stubborn, and just about the most impolite poster I have seen on this Forum.

RGV

What have I done? I don't mean to be this rude, am only asking for help please!

 

[Mark44]

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

2(2^x)²+9(2^x)+4=0

(2^x)²+(9/2)(2^x)=-2

(2^x)²+(9/2)(2^x) +(9/4)²=-2+(9/4)²

If 2^x confuse you, , replace it with y. Remember to replace it to original value in the final calculation.

Sorry please,I don't mean to be this rude. Is just that the problem is still remaining one.

Folks have been helping out on this. You would help yourself by paying attention to what these people have been saying throughout this thread.

If you don't understand what eumyang and aziz are saying above, then your next step should be to talk to your teacher, which has been suggested before in this thread.

 

[micromass]

Is everybody tired? Yet the problem is not yet finished. I brought two problem here, one has been dealt with, what about the other one? Is that the other is beyond what you can handle? Don't make me think like that please!

Drop the attitude please.

Folks have been helping out on this. You would help yourself by paying attention to what these people have been saying throughout this thread.

If you don't understand what eumyang and aziz are saying above, then your next step should be to talk to your teacher, which has been suggested before in this thread.

I agree. There's nothing more that we can do here. And since the OP has an attitude problem, I'm going to lock this.