Cannot follow the algebra

Question

The following equality is stated in my text book and I cannot follow the algebra that makes it true. Please help me step through this to show how

$$\frac{4^x}{3^{x-1}} = 4 \left(\frac{4}{3}\right)^{x-1}$$

Answer 1

$\frac{4^x}{3^{x-1}}\equiv\frac{4^1 \cdot 4^{x-1}}{3^{x-1}}\equiv \underbrace{4 \left[\frac{4^{x-1}}{3^{x-1}}\right]\equiv 4\left[\frac{4}{3}\right]^{x-1}}_{\text{since} \quad\frac{a^n}{b^n} \equiv \left[\frac{a}{b}\right]^n}$

Answer 2

First of all:

$$\frac{a^t}{b^t} = \left(\frac{a}{b}\right)^t$$

So

$$\frac{4^x}{3^{x-1}} = \frac{4·4^{x-1}}{3^{x-1}} = 4\frac{4^{x-1}}{3^{x-1}} = 4\left(\frac{4}{3}\right)^{x-1}$$

Answer 3

$$ 4\left(\frac43\right)^{x-1}=4\frac{4^{x-1}}{3^{x-1}}=\frac{4^x}{3^{x-1}} $$