Applying Combined Exponents Rules: Variable in the exponent of the power

Let's use the law of exponents for negative exponents:

$a^{-n}=\frac{1}{a^n}$and apply it to our problem:

$$$$$8^{-2x}=\frac{1}{8^{2x}}$Next, we'll use the law of exponents for power of a power:

$(a^m)^n=a^{m\cdot n}$and apply this law to the denominator in the expression we got:

$\frac{1}{8^{2x}}=\frac{1}{(8^2)^x}=\frac{1}{64^x}$where we actually used the above law in the opposite direction, meaning instead of expanding the parentheses and multiplying by the power exponent, we interpreted the multiplication by the power exponent as a power of a power, and in the final stage we calculated the power inside the parentheses in the denominator.

Let's summarize the solution steps, we got that:

$8^{-2x}= \frac{1}{8^{2x}}=\frac{1}{64^x}$

Therefore, the correct answer is answer D.

Let's handle each term in the initial expression separately:

a. We'll start with the leftmost term, meaning the exponent on the multiplication in parentheses,

We'll use the power rule for exponents on multiplication in parentheses:

$(z\cdot t)^n=z^n\cdot t^n$

which states that when an exponent applies to a multiplication in parentheses, it applies to each term in the multiplication when opening the parentheses,

Let's apply this to our problem for the leftmost term:

$(g\cdot a\cdot x)^4=g^4\cdot a^4\cdot x^4=g^4a^4x^4$

where in the final step we dropped the multiplication sign and switched to the conventional multiplication notation by placing the terms next to each other.

We're done with the leftmost term, let's move on to the next term.

b. Let's continue with the second term from the left, using the power rule for exponents:

$(b^m)^n=b^{m\cdot n}$

Let's apply this rule to the second term from the left:

$(4^a)^x=4^{ax}$

and we're done with this term as well,

Let's summarize the results from a and b for the two terms in the initial expression:

$(g\cdot a\cdot x)^4+(4^a)^x=g^4a^4x^4+4^{ax}$

Therefore, the correct answer is c.

Notes:

a. For clarity and better explanation, in the solution above we handled each term separately. However, to develop proficiency and mastery in applying exponent rules, it is recommended to solve the problem as one unit from start to finish, where the separate treatment mentioned above can be done in the margin (or on a separate draft) if unsure about handling a specific term.

b. From the stated power rule for parentheses mentioned in solution a, it might seem that it only applies to two terms in parentheses, but in fact, it is valid for any number of terms in a multiplication within parentheses, as demonstrated in this problem and others,

It would be a good exercise to prove that if this rule is valid for exponents on multiplication of two terms in parentheses (as stated above), then it is also valid for exponents on multiplication of multiple terms in parentheses (for example - three terms, etc.).

Let's start with multiplying the fractions, remembering that multiplication of fractions is performed by multiplying numerator by numerator and denominator by denominator:

$\frac{a^{20b}}{a^{15b}}\cdot\frac{a^{3b}}{a^{2b}}=\frac{a^{20b}\cdot a^{3b}}{a^{15b}\cdot a^{2b}}$

Next, we'll note that both in the numerator and denominator, multiplication occurs between terms with identical bases, so we'll use the power law for multiplying terms with identical bases:

$c^m\cdot c^n=c^{m+n}$

We emphasize that this law can only be used when multiplication is performed between terms with identical bases.

From this point forward, we will no longer use the multiplication sign, but instead use the conventional notation where placing terms next to each other implies multiplication.
Let's return to the problem and apply the above power law separately to the fraction's numerator and denominator:

$\frac{a^{20b}a^{3b}}{a^{15b}a^{2b}}=\frac{a^{20b+3b}}{a^{15b+2b}}=\frac{a^{23b}}{a^{17b}}$

where in the final step we calculated the sum of the exponents in the numerator and denominator.

Now we notice that we need to perform division between two terms with identical bases, so we'll use the power law for dividing terms with identical bases:

$\frac{c^m}{c^n}=c^{m-n}$

We emphasize that this law can only be used when division is performed between terms with identical bases.

Let's return to the problem and apply the above power law:

$\frac{a^{23b}}{a^{17b}}=a^{23b-17b}=a^{6b}$

where in the final step we calculated the subtraction between the exponents.

We have obtained the most simplified expression and therefore we are done.

Therefore, the correct answer is D.

We use the exponential property of a negative exponent:

$b^{-n}=\frac{1}{b^n}$We apply it to the problem:

$$$x^{-a}=\frac{1}{x^a}$Therefore, the correct answer is option C.