When we encounter an expression with a quotient (or division) inside parentheses and the entire expression is raised to a certain exponent, we can take the exponent and apply it to each of the terms in the expression. Let's not forget to maintain the fraction bar between the terms. Formula of the property: $(\frac {a}{b})^n=\frac {a^n}{b^n}$ This property is also relevant to algebraic expressions.

Let's observe that we have a power of a quotient, and the exponent$2$ is applied to the entire expression within the parentheses. Therefore, we can apply it to each of the terms while maintaining the fraction bar between them. We will obtain: $\frac{3^2}{4^2}=\frac{9}{16}$

If we have an exercise in which there is a quotient (or a fraction) and an exponent assigned to the entire fraction, through parentheses, we will apply the exponent to the entire expression in the numerator and in the denominator separately.

Let's look at some examples

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

We will see that the exponent $4$ is placed over the entire fraction, therefore, we can apply it to all terms, both in the numerator and in the denominator.

$\frac{\left(2^3\right)^4}{\left(x\right)^4}$

We will realize that in the numerator we have a base with exponent $3$, then, we will apply the external exponent on $2^3$ and not only on the $2$.

We will obtain:

$\frac{2^{12}}{X^4}$

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Now let's move on to a slightly more complex exercise.

$(\frac{(X^3\cdot2^4)}{4^2})^3=$

Don't be scared. If you work according to the rules, you can solve the exercise very easily.

First, we will notice that the external exponent to the parentheses applies to the whole fraction, so we apply the exponent to both the numerator and the denominator.

In the numerator, we obtain a power of a multiplication, so we apply the power to each one, maintaining the multiplication sign between them and we get:

Let's remember the following property that serves to solve a power of a power:

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

We will do it and we will obtain:

$(\frac{X^9\cdot2^{12}}{4^6})=$

Great! Now... We will notice that we can express the $4$ in the denominator as $2^2$ and thus we can apply the law of the quotient of powers of equal base.

We will do it and we will obtain:

$(\frac{X^9\cdot2^{12}}{(2^2)^6})=$

Now let's apply again the law of power of a power to the denominator and we will obtain:

$(\frac{X^9\cdot2^{12}}{2^{12}})=$

Great! This allows us to reduce very easily, which will give us:

$X^9$

Let's move on to an example that only includes unknowns.

$(\frac{Y^4}{X})^2\cdot(\frac{X}{Y})^5=$

We will notice that the exponents that are outside of the parentheses act on the entire fraction, therefore, we will apply them to each term in the numerator and in the denominator.

We will obtain:

$\frac{Y^8}{X^2}\cdot\frac{X^5}{Y^5}=$

Now, as we have a multiplication between fractions, we can multiply the numerator by the numerator and the denominator by the denominator and thus combine the two fractions into only one in the following way:

$\frac{Y^8\cdot X^5}{X^2\cdot Y^5}=$

Great! We have a single fraction. According to the property of the quotient of the same base, we can subtract the exponents (exponent of the numerator minus exponent of the denominator of the same base) and keep one base and one exponent.

Let's apply this property and we will obtain:

$Y^3\cdot X^3=$

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In this task, there is a combination of two properties, the power of a quotient rule, which allows us to apply the power to both the numerator and denominator in the fraction, along with the power of a power rule, which allows us to multiply the outer power by the inner one.

According to the power law, when we encounter an expression in which a power operates on an expression that contains a base and its own power (in parentheses), we can multiply the powers and convert the product obtained into a new power that we apply to the base number.

This task involves the use of two laws, both the power of a product and the power of a power. Each of the factors inside the parentheses receives the external power, since they have different bases and a multiplication operation between them. The power inside the parentheses is multiplied by the power outside of it, according to the property of the power of a power.

Even when it comes to unknowns, the property remains valid. When we encounter an expression in which the value of the power appears in the coefficient or throughout the exercise where there are only division operations among the fractions of the extremes (using parentheses throughout the expression), we can take the value of the power and apply it to each of the products

Answer:

$4^{xy}$

Exercise 6

Solve the following exercise:

$\left(4^2\right)^3+\left(g^3\right)^4=$

Solution:

When there are different products joined together in a sum, and each of them has parentheses with an external power, it is possible to use the property twice for each of the products separately.

When we have a power of a quotient, we can apply the power to the numerator and the denominator separately, maintaining the quotient operation.

How to calculate the power of a quotient?

Independently, the numerator and the denominator are raised to the indicated power, and the quotient is maintained.

What is a power definition?

A power is an abbreviation of an operation in which a number called the base is multiplied by itself as many times as indicated by another number called the exponent.

How is the quotient of a power with the same base calculated?

When we have a quotient with the same base, we can simplify the expression by subtracting the exponent of the denominator from the exponent of the numerator. The base is placed and the result of the subtraction is placed as the exponent.

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:

$8^3\times1=8^3$

Answer

$8^3$

Exercise #5

$5^4\cdot(\frac{1}{5})^4=\text{?}$

Video Solution

Step-by-Step Solution

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.