# X'root homework problem help

1. Homework Statement
(9^(3/x))^4
(without calculator)

2. Homework Equations
a^n/m = m'root'a^n

3. The Attempt at a Solution

adding the powers: 3/x*4/1 = 12/x

9^12/x

x'root'9^12 = ?

----------------------

Ok, it's supposed to be 27, but how can you know something that are being rooted from x?

The calculator says it's 2.655 when I use the x'root'9^12 (It's incorrect and I'm not allowed to do that)

How does it get that from x? And what is the correct way to find it?

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berkeman
Mentor
I don't understand your notation, and it does seem like you would need x in order to work out the problem. What is the full problem statement? Does it say "find x for each of the following", or something? Or given that x=(something), work out each of the following...

Is this what you are trying to write? $$( {9^\frac{3}{x}} )^4$$

That's excactly what I try to write, Thanks, and no, it doesn't say anything more than:

a)
b)
c)
d)
e)
f) $$( {9^\frac{3}{x}} )^4$$

(plus the relevant equation at the top of the page of a^n/m = m'root'a^n (I don't know the equation code for that)

berkeman
Mentor
Well, you can certainly calculate it for x = 1, 3, and 6, and all give different answers. So the answer is not independent of x. Given the relevant equation that you've listed (which must be part of the subject matter for the section you are working on), they must want you to just re-write the number... ? Which is what you did. I don't see how you can go farther without knowing x, with or without a calculator.

$$\sqrt[x]{9^{12}}$$

Thanks, I will tell my teacher this tomorrow.

(still don't know why my calculator came up with his wierd answer, but I will ask him that too)

berkeman
Mentor
Thanks, I will tell my teacher this tomorrow.

(still don't know why my calculator came up with his wierd answer, but I will ask him that too)
Oh great! Gonna get me in trouble with the prof, eh? :rofl:

BTW, your calculator gave you an answer because there was already something in the x register, most likely. Just left over from whatever your previous calculation was. Try it again, this time explicitly putting in the values 1, 3, 6, and see what the three answers are that you get (hopefully they'll be the same ones you worked out in your head).

You are safe behind the internet wall, so don't worry :)
He is a smart guy though....

EDIT: This is kind of embarrasing.... It might be that the machine that printed out this question have accidentaly cut away the top and bottom of the number "8", so it showed X instead. It's probably supposed to be 8, and that's is correct too... :grumpy:

See how much a little mistake can make lots of useless work for a poor fellow??

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BobG
Homework Helper
1. Homework Statement
(9^(3/x))^4
(without calculator)

2. Homework Equations
a^n/m = m'root'a^n

3. The Attempt at a Solution

adding the powers: 3/x*4/1 = 12/x

9^12/x

x'root'9^12 = ?

----------------------

Ok, it's supposed to be 27, but how can you know something that are being rooted from x?

The calculator says it's 2.655 when I use the x'root'9^12 (It's incorrect and I'm not allowed to do that)

How does it get that from x? And what is the correct way to find it?
What's supposed to be 27.

If x is 27, and
$$9^{(\frac{12}{x}) } = y$$
then y does equal 2.655.

If
$$9^{(\frac{12}{x}) } = 27$$
then you can solve for x.

The second makes more sense if you're supposed to solve this without a calculator.

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Oh, we posted at the same time....

I understand how we can find X, and It's probably going to be "8" if you read the post I made...

sorry about this.... berkeman
Mentor
It might be that the machine that printed out this question have accidentaly cut away the top and bottom of the number "8", so it showed X instead. It's probably supposed to be 8, and that's is correct too... :grumpy:
Oh my gosh, that's funny in a weird way. I think that's a new one -- a printing error that totally changes the meaning of the problem. Even if I had solved for 8 as the only possibility for x, I don't think I would have seen the geometric clue that x was really an 8. Oh well, all's well that ends well.