Question

I have a question (In comments)

Answer 1

\[(2n)^4 (\frac{ 3 }{ 2 }n)^3\]

Answer 2

How do I solve that?

I got \[16n^4 \frac{ 27 }{ 16 }n\] but what do I do next?

Answer 3

The (2n) ^4 = 16 n^4 part is correct.

Answer 4

The (3/2 n) ^3 needs some work.

Answer 5

You have some mistake there: \((2n)^4 = 16n^4\) Looks fine.
But \((\cfrac{3}{2} n)^3 = \cfrac{3^3}{2^3} n^3 = \cfrac{ 27}{8} n^3 \ne \cfrac{27}{16} n \)

Answer 6

The 27 is okay. The 16 is not. And, there's an exponent for the n. @JamesR4494

Answer 7

@JamesR4494 have you got your mistake?

Answer 8

yea I see my mistake, what do I do next though?

Answer 9

We have : \(16 \times \cfrac{27}{8} n^3\)

\(\cancel{16}^2 \cfrac{27}{\cancel{8}^1} n^3\)

I canceled 16 and 8 ... by getting 2 and 1 respectively

Answer 10

Let's go back to the second factor of (3/2 n ) ^3.

What is 3/2 * 3/2* 3/2 = ?

Answer 11

27/ ?

Answer 12

27/8

Answer 13

Sheldon, How do you cancel 16 and 8?

Answer 14

@JamesR4494 \(\cfrac{16}{8} = \cfrac{\cancel{16} 2}{\cancel{8} 1} \)

Answer 15

Oh I see now