Powers of a Fraction: Complete the equation

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.