We address the problem:

$\big( \big(\frac{1}{5} \big)^2 \big)^?:5=125$as an equation for all **(and of course it is indeed an equation)**,

**Therefore, we replace the problem sign in the unknown x and solve it:**

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

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

**Now we briefly discuss the solution technique:**

In quite a general way, 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 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 development of one of the sides with the help of power properties and algebra) and arrive at the following situation:

$b^{m(x)}=b^{n(x)}$when $m(x),\hspace{4pt}n(x)$**Algebraic expressions (actually functions of the unknown $x$) that can also exclude the unknowns ($x$) that we 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 we obtain,**

__We go back 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 properties 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:

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

**First we will arrive at a simple presentation of 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 property with negative exponent specified in A above **and we represent this fraction as a term with negative power** and in the next step we will apply the power property for a power of an exponent raised to another exponent specified in B above** and we will get rid of the parentheses, another step starting from the inner parenthesis to the outer, **we will do this, step by step below:

$\big( \big(\frac{1}{5} \big)^2 \big)^x\cdot \frac{1}{5}=125 \\
\big( (5^{-1})^2 \big)^x\cdot 5^{-1}=125 \\
(5^{(-1)\cdot 2} )^x\cdot 5^{-1}=125 \\
5^{(-1)\cdot 2\cdot x} \cdot 5^{-1}=125 \\
5^{-2x} \cdot 5^{-1}=125 \\$When we carry out the development of the left side of the equation as described above, we initially apply the power property with negative exponent mentioned above in A and in the following steps we apply the power property for a power of an exponent raised to another exponent mentioned above in B and we got rid of 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:

$a^m\cdot a^n=a^{m+n}$And we will apply this law to the left side of the equation we obtained in the last step, **this is to have a term on this side**, we will do this:

$5^{-2x} \cdot 5^{-1}=125 \\
5^{-2x+(-1)}=125 \\
5^{-2x-1}=125 \\$When 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 would like to obtain the same base on both sides of the equation**, the best way to achieve this is **by decomposing each and every one 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 go back to the equation we received 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 received 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 to solve the resulting equation for the unknown, we will do this:

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

$-2x-1=3 \\
-2x=3+1\\
-2x=4 \hspace{8pt}\text{/:}(-2) \\
\frac{\not{-2}x}{\not{-2}}=\frac{4}{-2}\\
x=-\frac{4}{2}\\
\bm{x=-2 }$When 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, **we briefly summarize** the solution steps:

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