Simplify $\frac {3^{(-3+x)}6^{(3-x)}}{3\cdot4^x}$

Question

$$\frac {3^{(-3+x)}6^{(3-x)}}{3\cdot4^x}$$

What is the simplest form?

Answer 1

$$ \frac{3^{x-3}\cdot 2^{3-x} \cdot 3^{3-x} }{3\cdot 2^{2x}} =2^{3-x-2x}\cdot3^{x-3+3-x-1}=\frac{2^{3-3x}}{3}.$$

Answer 2

You should know two facts.

$$\begin{split} a^x \cdot a^y &= a^{x+y}\\ (a\cdot b)^x &= a^x \cdot b^x \end{split}$$

---

Then you can easily obtain next two facts $$\begin{split} \left(a^b\right)^c &= \underbrace{a^b \cdot a^b \cdot ~\dots~ \cdot a^b}_{\text{c times}} = a^{b+b+\dots+b} &= a^{b\cdot c}\\ \frac{a^x}{a^y} &= a^x \cdot \left(\frac{1}{a}\right)^{y} = a^x \cdot a^{-y} &= a^{x-y} \end{split}$$

---

Hint: $$4^{x} = \left(2^2\right)^{x} = 2^{2x}$$

Hint2:

$$ 6^{3-x} = (3\cdot2)^{3-x} = 3^{3-x} \cdot 2^{3-x}$$

---

Solution:

$$\begin{split} \frac{3^{(-3)+x}\cdot 6^{3-x}}{3 \cdot 4^{x}} &= \frac{\left(3^{3-x}\right)^{-1}\cdot 3^{3-x}\cdot 2^{3-x}}{3 \cdot 2^{2x}}\\ &=\frac{\left(3^{-1}\right)^{3-x}\cdot 3^{3-x}}{3} \cdot \frac{2^{3-x}}{\cdot 2^{2x}}\\ &=\frac{\left(\frac{1}{3} \cdot 3\right)^{3-x}}{3} \cdot 2^{(3-x)-(2x)}\\ &= \frac{1}{3} \cdot 2 ^{3\cdot (1-x)} = \frac{1}{3 \cdot8^{x-1}} \end{split}$$

Answer 3

$$\frac{3^{-3+x}\cdot2^{3-x}\cdot 3^{3-x}}{3\cdot2^{2x}}=\frac{2^{3-3x}}{3}=\frac{8^{1-x}}{3}$$