# Is this expression correct? (12^3 M^33/4^-3)

#### Lii

##### New member

Hello, as a title
I posted on discors but I didn't get the given explanation..
So, is it correct? And if not, what I'm doing wrong, and why I cannot do so?

#### Subhotosh Khan

##### Super Moderator
Staff member

Hello, as a title
I posted on discors but I didn't get the given explanation..
So, is it correct? And if not, what I'm doing wrong, and why I cannot do so?
View attachment 20057
Almost correct. You have:

$$\displaystyle \left [ \frac{12^3}{4^{-3}}\right ]^2$$

$$\displaystyle =\frac{12^6}{4^{-6}}$$

$$\displaystyle = 12^6 * 4^6$$..................... because we know $$\displaystyle \frac{1}{4^{-6}} \ = \ 4^{6}$$

$$\displaystyle = (12 * 4)^6$$

$$\displaystyle = 48^6$$

$$\displaystyle = 3^6 * 2^{24}$$

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#### Lii

##### New member
Almost correct. You have:

$$\displaystyle \left [ \frac{12^3}{4^{-3}}\right ]^2$$

$$\displaystyle =\frac{12^6}{4^{-6}}$$

$$\displaystyle = 12^6 * 4^6$$..................... because we know $$\displaystyle \frac{1}{4^{-6}} \ = \ 4^{6}$$

$$\displaystyle = (12 * 4)^6$$

$$\displaystyle = 48^6$$

$$\displaystyle = 3^6 * 2^{24}$$
Thanks
But there is a reason on why divide 12^6 for 4^-6 ?
So 6-(-6) = 12; 12/4 = 3; 3^12

So I will not redo the same kind of error? Or should I always apply the powers rule that says when do I have the same exponent should I mantein that a do the rest with the base?

#### JeffM

##### Elite Member
Your question is not in good English so it is hard to know how to answer.

$$\displaystyle y = \left ( \dfrac{(12^3m^{33}}{4^{-3}} \right )^2.$$

is a valid equation, but it can be expresed in ways that may be easier to understand. This is called "simplification" in English. There is no objective meaning to the term. It depends partially on what you need to do later and partly on your personal feeling about what is easy to understand.

In this case, you can get rid of the parentheses by doing the exponentiation as follows:

$$\displaystyle y = \left ( \dfrac{12^3m^{33}}{4^{-3}} \right )^2 = \dfrac{12^{2*3}m^{2*33}}{4^{2*(-3)}} = \dfrac{12^6m^{66}}{4^{-6}}.$$

That still looks rather formidable to me. We can get rid of the fraction by eliminating the negative exponent:

$$\displaystyle y = \dfrac{12^6m^{66}}{4^{-6}} = \dfrac{\dfrac{12^6m^{66}}{1}}{\dfrac{1}{4^6}} = \dfrac{12^6m^{66}}{1} * \dfrac{4^6}{1} = 4^6 * 12^6m^{66}.$$

That looks simpler to me than what we started with. Depending on what we want to do, there are now two ways to go. One way is to combine the product of numbers with like exponents:

$$\displaystyle y = 4^6 * 12^6m^{66} = (4 * 12)^6m^{66} = 48^6m^{66}.$$

Does that look simpler to you than what we started with? It does to me.

However, if we know that all the numbers are integers, it might be more convenient for future work to simplify to the smallest integers possible. In that case we might go:

$$\displaystyle y = 4^6 * 12^6m^{66} = (2^2)^6 * (2 * 2 * 3)^6m^{66} = \\ 2^{12} * 2^6 * 2^6 * 3^6m^{66} = 2^{24} * 3^6m^{66}.$$
That involves more terms than our previous answer, but two and three are easier numbers to imagine than forty-eight. (In fact, I cannot "see" in my mind 48 of anything, but I can "see" the difference in my mind between two cats and three cats.)

To summarize, all the expressions that we got for y are correct in the sense that they are valid arithmetically, but only two are as easy to understand as possible.

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#### Lii

##### New member
@JeffM thanks

Anyway, I'm sorry for the previous post, and thanks for the exhaustive reply

#### HallsofIvy

##### Elite Member
Thanks
But there is a reason on why divide 12^6 for 4^-6 ?
Because that was what the problem said!

The title had "(12^3 M^33/4^-3)^2" which, as you were told, is 12^6 M^66/4^-6
since the square means that you multiply each exponent by 2.
So, putting the "M^66" aside for the moment we have 12^6/4^-6, 12^6 divided by 4^-6.

So 6-(-6) = 12; 12/4 = 3; 3^12
Where did you get "6- (-6)"? Because of the "6" power in the numerator and "-6" power in the denominator? It doesn't work that way because there are different bases! It is true that a^p/a^q= a^(p- q) but that requires the same base, a. In general "a^p/b^q" cannot be reduced unless there is some relationship between a and b

Here, "a" is 12 and "b" is 4. 12= 3(4) so we can write 12^6 as (3^6)(4^6). NOW we can say 12^6/4^=6= (3^6)(4^6)/4^-6= (3^6)(4^(6-(-6))= (3^6)(4^12).

So I will not redo the same kind of error? Or should I always apply the powers rule that says when do I have the same exponent should I mantein that a do the rest with the base?
The "power rule" is that (a^x)(a^y)= a^(x+y). That requires that the bases be the same.