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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

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

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up vote 6 down vote accepted

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

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1  
$4^x = 4*4^{x-1}$ That is what I needed. Thank you!!! – Jessie Jul 27 '14 at 13:46

First of all:

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

So

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

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

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