Applying Combined Exponents Rules: More than one unknown

We use the power property to multiply terms with identical bases:

$a^m\cdot a^n=a^{m+n}$It is important to note that this property is only valid for terms with identical bases,

We return to the problem

We notice that in the problem there are two types of terms with different bases. First, for the sake of order, we will use the substitution property of multiplication to rearrange the expression so that the two terms with the same base are grouped together. Then, we will proceed to work:

$a\cdot b\operatorname{\cdot}a\operatorname{\cdot}b\operatorname{\cdot}a^2=a\cdot a\cdot a^2\cdot b\cdot b$Next, we apply the power property for each type of term separately,

$a\cdot a\cdot a^2\cdot b\cdot b=a^{1+1+2}\cdot b^{1+1}=a^4\cdot b^2$

$$We apply the power property separately - for the terms whose bases are$a$and then for the terms whose bases are$b$and we add the exponents and simplify the terms.

$$Therefore, the correct answer is option c.

Note:

We use the fact that:

$a=a^1$and the same for $b$.

We use the power property to multiply terms with identical bases:

$a^m\cdot a^n=a^{m+n}$It should be noted that this property is only valid for terms with identical bases,

We return to the problem

We notice that in the problem there are two types of terms with different bases. First, for the sake of order, we will use the substitution property of multiplication to rearrange the expression so that the two terms with the same base are grouped together. Then, we will proceed to work:

$E^6\cdot F^{-4}\cdot E^0\cdot F^7\cdot E=E^6\cdot E^0\cdot E\cdot F^{-4}\cdot F^7$Next, we apply the power property for each type of term separately,

$E^6\cdot E^0\cdot E\cdot F^{-4}\cdot F^7=E^{6+0+1}\cdot F^{-4+7}=E^7\cdot F^3$

$$We apply the power property separately - for the terms whose bases are$E$and for the terms whose bases are$F$and we add the exponents and simplify the terms with the same base.

The correct answer is then option d.

Note:

We use the fact that:

$E=E^1$.