Solve (7/8)^(-4) · 8/7 + 7/8 · (8/7)^(-3): Complete Fraction Expression

We will use the following law of negative exponents:

$a^{-n}=\frac{1}{a^n}$Before we approach solving the problem we will understand this law in a slightly different way:

Note that if we treat this law as an equation (and it is indeed an equation in every sense), and multiply both sides of the equation by the common denominator which is:

$a^n$we get:

$a^{-n}=\frac{1}{a^n}\\ \frac{a^n}{1} =\frac{1}{a^n}\hspace{8pt} \text{/}\cdot a^n\\ a^n\cdot a^{-n}=1$In the first stage we remembered that any number can be made into a fraction simply by dividing by 1, we applied this to the left side of the equation, then we multiplied by the common denominator. To know by how much we need to multiply each numerator (after reduction with the common denominator) we ask the question "By how much did we multiply the original denominator to get the common denominator?".

Let's pay attention to the result we got:

$a^n\cdot a^{-n}=1$meaning that $a^n,\hspace{4pt}a^{-n}$are reciprocal numbers to each other, or in other words:

$a^n$is reciprocal to $a^{-n}$(and vice versa),

and in particular:

$a,\hspace{4pt}a^{-1}$are reciprocal to each other,

We can apply this understanding to the problem if we also remember that we get the reciprocal number of a fraction by swapping the numerator and denominator, meaning that the fractions:

$\frac{a}{b},\hspace{4pt}\frac{b}{a}$are reciprocal fractions to each other - which can be easily checked, since multiplying them will clearly give us 1.

If we combine this with the previous understanding, we can conclude that:

$\big(\frac{a}{b}\big)^{-1}=\frac{b}{a}$meaning that raising a fraction to the power of minus one will always give the reciprocal fraction, obtained by swapping the numerator and denominator.

Let's return to the problem and apply these understandings.

In addition we'll remember the law of multiplying exponents, but in the opposite direction:

$(a^m)^n=a^{m\cdot n}$We'll also apply this law to the problem:

$\big (\frac{7}{8} \big )^{-4}\cdot\frac{8}{7}+\frac{7}{8}\cdot \big (\frac{8}{7} \big )^{-3}=\text{?}$where we'll deal with each of the terms in the given sum separately:

a. We'll start by dealing separately with the first term from the left in the sum in the problem:

$\big (\frac{7}{8} \big )^{-4}\cdot\frac{8}{7}= \big (\frac{7}{8} \big )^{-1\cdot 4}\cdot\frac{8}{7}= \big (\big (\frac{7}{8}\big )^{-1} \big )^{4}\cdot\frac{8}{7}$In the first step, we presented the exponent expression as a multiplication, in the second step we applied the law of multiplying exponents in the opposite direction.

Next, we'll recall that raising a fraction to the power of negative one will always give the reciprocal fraction. We'll apply this to the first term in the multiplication expression we got in the last stage:

$\big (\big (\frac{7}{8}\big )^{-1} \big )^{4}\cdot\frac{8}{7}= \big (\frac{8}{7} \big )^{4}\cdot\frac{8}{7}$From here we notice that all the terms in the multiplication expression we got in the last stage have identical bases:

$\frac{8}{7}$So we'll remember the law of multiplying exponents with identical bases:$a^m\cdot a^n=a^{m+n}$and we'll apply this law to the expression we got in the last stage:

$\big (\frac{8}{7} \big )^{4}\cdot\frac{8}{7} =\big (\frac{8}{7} \big )^{4+1} =\big (\frac{8}{7} \big )^{5}$ In the first stage we applied the above-mentioned law of exponents while remembering that: $\frac{8}{7}=\big(\frac{8}{7}\big)^1$and in the following stages we simplified the expression in the exponent

Let's summarize what we've done so far:

For the first term in the sum in the problem, we got that:

$\big (\frac{7}{8} \big )^{-4}\cdot\frac{8}{7}= \big (\frac{8}{7} \big )^{4}\cdot\frac{8}{7} =\big (\frac{8}{7} \big )^{5}$

b. Now we'll move on to dealing with the second term:

$\frac{7}{8}\cdot \big (\frac{8}{7} \big )^{-3}$

We'll use the commutative law of multiplication and swap, for convenience, between the two terms in the multiplication expression we're dealing with now, then we'll apply (again) the law of multiplying exponents, but in its opposite direction:

$(a^m)^n=a^{m\cdot n}$ and then - we'll treat this term in the same way as we did for the first term.

$\big (\frac{8}{7} \big )^{-3}\cdot \frac{7}{8}= \big (\frac{8}{7} \big )^{-1\cdot 3}\cdot\frac{7}{8}= \big (\big (\frac{8}{7}\big )^{-1} \big )^{3}\cdot\frac{7}{8}$In the first step we present the exponent expression as a multiplication, in the second step we apply the law of multiplying exponents in its opposite direction.

Next, we'll apply the understanding that raising a fraction to the power of negative one will always give the reciprocal fraction.

We'll apply this to the first term in the multiplication expression we got in the last step:

$\big (\big (\frac{8}{7}\big )^{-1} \big )^{3}\cdot\frac{7}{8} = \big (\frac{7}{8} \big )^{3}\cdot\frac{7}{8}$From here we notice that all the terms in the multiplication expression we got in the last stage have identical bases:

$\frac{7}{8}$So we'll apply the law of multiplying exponents with identical bases:$a^m\cdot a^n=a^{m+n}$and we'll apply this law to the expression we got in the last stage:

$\big (\frac{7}{8} \big )^{3}\cdot\frac{7}{8} =\big (\frac{7}{8} \big )^{3+1} =\big (\frac{7}{8} \big )^{4}$In the first stage we applied the above-mentioned law of exponents while remembering that: $\frac{7}{8}=\big(\frac{7}{8}\big)^1$and in the following stages we simplified the expression in the exponent.

Let's summarize what we've got so far for the second term from the left in the problem.

We got that:

$\big (\frac{8}{7} \big )^{-3}\cdot \frac{7}{8} = \big (\frac{7}{8} \big )^{3}\cdot\frac{7}{8} =\big (\frac{7}{8} \big )^{4}$

Now let's return to the original problem and swap the original terms with solution a and b .

We got that:

$\big (\frac{7}{8} \big )^{-4}\cdot\frac{8}{7}+\frac{7}{8}\cdot \big (\frac{8}{7} \big )^{-3}= \big (\frac{8}{7} \big )^{4}\cdot\frac{8}{7} + \big (\frac{7}{8} \big )^{3}\cdot\frac{7}{8} =\big (\frac{8}{7} \big )^{5}+ \big (\frac{7}{8} \big )^{4}$

Therefore, the correct answer is answer a.