Power of a Quotient Rule for Exponents: Numbers as coefficients

Due to the fact that the numerator and the denominator of the fraction have terms with identical bases, we will begin by applying the law of exponents for the division of terms with identical bases:

$\frac{b^m}{b^n}=b^{m-n}$We apply it to the problem:

$\frac{3a^2}{2a}=\frac{3}{2}\cdot a^{2-1}=\frac{3}{2}\cdot a^1$In the first step we simplify the numerical part of the fraction. This is a simple and intuitive step which makes it easier to work with the said fraction.

$\frac{3a^2}{2a}=\frac{3}{2}\cdot\frac{a^2}{a}=\frac{3}{2}\cdot a^{2-1}=\ldots$Let's return to the problem, remember that any number raised to 1 is equal to the number itself, that is:

$b^1=b$Thus we apply it to the problem:

$\frac{3}{2}\cdot a^1=\frac{3}{2}\cdot a=1\frac{1}{2}a$In the last step we convert the fraction into a mixed fraction.

$$Therefore, the correct answer is option D.

Due to the fact that the numerator and the denominator of the fraction have terms with identical bases, we will begin by applying the law of exponents for the division of terms with identical bases:

$\frac{b^m}{b^n}=b^{m-n}$We begin by applying the formula to the given problem:

$\frac{4a^5}{2a^3}=2\cdot a^{5-3}=2\cdot a^2$In the first step we simplify the numerical part of the fraction. This is simple to do and makes it easier to work with the said fraction.

$\frac{4a^5}{2a^3}=\frac{4}{2}\cdot\frac{a^5}{a^3}=2\cdot a^{5-3}=\ldots$We obtain the following answer:

$2a^2$

$$Therefore, the correct answer is option A.

Due to the fact that the numerator and the denominator of the fraction have terms with identical bases, we will begin by applying the law of exponents for the division of terms with identical bases:

$\frac{b^m}{b^n}=b^{m-n}$We apply it to the problem:

$\frac{14a^{-3}}{7a^{-3}}=2a^{-3-(-3)}=2a^{-3+3}=2a^0$In the first step we simplify the numerical part of the fraction. This is a simple and intuitive step as it makes it easier to work with the said fraction.:

$\frac{14a^{-3}}{7a^{-3}}=\frac{14}{7}\cdot\frac{a^{-3}}{a^{-3}}=2a^{-3-(-3)}=\ldots$We then return to the problem and remember that any number raised to the 0th power is 1, that is:

$b^0=1$Thus, in the problem we obtain the following:

$2a^0=2\cdot1=2$Therefore, the correct answer is option B.

Let's consider that the numerator and the denominator of the fraction have terms with identical bases, therefore we will use the law of exponents for the division of terms with identical bases:

$\frac{c^m}{c^n}=c^{m-n}$We apply it to the problem:

$\frac{12b^4}{4b^{-5}}=3\cdot b^{4-(-5)}=3\cdot b^{4+5}=3b^9$When in the first step we simplify the numerical part of the fraction. This operation is intuitive as well as correct since it is possible to write down in advance the said fraction as a product of fractions and reduce:

$\frac{12b^4}{4b^{-5}}=\frac{12}{4}\cdot\frac{b^4}{b^{-5}}=3\cdot b^{4-(-5)}=\ldots$We return once again to the problem. The simplified expression obtained is as follows:

$3b^9$

$$Therefore, the correct answer is option D.

Due to the fact that the numerator and the denominator of the fraction have terms with identical bases, we will begin by applying the law of exponents for the division of terms with identical bases:

$\frac{b^m}{b^n}=b^{m-n}$We apply it to the problem:

$\frac{-3a^{-2}}{-6a^{-6}}=\frac{1}{2}\cdot a^{-2-(-6)}=\frac{1}{2}\cdot a^{-2+6}=\frac{1}{2}\cdot a^4$In the first step we simplify the numerical part of the fraction. This is a simple and intuitive step as it makes it easier to work with the said fraction.

$\frac{-3a^{-2}}{-6a^{-6}}=\frac{-3}{-6}\cdot\frac{a^{-2}}{a^{-6}}=\frac{1}{2}\cdot\frac{a^{-2}}{a^{-6}}=\ldots$We then return to the problem and subsequently obtain the following expression:

$\frac{1}{2}\cdot a^4$Therefore, the correct answer is option C.