In general, you make use of the rules of exponents. It can be helpful to understand where they come from.
An exponent signifies repeated multiplication.
[tex]x\cdot x\cdot x=x^{3}\qquad\text{the exponent 3 means x is a factor 3 times}[/tex]
When you multiply, you add exponents.
[tex](x\cdot x\cdot x)\times (x\cdot x)=(x\cdot x\cdot x\cdot x\cdot x)\\\\x^{3}\times x^{2}=x^{(3+2)}=x^{5}[/tex]
Likewise, when you divide, you subtract exponents. You can also think of this as adding the opposite of exponents that are in the denominator.
[tex]\dfrac{x\cdot x\cdot x}{x\cdot x}=\dfrac{x\cdot x}{x\cdot x}\times x=x\\\\\dfrac{x^{3}}{x^{2}}=x^{(3-2)}=x^{1}=x[/tex]
It should be no surprise then that if there are excess factors in the denominator, they can be expressed using a negative exponent.
[tex]\dfrac{x\cdot x}{x\cdot x\cdot x}=\dfrac{1}{x}\\\\\dfrac{x^{2}}{x^{3}}=x^{(2-3)}=x^{-1}\qquad\text{using exponents}[/tex]
The idea of using multiplication to show repeated addition applies to exponents as well.
[tex](x\cdot x)\times (x\cdot x)\times (x\cdot x)=(x\cdot x\cdot x\cdot x\cdot x\cdot x)\\\\=x^{2}\cdot x^{2}\cdot x^{2}=x^{(2+2+2)}\\\\=\left(x^{2}\right)^{3}=x^{2\cdot 3}=x^{6}[/tex]
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With these ideas in mind ...
[tex]\frac{2}{3}ab^{3}a^{4}=\frac{2}{3}a^{(1+4)}b^{3}=\frac{2}{3}a^5b^3[/tex]
[tex]\dfrac{8xy^{8}}{16y^{6}}=\frac{8}{16}xy^{(8-6)}=\frac{1}{2}xy^{2}[/tex]
