Powers of a Fraction - Examples, Exercises and Solutions

We use the formula:

$(\frac{a}{b})^n=\frac{a^n}{b^n}$

$(\frac{2}{6})^3=(\frac{2}{2\times3})^3$

We simplify:

$(\frac{1}{3})^3=\frac{1^3}{3^3}$

$\frac{1\times1\times1}{3\times3\times3}=\frac{1}{27}$

$(\frac{4^2}{7^4})^2=\frac{4^{2\times2}}{7^{4\times2}}=\frac{4^4}{7^8}$

We use the formula:

$(\frac{a}{b})^{-n}=(\frac{b}{a})^n$

Therefore, we obtain:

$(\frac{3}{2})^4$

We use the formula:

$(\frac{b}{a})^n=\frac{b^n}{a^n}$

Therefore, we obtain:

$\frac{3^4}{2^4}=\frac{3\times3\times3\times3}{2\times2\times2\times2}=\frac{81}{16}$

We use the formula:

$(\frac{a}{b})^n=\frac{a^n}{b^n}$

We decompose the fraction inside of the parentheses:

$(\frac{1}{7})^4=\frac{1^4}{7^4}$

We obtain:

$7^4\times8^3\times\frac{1^4}{7^4}$

We simplify the powers: $7^4$

We obtain:

$8^3\times1^4$

Remember that the number 1 in any power is equal to 1, thus we obtain:

$8^3\times1=8^3$

This problem can be solved using the Law of exponents power rules for a negative power, power over a power, as well as the power rule for the product between terms with identical bases.

However we prefer to solve it in a quicker way:

To this end, the power by power law is applied to the parentheses in which the terms are multiplied, but in the opposite direction:

$x^n\cdot y^n=(x\cdot y)^n$Since in the expression in the problem there is a multiplication between two terms with identical powers, this law can be used in its opposite sense.

$5^4\cdot(\frac{1}{5})^4=\big(5\cdot\frac{1}{5}\big)^4$Since the multiplication in the given problem is between terms with the same power, we can apply this law in the opposite direction and write the expression as the multiplication of the bases of the terms in parentheses to which the same power is applied.

We continue and simplify the expression inside of the parentheses. We can do it quickly if inside the parentheses there is a multiplication between two opposite numbers, then their product will give the result: 1, All of the above is applied to the problem leading us to the last step:

$\big(5\cdot\frac{1}{5}\big)^4 = 1^4=1$We remember that raising the number 1 to any power will always give the result: 1, which means that:

$1^x=1$Summarizing the steps to solve the problem, we obtain the following:

$5^4\cdot(\frac{1}{5})^4=\big(5\cdot\frac{1}{5}\big)^4 =1$Therefore, the correct answer is option b.

In this problem, we are asked to determine whether it's an equality or an inequality, and if it's an inequality - what is its direction?

To do this, we will first use the law of exponents for negative exponents:

$x^{-n}=\frac{1}{x^n}$Before we start solving the problem, let's understand this law in a slightly different way:

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

$x^n$We get:

$x^{-n}=\frac{1}{x^n}\\ \frac{x^n}{1} =\frac{1}{x^n}\hspace{8pt} \text{/}\cdot x^n\\ x^n\cdot x^{-n}=1$In the first part we recall that any number can be represented as itself divided by 1. We apply this to the left side of the equation, then we multiply 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 current denominator to get the common denominator?".

Let's see the result we got:

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

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

And in particular:

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

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

$\frac{z}{w},\hspace{4pt}\frac{w}{z}$are reciprocal fractions to each other - which can be understood logically, as their multiplication will clearly give the result 1.

And if we combine this with the previous understanding, we can easily conclude that:

$\big(\frac{z}{w}\big)^{-1}=\frac{w}{z}$Meaning that raising a fraction to the power of negative one will give us the reciprocal fraction, obtained by swapping the numerator and denominator.

Let's return to the problem and apply these understandings, in addition we'll recall the law of multiplying exponents, but in the opposite direction:

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

We'll also apply this law to the problem, we'll first deal with the left term:

$\big(\frac{a}{b}\big)^{-7}\cdot b^8$We'll start with the first term in the expression:

$\big(\frac{a}{b}\big)^{-7}= \big (\frac{a}{b} \big )^{-1\cdot 7}= \big (\big (\frac{a}{b}\big )^{-1} \big )^{7}$In the first part we present the exponent expression as a multiplication between two numbers, in the second part 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, obtained by swapping the numerator with the denominator: we'll apply this to the first term in the expression we got in the last part:

$\big (\big (\frac{a}{b}\big )^{-1} \big )^{7} = \big (\frac{b}{a} \big )^{7}$Let's summarize. We got that:

$\big(\frac{a}{b}\big)^{-7}\cdot b^8 = \big (\big (\frac{a}{b}\big )^{-1} \big )^{7}\cdot b^8= \big (\frac{b}{a} \big )^{7} \cdot b^8$

Now let's return to the problem and examine what we have:

$\big (\frac{b}{a} \big )^{7} \cdot b^8 \text{ }_{—\text{ }}\big(\frac{b}{a}\big)^7 \cdot b^{-4}$We use the distributive property and rearrange the right-side expression.

Note that, on both sides, the first expression (i.e., the fraction with the exponent) is identical. However, the second term in the multiplication is different on both sides, and this is because it's given that:

b>1 (If it could also be equal to one, we could argue that maybe these terms could be equal, but it's given that it's greater than one and therefore these terms are certainly different).

Therefore we can conclude that this is not an equality but an inequality, and we need to determine its direction.

Next, let's note that since it's also given that:

a>0 We can conclude that:

\big (\frac{b}{a} \big )^{7} >0 And this is because both the numerator of the fraction and the denominator of the fraction are positive numbers,

And therefore the direction of the inequality is not dependent on this term, (if we didn't know the sign of this term for certain, we wouldn't be able to determine the direction of the inequality later on)

Meaning-

The term that will determine the direction of the inequality is the second term in the multiplication on both sides, meaning- we need to find the direction between the terms:

$b^8 \text{ }_{—\text{ }}b^{-4}$We keep the original sides these terms were on.

It will be enough to answer the given problem.

For this, we'll remember the rules of inequality for exponential expressions, which simply state that the direction of inequality between exponential expressions with equal bases will be determined both by the value of the bases and by the exponents in the following way:

For a base greater than one, the direction of inequality between the exponential expressions will maintain the direction of inequality between the exponents, meaning- for a base: $x$, such that:

$$ x>1 (The base is always defined to be a positive number)

And exponents $z,\hspace{4pt}w$such that: z>w It holds that:

x^z>x^w

And for a base smaller than 1 and greater than 0, the direction of inequality between the exponential expressions will be opposite to the direction of inequality between the exponents, meaning- for a base: $x$, such that:

$$ 1 >x>0 (The base is always defined to be a positive number)

And exponents $z,\hspace{4pt}w$such that: z>w It holds that:

x^w >x^z

Let's return then to the problem:

We are required to determine the direction of inequality between the expressions:

$b^8 \text{ }_{—\text{ }}b^{-4}$From what's given in the problem b>1 Meaning greater than one, and therefore the direction of inequality between the expressions will be the same as the direction of inequality that between the exponents.

Therefore, we'll examine the exponents of the expressions in question here.

Since it's clear that:

8>-4 Then it holds that:

b^8 \text{ }>{\text{ }}b^{-4}

And therefore the correct answer is answer B.

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.

We begin by addressing the problem as an equation:

$9^?(\frac{1}{2})^{-4}=\frac{16}{3}$

Therefore, we replace the question mark with an x and proceed to solve it:

$9^x(\frac{1}{2})^{-4}=\frac{16}{3}$

Let's briefly discuss the solution technique: Generally speaking the goal when solving exponential equations is to reach a situation whereby there is a term on each side of the equation so that both sides have the same base, in such a situation we can unequivocally state that the powers exponents on both sides of the equation are equal, and solve a simple equation for the unknown,

Mathematically, we will perform a mathematical manipulation (according to the laws of course) on both sides of the equation (or instead the development of one side of the equation with the help of power rules and algebra) Using the above methods we should reach the following stage as shown below.

$b^{m(x)}=b^{n(x)}$when$m(x),\hspace{4pt}n(x)$Algebraic expressions (actually functions of the unknown$x$) that can also exclude the unknowns ($x$) that we try to find in the problem, which is ultimately the solution to the equation,

It is then stated that:

$m(x)=n(x)$and we proceed to solve the simple equation that we obtained.

We solve the equation in the given problem again:

$9^x(\frac{1}{2})^{-4}=\frac{16}{3}$In solving this equation, various power rules are used:

a. Power rule for a negative exponent:

$a^{-n}=\frac{1}{a^n}$b. Power rule for a power of an exponent raised to another exponent:

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

First we will reach a simple presentation of the terms of the equation, that is, "eliminate" fractions and roots (if there are any in the problem, in this case there are none)

To do this, we start by treating the fraction on the right side of the equation, this is carried out by using the power rule of a negative exponent specified in A above and represent this fraction (in parentheses) as a term with a negative exponent:

$9^x(\frac{1}{2})^{-4}=\frac{16}{3} \\ 9^x(2^{-1})^{-4}=\frac{16}{3}\\ 9^x2^{(-1)\cdot(-4)}=\frac{16}{3}\\ 9^x2^{4}=\frac{16}{3}\\$We then proceed to the development on the left side of the equation as described above, and in the last step simplify the expression in the power exponent on the left side of the equation,

We then want to be able to obtain an identical base on both sides of the equation, the best way to achieve this is by decomposing all of the numbers in the problem into prime factors (using powers as well)

In the given problem the following numbers are present:

$16,\hspace{4pt}9,\hspace{4pt}3,\hspace{4pt}2$The numbers: 2, 3 are prime, so we will not touch them. Take note that the number 16 is a power of the number 2 and that the number 9 is a power of the number 3, that is:

$16=2^4\\ 9=3^2$This is the presentation (decomposition) of the numbers 16 and 9 with the help of their prime factors. Thus we refer back to the equation we obtained in the previous step and replace these numbers with the decomposition of their prime factors:

$9^x2^{4}=\frac{16}{3}\\ (3^2)^x2^{4}=\frac{2^4}{3}\\$Now we notice that we can eliminate the term.$2^4$By dividing both sides of the equation by it, we also notice that this term does not depend on the unknown and is different from zero and therefore there is no limitation that says it is forbidden to divide it,

Hence we proceed to do just this:

$(3^2)^x2^{4}=\frac{2^4}{3} \hspace{8pt}\text{/:}2^{4}\\ \frac{(3^2)^x\cdot\not{2^4}}{\not{2^4}}=\frac{\not{2^4}}{3\cdot\not{2^4}} \\ (3^2)^x=\frac{1}{3}$In the first step we divide both sides of the equation by the term we want to eliminate and then proceed to simplify the fractions obtained on both sides of the equation,

Now we return once more to the power laws that we have already used and that were mentioned before:

a. Power rule with negative exponent:

$a^{-n}=\frac{1}{a^n}$b. Power rule for a power of an exponent raised to another exponent:

$(a^m)^n=a^{m\cdot n}$In the next step, we apply the power raised to another power law specified in B above on the left side in order to eliminate the parentheses.

In the next step we deal with the right side with the goal of breaking down the fraction.

For this purpose we use the power property with a negative exponent specified in A above .

Below one can visualise this step by step:

$(3^2)^x=\frac{1}{3} \\ 3^{2x}=\frac{1}{3} \\ 3^{2x}=3^{-1}$Thus we have reached our goal. We obtained an equation in which both sides have terms with the same base, therefore we can affirm that the power exponents of the terms on both sides are equal, and in order to solve the resulting equation for the unknown, we do the following:

$3^{2x}=3^{-1} \\ \downarrow\\ 2x=-1$We then proceed to solve the resulting equation.

We can do this by isolating the unknown on the left side ,by dividing both sides of the equation by its coefficient:

$2x=-1 \hspace{8pt}\text{/:}2 \\ \bm{x=-\frac{1}{2} }$We have thus solved the given equation, we will briefly summarize the solution steps as shown below:$9^x(\frac{1}{2})^{-4}=\frac{16}{3} \\ 9^x2^{4}=\frac{16}{3}\\ (3^2)^x2^{4}=\frac{2^4}{3}\hspace{8pt}\text{/:}2^{4}\\ (3^2)^x=\frac{1}{3} \\ 3^{2x}=3^{-1} \\ \downarrow\\ 2x=-1\hspace{8pt}\text{/:}2 \\ \bm{x=-\frac{1}{2} }$Therefore, the correct answer is option c.

Let us begin by addressing the given problem as an equation:

$\big( \big(\frac{1}{5} \big)^2 \big)^?:5=125$Therefore, we shall replace the question mark with an x and proceed to solve it:

$\big( \big(\frac{1}{5} \big)^2 \big)^x:5=125$ Remember that dividing by a certain number is equivalent to multiplying by its inverse, so we will rewrite the given equation bearing this in mind:

$\big( \big(\frac{1}{5} \big)^2 \big)^x\cdot \frac{1}{5}=125$

Let's briefly discuss the solution technique:

Generally speaking the goal when solving exponential equations is to achieve a situation where there is a term on each of the two sides of the equation so that both sides have the same base. In such a situation we can unequivocally state that the power exponents on both sides of the equation are equal, and thus solve a simple equation for the unknown.

Mathematically, we will perform a mathematical manipulation (according to the laws of equation manipulation) on both sides of the equation. Or we will concentrate on the development of one of the sides of the equation with the help of power rules and algebra in order to reach the following situation:

$b^{m(x)}=b^{n(x)}$when $m(x),\hspace{4pt}n(x)$Algebraic expressions ( functions of the unknown $x$) that can also exclude the unknowns ($x$) that we are trying to find in the problem, which is the solution to the equation,

It is then stated that:

$m(x)=n(x)$and we solve the simple equation that we obtained.

We return to solving the equation in the given problem:

$\big( \big(\frac{1}{5} \big)^2 \big)^x\cdot \frac{1}{5}=125$In solving this equation, various power rules are used:

a. Power property with negative exponent:

$a^{-n}=\frac{1}{a^n}$b. Power property for a power of an exponent raised to another exponent:

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

Our initial goal is to simplify the terms of the equation, that is, "eliminate" fractions and roots (if there are any in the problem, there are none here)

To do this, we will start by dealing with the fraction on the left side of the equation:

$\frac{1}{5}$That is, both the fraction inside the parenthesis and the fraction outside the parenthesis, this is done with the help of the power rule for a negative exponent specified in A above. We then represent this fraction as a term with a negative power and in the next step we apply the power rule for a power of an exponent raised to another exponent specified in B above. We then are able to remove the parentheses starting from the inner parenthesis to the outer ones. This is shown below step by step:

$\big( \big(\frac{1}{5} \big)^2 \big)^x\cdot \frac{1}{5}=125 \\ \big( (5^{-1})^2 \big)^x\cdot 5^{-1}=125 \\ (5^{(-1)\cdot 2} )^x\cdot 5^{-1}=125 \\ 5^{(-1)\cdot 2\cdot x} \cdot 5^{-1}=125 \\ 5^{-2x} \cdot 5^{-1}=125 \\$When we carry out the development of the left side of the equation as described above, we initially apply the power rule for a negative exponent mentioned above in A.

In the following steps we apply the power rule for a power of an exponent raised to another exponent as mentioned above in B. We remove the parentheses: starting from the inner parenthesis to the outer. In the last step we simplify the expression in the power exponent on the left side of the equation,

c. Later we remember the power property for multiplying terms with identical bases:

$a^m\cdot a^n=a^{m+n}$Thus we apply this law to the left side of the equation that we obtained in the last step.

$5^{-2x} \cdot 5^{-1}=125 \\ 5^{-2x+(-1)}=125 \\ 5^{-2x-1}=125 \\$In the first step we apply the aforementioned power law to the product between members with identical bases mentioned above in C and in the following steps we simplify the expression in the power exponent on the left side,

Next, we seek to obtain the same base on both sides of the equation, the best way to achieve this is by decomposing each of the numbers in the problem into prime factors (using powers as well), you will notice that the number 125 is a power of the number 5, that is:

$125=5^3$This is the presentation (factorization) of the number 125 using its prime factor, which is the number 5.

So we return to the equation we obtained in the previous step and replace this number with its decomposition into prime factors:

$5^{-2x-1}=125 \\ 5^{-2x-1}=5^3 \\$We have reached our goal, we have obtained an equation in which both sides have terms with the same base, therefore we can state that the power exponents of the terms on both sides are equal, and in order solve the resulting equation for the unknown, we proceed as follows:

$5^{-2x-1}=5^3 \\ \\ \downarrow\\ -2x-1=3$We will continue to solve the resulting equation by isolating the unknown on the left side. We can achieve this in an usual way, by moving the sections and dividing the final equation by the unknown's coefficient:

$-2x-1=3 \\ -2x=3+1\\ -2x=4 \hspace{8pt}\text{/:}(-2) \\ \frac{\not{-2}x}{\not{-2}}=\frac{4}{-2}\\ x=-\frac{4}{2}\\ \bm{x=-2 }$In the first step we simplify the equation by moving the sides, remembering that when a term is moved its sign changes, then we complete the isolation by nullifying dividing both sides of the equation by its coefficient. In the last steps, we simplify the expression obtained by reducing the fractions,

We have thus solved the given equation. Below is a brief step by step summary of the solution:

$\big( \big(\frac{1}{5} \big)^2 \big)^x\cdot \frac{1}{5}=125 \\ \big( (5^{-1})^2 \big)^x\cdot 5^{-1}=125 \\ 5^{-2x} \cdot 5^{-1}=125 \\ 5^{-2x-1}=5^3 \\ \downarrow\\ -2x-1=3 \\ -2x=4 \hspace{8pt}\text{/:}(-2) \\ \bm{x=-2 }$Therefore, the correct answer is option a.