Applying Combined Exponents Rules: Using multiple rules

First we rewrite the first expression on the left of the problem as a fraction:

$\frac{a^2}{a}+a^3\cdot a^5$Then we use two properties of exponentiation, to multiply and divide terms with identical bases:

1.

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

2.

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

Returning to the problem and applying the two properties of exponentiation mentioned earlier:

$\frac{a^2}{a}+a^3\cdot a^5=a^{2-1}+a^{3+5}=a^1+a^8=a+a^8$

Later on, keep in mind that we need to factor the expression we obtained in the last step by extracting the common factor,

Therefore, we extract from outside the parentheses the greatest common divisor to the two terms which are:

$a$ We obtain the expression:

$a+a^8=a(1+a^7)$ when we use the property of exponentiation mentioned earlier in A.

$$$a^8=a^{1+7}=a^1\cdot a^7=a\cdot a^7$

Summarizing the solution to the problem and all the steps, we obtained the following:

$\frac{a^2}{a}+a^3\cdot a^5=a(1+a^{7)}$

Therefore, the correct answer is option b.

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

We use the formula:

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

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

We simplify:

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

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

We will solve the problem in two steps, in the first step we will use the power of a product rule:

$(z\cdot t)^n=z^n\cdot t^n$The rule states that the power affecting a product within parentheses applies to each of the elements of the product when the parentheses are opened,

We begin by applying the law to the given problem:

$(y^3\cdot x^2)^4=(y^3)^4\cdot(x^2)^4$When we open the parentheses, we apply the power to each of the terms of the product separately, but since each of these terms is already raised to a power, we must be careful to use parentheses.

We then use the power of a power rule.

$(b^m)^n=b^{m\cdot n}$We apply the rule to the given problem and we should obtain the following result:

$(y^3)^4\cdot(x^2)^4=y^{3\cdot4}\cdot x^{2\cdot4}=y^{12}\cdot x^8$When in the second step we perform the multiplication operation on the power exponents of the obtained terms.

Therefore, the correct answer is option d.

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.

First we will rearrange the expression and use the fact that multiplying a fraction means multiplying the numerator of the fraction, and the distributive property of multiplication:

$x^3\cdot x^4\cdot\frac{2}{x^3}\cdot x^{-8}=x^3\cdot x^4\cdot 2\cdot \frac{1}{x^3}\cdot x^{-8}=2\cdot x^3\cdot x^4\cdot \frac{1}{x^3}\cdot x^{-8}$$$Next, we'll use the law of exponents for negative exponents:

$a^{-n}=\frac{1}{a^n}$We'll apply the law of exponents to the expression in the problem:

$2\cdot x^3\cdot x^4\cdot\frac{1}{x^3}\cdot x^{-8} =2\cdot x^3\cdot x^4\cdot x^{-3}\cdot x^{-8}$When we applied the above law of exponents for the fraction in the multiplication,

From now on, we will no longer use the multiplication sign and will switch to the conventional notation where juxtaposition of terms means multiplication between them,

Now we'll recall the law of exponents for multiplying terms with the same base:

$a^m\cdot a^n=a^{m+n}$And we'll apply this law of exponents to the expression we got in the last step:

$2x^3x^4x^{-3}x^{-8} =2x^{3+4+(-3)+(-8)} =2x^{3+4-3-8}=2x^{-4}$When in the first stage we applied the above law of exponents and in the following stages we simplified the expression in the exponent,

Let's summarize the solution steps so far, we got that:
$x^3 x^4\cdot\frac{2}{x^3}\cdot x^{-8}= 2 x^3x^4 x^{-3} x^{-8} =2x^{-4}$

Now let's note that there is no such answer in the given options, a further check of what we've done so far will also reveal that there is no calculation error,

Therefore, we can conclude that additional mathematical manipulation is required to determine which is the correct answer among the suggested answers,

Let's note that in answers A and B there are similar expressions to the one we got in the last stage, however - we can directly rule out the other two options since they are clearly different from the expression we got,

Furthermore, we'll note that in the expression we got, x is in a negative exponent and is in the numerator (Note at the end of the solution on this topic), whereas in answer B it is in a positive exponent and in the denominator (and both are in the numerator - Note at the end of the solution on this topic), so we'll rule out this answer,

If so - we are left with only one option - which is answer A', however we want to verify (and need to verify!) that this is indeed the correct answer:

Let's note that in the expression we got x is in a negative exponent and is in the numerator (Note at the end of the solution on this topic), whereas in answer B it is in a positive exponent and in the denominator , which reminds us of the law of exponents for negative exponents mentioned at the beginning of the solution,

In addition, let's note that in answer B x is in the second power but inside parentheses that are also in the second power, whereas in the expression we got in the last stage of solving the problem x is in the fourth power which might remind us of the law of exponents for power to a power,

We'll check this, starting with the law of exponents for negative exponents mentioned at the beginning of the solution, but in the opposite direction:

$\frac{1}{a^n} =a^{-n}$Next, we'll represent the term with the negative exponent that we got in the last stage of solving the problem, as a term in the denominator of the fraction with a positive exponent:

$2x^{-4}=2\cdot\frac{1}{x^4}$When we applied the above law of exponents,

Next, let's note that using the law of exponents for power to a power, but in the opposite direction:

$a^{m\cdot n}= (a^m)^n$We can conclude that:

$x^4=x^{2\cdot2}=(x^2)^2$Therefore, we'll return to the expression we got in the last stage and apply this understanding:

$2\cdot\frac{1}{x^4} =2\cdot\frac{1}{(x^2)^2}$ Let's summarize then the problem-solving stages so far, we got that:

$x^3 x^4\cdot\frac{2}{x^3}\cdot x^{-8}=2x^{-4} =2\cdot\frac{1}{(x^2)^2}$Let's note that we still haven't got the exact expression suggested in answer A, but we are already very close,

To reach the exact expression claimed in answer A, we'll recall another important law of exponents, and a useful mathematical fact:

Let's recall the law of exponents for exponents applying to terms in parentheses, but in the opposite direction:

$\frac{a^n}{c^n}=\big(\frac{a}{c}\big)^n$And let's also recall the fact that raising the number 1 to any power will yield the result 1:

$1^{x}=1$And therefore we can write the expression we got in the last stage in the following way:

$2\cdot\frac{1}{(x^2)^2}=2\cdot\frac{1^2}{(x^2)^2}$And then since in the numerator and denominator of the fraction there are terms with the same exponent we can apply the above law of exponents, and represent the fraction whose numerator and denominator are terms with the same exponent as a fraction whose numerator and denominator are the bases of the terms and it is raised to the same exponent:

$2\cdot\frac{1^2}{(x^2)^2}=2\cdot\big(\frac{1}{x^2}\big)^2$ Let's summarize then the solution stages so far, we got that:

$x^3 x^4\cdot\frac{2}{x^3}\cdot x^{-8}=2x^{-4}=2\cdot\frac{1}{(x^2)^2}=2\cdot\big(\frac{1}{x^2}\big)^2$And therefore the correct answer is indeed answer A.

Note:

When it's written "the number in the numerator" despite the fact that there is no fraction in the expression at all, 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, that is, 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.

$\left(8by\right)^{3\cdot y}+3^{x\cdot a}$

We begin by applying the following rule:

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

We then open the parentheses according to the above rule.

$\left(abc\right)^x=a^x\cdot b^x\cdot c^x$

$8^{3y}\cdot b^{3y}\cdot y^{3y}+3^{xa}$

In solving the problem, we use two laws of exponents, which we will mention:

a. The law of exponents for multiplying powers with the same bases:

$a^m\cdot a^n=a^{m+n}$b. The law of exponents for a power of a power:

$(a^m)^n=a^{m\cdot n}$We will apply these two laws of exponents in solving the problem in two steps:

Let's start by applying the law of exponents mentioned in a' to the first expression on the left side of the problem:

$10^{-3}\cdot10^4=10^{-3+4}=10^1=10$When in the first step we applied the law of exponents mentioned in a' and in the following steps we simplified the expression that was obtained,

We continue to the next step and apply the law of exponents mentioned in b' and handle the third expression on the left side of the problem:

$(4^2)^5=4^{2\cdot5}=4^{10}$When in the first step we applied the law of exponents mentioned in b' and in the following steps we simplified the expression that was obtained,

We combine the two steps detailed above to the complete problem solution:

$10^{-3}\cdot10^4-(7\cdot9\cdot5)^3+(4^2)^5= 10-(7\cdot9\cdot5)^3+4^{10}$In the next step we calculate the result of multiplying the numbers inside the parentheses in the second expression on the left:

$10-(7\cdot9\cdot5)^3+4^{10}= 10-315^3+4^{10}$Therefore, the correct answer is answer b'.

We use the three appropriate power properties to solve the problem:

1. Power law for multiplication between terms with identical bases:

$a^m\cdot a^n=a^{m+n}$ 2. Power law for an exponent raised to another exponent:

$(a^m)^n=a^{m\cdot n}$ 3. Power law for the division of terms with identical bases:

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

We continue and apply the three previous laws to the problem:

$2^3\cdot2^4+(4^3)^2+\frac{2^5}{2^3}=2^{3+4}+4^{3\cdot2}+2^{5-3}=2^7+4^6+2^2$

In the first step we apply the power law mentioned in point 1 to the first expression on the left, the power law mentioned in point 2 to the second expression on the left, and the power law mentioned in point 3 to the third expression on the left, separately. In the second step, we simplify the expressions by exponents possession of the received terms,

Then,after using the substitution property for addition, we find that the correct answer is D.

In order to solve the problem we must use two power laws, as shown below:

A. Power property for terms with identical bases:

$\frac{a^m}{a^n}=a^{m-n}$B. Power property for an exponent raised to another exponent:

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

We will apply these two power laws to the problem in two steps:

Let's start by applying the power law specified in A to the second term from the left in the given problem:

$\frac{2^7}{2^4}=2^{7-4}=2^3$In the first step we apply the power law specified in A and then proceed to simplify the resulting expression,

We then advance to the next step and apply the power law specified in B to the third term from the left in the given problem :

$(8^2)^5=8^{2\cdot5}=8^{10}$In the first stage we apply the power law specified in B and then proceed to simplify the resulting expression,

Let's summarize the two steps listed above to solve the general problem:

$(4\cdot7)^9+\frac{2^7}{2^4}+(8^2)^5= (4\cdot7)^9+2^3+8^{10}$In the final step, we calculate the result of multiplying the terms within the parentheses in the first term from the left:

$(4\cdot7)^9+2^3+8^{10}=28^9+2^3+8^{10}$Therefore, the correct answer is option c.

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.

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.