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.
![x\cdot x\cdot x=x^(3)\qquad\text{the exponent 3 means x is a factor 3 times}](https://img.qammunity.org/2019/formulas/mathematics/middle-school/b995tde74emlcpeudns4t5pwvfyuv9m539.png)
When you multiply, you add exponents.
![(x\cdot x\cdot x)* (x\cdot x)=(x\cdot x\cdot x\cdot x\cdot x)\\\\x^(3)* x^(2)=x^((3+2))=x^(5)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/zm8n6tiwlzryvc442xs06nf7hlxxzkchyn.png)
Likewise, when you divide, you subtract exponents. You can also think of this as adding the opposite of exponents that are in the denominator.
![(x\cdot x\cdot x)/(x\cdot x)=(x\cdot x)/(x\cdot x)* x=x\\\\(x^(3))/(x^(2))=x^((3-2))=x^(1)=x](https://img.qammunity.org/2019/formulas/mathematics/middle-school/jphw0nonv3saxk61hyq3sepwkrqnz2p2jr.png)
It should be no surprise then that if there are excess factors in the denominator, they can be expressed using a negative exponent.
![(x\cdot x)/(x\cdot x\cdot x)=(1)/(x)\\\\(x^(2))/(x^(3))=x^((2-3))=x^(-1)\qquad\text{using exponents}](https://img.qammunity.org/2019/formulas/mathematics/middle-school/pyuw71fl7ltu1rkf1qao1m03v2ms1cbj76.png)
The idea of using multiplication to show repeated addition applies to exponents as well.
![(x\cdot x)* (x\cdot x)* (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)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/y43r66f5jerdaifpjxfsh7pckeqvmfxyl5.png)
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With these ideas in mind ...
![(2)/(3)ab^(3)a^(4)=(2)/(3)a^((1+4))b^(3)=(2)/(3)a^5b^3](https://img.qammunity.org/2019/formulas/mathematics/middle-school/xrsebpjmip0lahfgmlsnz3zgwknczc2625.png)
![(8xy^(8))/(16y^(6))=(8)/(16)xy^((8-6))=(1)/(2)xy^(2)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/54355pnbsmt14swhoyurr9o45sarsp0514.png)