Thread: Obvious but in need of confirmation!!

1. Obvious but in need of confirmation!!

Is it possible to have a power raised to a power E.g (Its attached)

I am in year nine N.S.W sydney mathematics, stage 5.2-5.3 I think, and in a disadvantaged school ( Chester Hill High to be exact, it has an Intensive English Centre in which students with little kowledge of english are separated and taught, then released into the rest of the school community i.e the one im in.)

2. Originally Posted by futsalfred
Is it possible to have a power raised to a power E.g (Its attached)

I am in year nine N.S.W sydney mathematics, stage 5.2-5.3 I think, and in a disadvantaged school ( Chester Hill High to be exact, it has an Intensive English Centre in which students with little kowledge of english are separated and taught, then released into the rest of the school community i.e the one im in.)
That is the same as
(2^3)^4
Meaning, the fourth power of 2-cube.

And that is simplified into
2^(3*4)

Another way of looking at it,
= (2^3)(2^3)(2^3)(2^3)
= 2^(3+3+3+3)
= 2^12.

3. It's quite possible but you have to think a little carefully about what you mean by $\displaystyle 2^{3^4}$, which I'll write on one line as 2^3^4. Do you mean (2^3)^4, ie take 2, raise that to the power 3, then raise the result (8) to the power 4, giving 4096; or do you mean 2^(3^4), ie take the 4th power of 3, then take that power (81) of 2 giving 2417851639229258349412352. These are different in general, and a way of saying that is that the ^ operator is not associative. You have actually seen this distinction before: addition and multiplication are associative, that is $\displaystyle a+(b+c) = (a+b)+c$ and $\displaystyle a \times (b \times c) = (a \times b) \times c$, whereas subtraction and division are not.

Anyway it wouldn't really matter which way of inserting the brackets you chose, provided that everyone else understood the same thing by it, were it not for one thing. The first way of reading $\displaystyle a^{b^c}$, as $\displaystyle \left(a^b\right)^c$ has another expression, since it is just $\displaystyle a^{(b\times c)}$. Since we don't really need two ways of writing the same thing, it makes sense for everyone to agree that $\displaystyle a^{b^c}$ should mean $\displaystyle a^{\left(b^c\right)}$.

So the short answer is: yes, $\displaystyle a^{b^c} = a^{\left(b^c\right)}$.

4. Originally Posted by rgep
It's quite possible but you have to think a little carefully about what you mean by $\displaystyle 2^{3^4}$, which I'll write on one line as 2^3^4. Do you mean (2^3)^4, ie take 2, raise that to the power 3, then raise the result (8) to the power 4, giving 4096; or do you mean 2^(3^4), ie take the 4th power of 3, then take that power (81) of 2 giving 2417851639229258349412352. These are different in general, and a way of saying that is that the ^ operator is not associative. You have actually seen this distinction before: addition and multiplication are associative, that is $\displaystyle a+(b+c) = (a+b)+c$ and $\displaystyle a \times (b \times c) = (a \times b) \times c$, whereas subtraction and division are not.

Anyway it wouldn't really matter which way of inserting the brackets you chose, provided that everyone else understood the same thing by it, were it not for one thing. The first way of reading $\displaystyle a^{b^c}$, as $\displaystyle \left(a^b\right)^c$ has another expression, since it is just $\displaystyle a^{(b\times c)}$. Since we don't really need two ways of writing the same thing, it makes sense for everyone to agree that $\displaystyle a^{b^c}$ should mean $\displaystyle a^{\left(b^c\right)}$.

So the short answer is: yes, $\displaystyle a^{b^c} = a^{\left(b^c\right)}$.
I just read that and went huh?
Anyway, I think you went too deep into the matter. I was doing a question and looked in the textbook for the answer (Not to cheat, more like see the answer and try to relate it to other questions), and the answer was exactly like the image I posted before. I asked my teacher about, and she said
it was too advanced and I don't have to worry about it now. I still don't get it, could you please show this in more depth (if possible?) Thanks in advance.

5. Originally Posted by futsalfred
I just read that and went huh?
Anyway, I think you went too deep into the matter. I was doing a question and looked in the textbook for the answer (Not to cheat, more like see the answer and try to relate it to other questions), and the answer was exactly like the image I posted before. I asked my teacher about, and she said
it was too advanced and I don't have to worry about it now. I still don't get it, could you please show this in more depth (if possible?) Thanks in advance.
I think what the others are trying to say is that there are two ways to interpret the expression you wrote:

1) $\displaystyle 2^{(3^4)}=2^{81}$

2) $\displaystyle (2^3)^4=2^{12}$

Both expressions are possible and there is apparently no convention in place to say which one you should be using.

-Dan

6. Using a calc, if you punch in those numbers >>

2^3^4, youll get 2^12

Unless it indicates that there are "()" in between the powers ur pretty much set.

2^a^b = 2^(a*b)
2^3^4 = 2^(3*4) = 2^(12)

now if the brackets do exsist then its diff

2^(3^4) = 2^(3*3*3*3) = 2^(81)

Well i hope that makes a lil clearer its the same thing that the others are saying pretty much hope it helped.

7. ?

I understand now! Thanks for help me and mm goy sai!