Solve: (7^8)/(7^-4 × 4^2) × 32 Using Exponent Rules

First, we rewrite the given expression as a multiplication of fractions by applying the fraction multiplication rule in reverse:

$\frac{a\cdot c}{b\cdot d}= \frac{a}{b}\cdot\frac{c}{d}$

Let's apply this rule to the fraction in the problem:

$\frac{7^8}{7^{-4}\cdot4^2}\cdot32=\frac{7^8\cdot1}{7^{-4}\cdot4^2}\cdot32=\frac{7^8}{7^{-4}}\cdot\frac{1}{4^2}\cdot32$$$

In the first step, we recalled that any number can be written as itself times 1. In the next step, we rewrote the fraction as a product of separate fractions, ensuring that each fraction contained terms with identical bases.

Next, let's recall two laws of exponents:

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

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

b. The law of exponents for negative exponents but in reverse:

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

Let's apply these two laws of exponents to the expression we got in the last step:

$\frac{7^8}{7^{-4}}\cdot\frac{1}{4^2}\cdot32=7^{8-(-4)}\cdot4^{-2}\cdot32=7^{8+4}\cdot4^{-2}\cdot32=7^{12}\cdot4^{-2}\cdot32$

In the first step, we applied the law of exponents for division of terms with the same base (rule a) to the first factor, and the law of negative exponents (rule b) to the second factor. We then simplified the resulting expression.

Let's summarize the solution steps so far, we obtained the following:

$\frac{7^8}{7^{-4}\cdot4^2}\cdot32=\frac{7^8}{7^{-4}}\cdot\frac{1}{4^2}\cdot32=7^{8-(-4)}\cdot4^{-2}\cdot32=7^{12}\cdot4^{-2}\cdot32$

Now let's note an important fact:

The number 4 is a power of 2 and also the number 32 is a power of 2:

$4=2^2, \hspace{8pt}32=2^{5}$

Therefore, we can substitute these values into the expression from the previous step and rewrite it using base-2 terms:

$7^{12}\cdot4^{-2}\cdot32=7^{12}\cdot(2^2)^{-2}\cdot2^5$

From here we'll continue and recall two more laws of exponents:

c. The law of exponents for power of a power:

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

d. The law of exponents for multiplication between terms with identical bases:

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

Let's return to the problem and apply these two laws:

$7^{12}\cdot(2^2)^{-2}\cdot2^5=7^{12}\cdot2^{2\cdot(-2)}\cdot2^5=7^{12}\cdot2^{-4}\cdot2^5=7^{12}\cdot2^{-4+5}=7^{12}\cdot2^1=7^{12}\cdot2$

Where in the first step we applied the law of power of a power mentioned in c. to the second term in the multiplication, then we simplified the resulting expression and in the third step we applied the law of exponents for multiplication between terms with identical bases mentioned in d. and again simplified the resulting expression,

Let's summarize the solution steps so far, we obtained the following:

$\frac{7^8}{7^{-4}\cdot4^2}\cdot32=7^{12}\cdot4^{-2}\cdot32 =7^{12}\cdot(2^2)^{-2}\cdot2^5=7^{12}\cdot2^{2\cdot(-2)}\cdot2^5=7^{12}\cdot2$

Therefore using the multiplication substitution law we can observe that the correct answer is answer c.