Powers of a Fraction: converting Negative Exponents to Positive Exponents

Let's start by emphasizing that this problem requires a different approach to applying the laws of exponents and is not as straightforward as many other problems solved so far. We should note that it's actually a very simplified expression, however, to understand which of the answers is correct, let's present it in a slightly different way,

Let's recall two of the laws of exponents:

a. The law of exponents raised to an exponent, but in the opposite direction:

$a^{m\cdot n} = (a^m)^n$b. The law of exponents applied to fractions, but in the opposite direction:

$\frac{a^n}{c^n} = \big(\frac{a}{c}\big)^n$

We'll work onthe two terms in the problem separately, starting with the first term on the left:

$\frac{z^{8n}}{m^{4t}}$

Note that both in the numerator and denominator, the number we are given in the exponents is a multiple of 4. Therefore, using the first law of exponents (in the opposite direction) mentioned above in a', we can represent both the term in the numerator and the term in the denominator as terms with an exponent of 4:

$\frac{z^{8n}}{m^{4t}}=\frac{z^{2n\cdot4}}{m^{t\cdot4}}=\frac{(z^{2n})^4}{(m^t)^4}$

First we see the exponents as a multiple of 4, and then we apply the law of exponents mentioned in a', to the numerator and denominator.

Next, we'll notice that both the numerator and the denominator are have the same exponent, and therefore we can use the second law of exponents mentioned in b', in the opposite direction:

$\frac{(z^{2n})^4}{(m^t)^4} =\big(\frac{z^{2n}}{m^t}\big)^4$

We could use the second law of exponents in its opposite direction because the terms in the numerator and denominator of the fraction have the same exponent.

Let's summarize the solution so far. We got that:

$\frac{z^{8n}}{m^{4t}}=\frac{(z^{2n})^4}{(m^t)^4}=\big(\frac{z^{2n}}{m^t}\big)^4$

Now let's stop here and take a look at the given answers:

Note that similar terms exist in all the answers, however, in answer a' the exponent (in this case its numerator and denominator are opposite to the expression we got in the last stage) is completely different from the exponent in the expression we got (that is - it's not even in the opposite sign to the exponent in the expression we got).

In addition, there's the coefficient 4 which doesn't exist in our expression, therefore we'll disqualify this answer,

Let's now refer to the proposed answer d' where only the first term from the multiplication in the given problem exists and it's clear that there's no information in the problem that could lead to the value of the second term in the multiplication being 1, so we'll disqualify this answer as well,

If so, we're left with answers b' or c', but the first term:

$(\frac{m^t}{z^{2n}})^{-4}$in them, is similar but not identical, to the term we got in the last stage:

$\big(\frac{z^{2n}}{m^t}\big)^4$The clear difference between them is in the exponent, which in the expression we got is positive and in answers b' and c' is negative,

This reminds us of the law of negative exponents:

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

Before we return to solving the problem let's understand this law in a slightly different, indirect way:

If we refer to this law as an equation (and it is indeed an equation for all intents and purposes), and multiply both sides of the equation by the common denominator which is:

$a^n$we'll get:

$a^{-n}=\frac{1}{a^n}\\ \frac{a^n}{1} =\frac{1}{a^n}\hspace{8pt} \text{/}\cdot a^n\\ a^n\cdot a^{-n}=1$

Here we remember that any number can be made into a fraction by writing it as itself divided by 1 , we applied this to the left side of the equation, then we multiplied by the common denominator and to know by how much we multiplied each numerator (after finding the common denominator) we asked the question "by how much did we multiply the current denominator to get the common denominator?".

Let's see 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).

We can apply this understanding to the problem if we also remember that the reciprocal number to a fraction is the number gotten 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 makes sense, since multiplying them will give us 1.

And 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 negative one will give a result that is the reciprocal fraction, gotten by swapping the numerator and denominator.

Let's return to the problem and apply these understandings. First we'll briefly review what we've done already:

We dealt with the first term from the left from the problem:

$\frac{z^{8n}}{m^{4t}}\cdot c^z=\text{?}$and after dealing with it using the laws of exponents we got that it can be represented as:

$\frac{z^{8n}}{m^{4t}}=\frac{(z^{2n})^4}{(m^t)^4}=\big(\frac{z^{2n}}{m^t}\big)^4$

Then after disqualifying answers a' and d' for the reasons mentioned earlier, we wanted to show that the term we got in the last stage:

$\big(\frac{z^{2n}}{m^t}\big)^4$is identical to the first term in the multiplication of terms in answers b' -c':

$(\frac{m^t}{z^{2n}})^{-4}$Now after we understood that raising a fraction to the power of $-1$will swap between the numerator and denominator, meaning that:

$\big(\frac{a}{b}\big)^{-1}=\frac{b}{a}$we can return to the expression we got for the first term in the multiplication , and present it as a term with a negative exponent and in the denominator of the fraction:

$\big(\frac{z^{2n}}{m^t}\big)^4 = \big(\big(\frac{m^t }{z^{2n}}\big)^{-1}\big)^4 = \big(\frac{m^t }{z^{2n}}\big)^{-1\cdot4}=\big(\frac{m^t }{z^{2n}}\big)^{-4}$

We applied the aforementioned understanding inside the parentheses and presented the fraction as the reciprocal fraction to the power of $-1$and in the next stage we applied the law of exponents raised to an exponent:

$(a^m)^n=a^{m\cdot n}$to the expression we got, then we simplified the expression in the exponent,

If so, we proved that the expression we got in the last step (the first expression in the problem) is identical to the first expression in the multiplication in answers b' and c',

We'll continue then and focus the choosing between these options for the second term in the problem.

The second term in the multiplication in the problem is:

$c^z$

$$Let's return to the proposed answers b' and c' (which haven't been disqualified yet) and note that actually only the second term in the multiplication in answer b' is similar to this term (and not in answer c'), except that it's in the denominator and has a negative exponent while in our case (the term in the problem) it's in the numerator (see note at the end of solution) and has a positive exponent.

This will again remind us of the law of negative exponents, meaning we'll want to present the term in the problem we're currently dealing with, as having a negative exponent and in the denominator, we'll do this as follows:

$c^z=c^{-(\underline{-z})}=\frac{1}{c^{\underline{-z}}}$$$

Here we present the term in question as having a negative exponent , using the multiplication laws, and then we applied the law of negative exponents:

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

Carefully - because the expression we're dealing with now has a negative sign (indicated by an underline , both in the law of exponents here and in the last calculation made)

Let's summarize:

$\frac{z^{8n}}{m^{4t}}\cdot c^z=\big(\frac{z^{2n}}{m^t}\big)^4\cdot c^z=\big(\frac{m^t }{z^{2n}}\big)^{-4}\cdot\frac{1}{c^{-z}}$And therefore the correct answer is answer b'.

Note:

When we see "the number in the numerator" when there's no fraction, it's because we can always refer to any number as a number in the numerator of a fraction if we remember that any number divided by 1 equals itself , meaning, we can always write a number as a fraction by writing it like this:

$X=\frac{X}{1}$and therefore we can actually refer to $X$as a number in the numerator of a fraction.

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.