Applying Combined Exponents Rules: Using variables

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.

We begin by using the negative exponent rule.

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

$$$$$a^{-4}=\frac{1}{a^4}$Therefore, the correct answer is option B.

This question is actually a proof of the law of exponents for negative exponents, we will prove it simply using two other laws of exponents:

a. The zero exponent law, which states that raising any number to the power of 0 (except 0) will give the result 1:

$X^0=1$

b. The law of exponents for division between terms with identical bases:

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

Let's return to the problem and pay attention to two things, the first is that in the denominator of the fraction there is a term with base $a$and the second thing is that according to the zero exponent law mentioned above in a' we can always write the number 1 as any number (except 0) to the power of 0, particularly in this problem, given that $a\neq0$ we can claim that:

$1=a^0$

Let's apply this to the problem:

$\frac{1}{a^n}=\frac{a^0}{a^n}$

Now that we have in the numerator and denominator of the fraction terms with identical bases, we can use the law of division between terms with identical bases mentioned in b' in the problem:

$\frac{a^0}{a^n}=a^{0-n}=a^{-n}$

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

$\frac{1}{a^n}=\frac{a^0}{a^n}=a^{-n}$

In other words, we proved the law of exponents for negative exponents and understood why the correct answer is answer c.

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.