Question

How to use 12, 3, 4, 9 to make 24? Can use multiplication, division, addition, and subtraction.

And:
There are two numbers whose 400th powers are equal to 9^1000. In other words, there are two numbers that can replace x in the equation x^400 = 9^1000, making the equation true.

What are those numbers? Explain the process by which you got your answer.

Answer 1

There are 60 different ways to solve the 24-problem.

If A = 3, B = 4, C = 9, D = 12, we have:

Dx(C-(B+A)) ((D-C)+A)xB D+(Cx(B/A)) Cx(A-(B/D)) ((D/B)xC)-A
Dx(C-(A+B)) ((D+A)-C)xB D+(Bx(C/A)) (D-B)x(C/A) ((C/B)xD)-A
Bx(D+(A-C)) ((A+D)-C)xB D+((CxB)/A) (C/A)x(D-B) ((D-B)xC)/A
Bx(A+(D-C)) ((A-C)+D)xB D+((BxC)/A) (CxA)-(D/B) ((D-B)/A)xC
Bx((D-C)+A) Bx(D-(C-A)) D+((C/A)xB) (AxC)-(D/B) D+(C/(A/B))
Bx((D+A)-C) Bx(A-(C-D)) D+((B/A)xC) Cx((D-B)/A) D+(B/(A/C))
Bx((A+D)-C) Dx((C-B)-A) (Cx(B/A))+D (Dx(C/B))-A (C/(A/B))+D
Bx((A-C)+D) Dx((C-A)-B) (Bx(C/A))+D (Cx(D/B))-A (B/(A/C))+D
(D+(A-C))xB (D-(C-A))xB ((CxB)/A)+D (Cx(D-B))/A C/(A/(D-B))
(A+(D-C))xB (A-(C-D))xB ((BxC)/A)+D (A-(B/D))xC (D-B)/(A/C)
(C-(B+A))xD ((C-B)-A)xD ((C/A)xB)+D ((DxC)/B)-A (D/(B/C))-A
(C-(A+B))xD ((C-A)-B)xD ((B/A)xC)+D ((CxD)/B)-A (C/(B/D))-A

How's that?

Answer 2

Thank you so so much! I really appreciate your help. Do you have any ideas on the second one?

Answer 3

Have you considered factoring the given equation?

Answer 4

I have, but I have no idea how to start, or even how to do it. See, I was sick today and I missed class, and my friend brought me homework. However, I am not even sure how to solve this!

Answer 5

Sure you do. Ever see a "Difference of Squares?

Answer 6

x^400 = 9^1000 ==> \(x^{400} - 9^{1000} = 0\)

or

\(\left(x^{200} + 9^{500}\right)\cdot\left(x^{200} - 9^{500}\right) = 0\)

Now what?

Answer 7

I'm so sorry, I feel so dumb. What you are saying makes sense, but I am at a loss at how I am supposed to use that in the equation. Perhaps I... make it 9x^400000? But how does that help me find the numbers?

Answer 8

Why are you thinking about anything but factoring? You don't have to know the final answer from the beginning. Just jump in and try it out. It's an equation. You won't break it.

There is clearly no solution for the first parentheses, this leaves only the second set of parentheses. Factor again.

\(\left(x^{100} + 9^{250}\right)\cdot\left(x^{100} - 9^{250}\right) = 0\)

Do it again!

Answer 9

x^400=9^1000
400 * log(x) = 1000 * log(9)
log(x) = 1000 * log(9) / 400
log(x) = 2.5 * log(9)
log(x) = 2.3856062736
x = 10^2.3856062736
x = 243

Answer 10

Thank you, but can you explain that in more detail wolf1728? I want to understand the answer, not have it given to me.

Answer 11

There's ONE of them. wolf just used logarithms. There is a little more work to do.

All will be explained if you keep factoring.

Answer 12

Thank you. So 243 is one of the answers, and I think I understand why a little. But I still need the other, and I need to be able to understand it to the extent that I can explain it in detail for my assignment.

Answer 13

That's why I want you to keep factoring.

Answer 14

But I don't understand! I'm so sorry, I sound like a petulant bratt, but I just don't understand what to do.

Answer 15

Well, then why don't you do it? I'm telling you that you WILL understand if you do it.

\(\left(x^{100} + 9^{250}\right)\cdot\left(x^{100} - 9^{250}\right)\) = 0

Do it again!

Answer 16

Okay let's take a simpler example.
x^8 = 16^2
8 * log(x) = 2 * log(16)
log(x) = 2*log(16) / 8
log(x) = .25*log(16)
log(x) = .25 * 1.2041199827
log(x) = 0.3010299957
x = 10^.3010299957
x = 2

Answer 17

I'm sorry wolf1728, I haven't learned logarithams yet.

Answer 18

I understand that part, it's the rest I don't understand.

Answer 19

Okay. but at least you can see from my second posting that 2^8 = 16^2 right?

Answer 20

Here's the hint to the problem:
Start by expressing 9^1000 as a power of an integer smaller than 9.

Answer 21

Yes

Answer 22

Factor more. You are not understanding how this will lead you to a solution. I am telling you that it is not necessary to understand that for now. Continue the34 process until you do understand. I'm telling you that you will. Keep factoring.

\(\left(x^{50} + 9^{125}\right)\cdot\left(x^{50} - 9^{125}\right) = 0\)

Now what?

Answer 23

Then (x to the 10+9 to the 25) (x to the tenth-9 to the 25)=0?

Answer 24

But they don't want me to factor it that way. They want me to raise the power and lower the integer. Look at the problem hint.

Answer 25

No, that doesn't quite work. Good try, though. Very good making an effort.

The problem statement and the hint says to solve the given equation.

This one's a little trickier.
\(\left(x^{50} - 9^{125}\right) = \left(x^{50} - 3^{250}\right) = \left(x^{25} + 3^{125}\right)\cdot\left(x^{25} - 3^{125}\right)\)

We're almost done.

Answer 26

Oh! I see now. So you factor out the 50 into 25 in order to divide the number. I get it! So now we start factoring down the exponents more, correct?

Answer 27

Excellent. Not that it will lead to a solution, but that you are thinking about it and trying things in your head.

That was a lot of factoring practice.

If we have \(x^{25} - 3^{125} = 0\), it must be so that \(x^{25} - \left(3^{5}\right)^{25} = 0\). It is just exponent rules. Do you see it?

Answer 28

Yes.

Answer 29

I now understand factoring, and for that I am grateful. But my homework... I am unsure on haw to use this to get to the answer. I was thinking it was possible I multiply both numbers by 1/3. Is that possible? And then with the exponents I could factor it down even further...?

Answer 30

I'm sorry if that seems stupid, I'm trying to think critically.

Answer 31

The factoring was just an exercise to see if we can find smaller exponents. You seemed to want to have smaller exponents. It's actually quite unnecessary to do so. We can work with the original equation.

\(x^{400} - 9^{1000} = x^{400} - \left(3^{2}\right)^{1000} = x^{400} - 3^{2000} = x^{400} - \left(3^{5}\right)^{400} = 0\)

And we see that \(x = 3^{5} = 243\) is a solution. We also see, with a little more looking around, that \(x = -3^{5} = -243\) is a solution, since the exponents were even.

Answer 32

Thank you so so much! I will ace the homework and the test now. Is there any way I can thank you besides giving you the medal?

Answer 33

Yes. Go ace that exam. Feel free to be the factoring guru in your class!

Answer 34

I will!