Indices Simplification from Fraction

Question

I am really struggling to simplify this equation.

Please explain how you would go about finding the answer:

The Equation

Answer 1

Well, we have:

$$\frac{\left(2\cdot\text{n}\right)^2\cdot3\cdot\text{n}^\frac{5}{4}}{6\cdot\text{n}^\frac{1}{2}\cdot\text{n}^3}\tag1$$

Now, we can use:

$$\left(2\cdot\text{n}\right)^2=2^2\cdot\text{n}^2=4\cdot\text{n}^2\tag2$$

So, we get:

$$\frac{4\cdot\text{n}^2\cdot3\cdot\text{n}^\frac{5}{4}}{6\cdot\text{n}^\frac{1}{2}\cdot\text{n}^3}\tag3$$

Now, we can use:

$$\frac{4\cdot3}{6}=2\tag4$$

So, we get:

$$\frac{2\cdot\text{n}^2\cdot\text{n}^\frac{5}{4}}{\text{n}^\frac{1}{2}\cdot\text{n}^3}\tag5$$

Now, we can use:

• $$\text{n}^2\cdot\text{n}^\frac{5}{4}=\text{n}^{2+\frac{5}{4}}=\text{n}^\frac{13}{4}\tag6$$
• $$\text{n}^\frac{1}{2}\cdot\text{n}^3=\text{n}^{\frac{1}{2}+3}=\text{n}^\frac{7}{2}\tag7$$

So, we get:

$$\frac{2\cdot\text{n}^\frac{13}{4}}{\text{n}^\frac{7}{2}}\tag8$$

Now, we can use:

$$\frac{\text{n}^\frac{13}{4}}{\text{n}^\frac{7}{2}}=\text{n}^{\frac{13}{4}-\frac{7}{2}}=\text{n}^{-\frac{1}{4}}\tag9$$

So, we get:

$$2\cdot\text{n}^{-\frac{1}{4}}=\frac{2}{\text{n}^\frac{1}{4}}\tag{10}$$