Powers of a Fraction - Examples, Exercises and Solutions

To solve the expression $\left(\frac{1}{3}\right)^3$, we need to apply the rule for exponents of a fraction, which states:

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

Using this property, we can rewrite the fraction with its exponent as follows:

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

Now, calculate the powers of the numerator and the denominator separately:

• $1^3 = 1$

• $3^3 = 27$

Thus, putting it all together, we have:

$\frac{1^3}{3^3} = \frac{1}{27}$

This shows that raising both the numerator and the denominator of a fraction to a power involves calculating the power of each part separately and then constructing a new fraction.

The solution to the question is: $\frac{1}{27}$

To solve the given power expression, we need to apply the formula for powers of a fraction. The expression we are given is:
$\left(\frac{2}{3}\right)^3$

Let's break down the steps:

• When we raise a fraction to a power, we apply the exponent to both the numerator and the denominator separately. This means raising both 2 and 3 to the power of 3.
• Thus, we calculate:
  $2^3 = 8$ and $3^3 = 27$.
• Therefore, $\left(\frac{2}{3}\right)^3 = \frac{2^3}{3^3} = \frac{8}{27}$.

So, the result of the expression $\left(\frac{2}{3}\right)^3$ is $\frac{8}{27}$.

We use the formula:

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

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

We simplify:

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

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

To solve the problem, we need to simplify the expression $\left(\frac{11\times9}{10\times12}\right)^{x+a}$ and write it in the form requested in the question.

We begin by using the exponent rule: $(\frac{a}{b})^n = \frac{a^n}{b^n}$. Applying this rule here:

$<span class="katex"> \left(\frac{11\times9}{10\times12}\right)^{x+a} = \frac{(11\times9)^{x+a}}{(10\times12)^{x+a}} </span>$

Next, we can simplify the expression further by applying the power over a product rule: $(ab)^n = a^n \times b^n$.

Applying this rule to both the numerator and denominator gives us:

Numerator: $(11\times9)^{x+a} = 11^{x+a} \times 9^{x+a}$

Denominator: $(10\times12)^{x+a} = 10^{x+a} \times 12^{x+a}$

Therefore, the entire expression becomes:

$<span class="katex"> \frac{11^{x+a} \times 9^{x+a}}{10^{x+a} \times 12^{x+a}} </span>$

This matches the given answer. Thus, the solution to the question is:

$\frac{11^{x+a}\times9^{x+a}}{10^{x+a}\times12^{x+a}}$

Let's start by examining the expression given in the question:

$\left(\frac{13}{7\times6\times3}\right)^{x+y}$

This expression is a power of a fraction. There is a general rule in exponents which states:

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

Using this rule, we will apply it to our original expression.

Given, $a = 13$, $b = 7\times6\times3$, and $n = x+y$, we can rewrite our expression as:

$\frac{13^{x+y}}{(7\times6\times3)^{x+y}}$

The solution to the question is:

$\frac{13^{x+y}}{(7\times6\times3)^{x+y}}$

Let's analyze the expression we are given:

$\left(\frac{2\times4\times6}{7\times8\times9}\right)^{3x}$

The expression is a power of a fraction. The rule for powers of a fraction is that each component of the fraction must be raised to the power separately. This can be expressed as:

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

Applying this rule to our expression, we have:

• The numerator inside the power: $2 \times 4 \times 6$
• The denominator inside the power: $7 \times 8 \times 9$

Therefore, raising each part to the power $3x$ gives us:

$\frac{(2\times4\times6)^{3x}}{(7\times8\times9)^{3x}}$

Thus, the simplified expression for the given equation is:

$\frac{\left(2\times4\times6\right)^{3x}}{\left(7\times8\times9\right)^{3x}}$

The solution to the question is: $\frac{\left(2\times4\times6\right)^{3x}}{\left(7\times8\times9\right)^{3x}}$

To solve this problem, our goal is to express the given quotient of powers in a simplified form using exponent laws.

• Step 1: Understand the original expression. We have $\frac{a^5 \times x^5}{7^5 \times b^5}$.
• Step 2: Recognize the structure. Notice that both the numerator and denominator are raised to the fifth power.
• Step 3: Apply the property of exponents for quotients and products, which states that $\left(\frac{m}{n}\right)^k = \frac{m^k}{n^k}$ and $(m \cdot n)^k = m^k \cdot n^k$.
• Step 4: Rewrite the expression as a single fraction raised to the power of 5. Since each term in the numerator and denominator is raised to the fifth power separately, we combine them under a single power:
• Numerator: $a^5 \times x^5 = (a \times x)^5$
• Denominator: $7^5 \times b^5 = (7 \times b)^5$
• Therefore, $\frac{a^5 \times x^5}{7^5 \times b^5} = \left(\frac{a \times x}{7 \times b}\right)^5$.

Thus, the expression can be written as: $\left(\frac{a \times x}{7 \times b}\right)^5$.

Now, comparing this with the answer choices provided:

• Choice 1: $\frac{(a \times x)^5}{7 \times b^5}$ - does not match, as it retains the separate powers incorrectly.
• Choice 2: $\left(\frac{a \times x}{7 \times b}\right)^5$ - matches perfectly as derived.
• Choice 3: $\frac{a \times x^5}{(7 \times b)^5}$ - incorrect form compared to derived structure.
• Choice 4: $7 \times \left(\frac{a \times x}{b}\right)^5$ - unrelated format, doesn't match.

The correct choice is therefore Choice 2. This matches our derived expression using the laws of exponents correctly.

The given expression is:
$\left(\frac{7\times11\times19}{3\times12\times15}\right)^{-4}$

To solve this expression, we need to apply the rules of exponents, specifically the rule for powers of a fraction. For any fraction$\left(\frac{a}{b}\right)^{-n}$, the expression is equivalent to$\left(\frac{b}{a}\right)^n$.
Therefore, negative exponents indicate that the fraction should be flipped and raised to the positive of that exponent.

Substitute the terms into this formula:
1. Flip the fraction: $\left(\frac{3\times12\times15}{7\times11\times19}\right)$
2. Raise both numerator and denominator to the power of 4:
Thus, we have:
$\left(\frac{3\times12\times15}{7\times11\times19}\right)^{4}$

Now evaluating each term individually:
- In the numerator:
- $3^4\times12^4\times15^4$
- In the denominator:
- $7^4\times11^4\times19^4$

Applying the negative exponent rule, each individual factor in both numerator and denominator should be inverted, altering the exponents to negative:
1. Numerator becomes: $3^{-4}\times12^{-4}\times15^{-4}$
2. Denominator becomes: $7^{-4}\times11^{-4}\times19^{-4}$

Rewriting the expression, we achieve:
$\frac{7^{-4}\times11^{-4}\times19^{-4}}{3^{-4}\times12^{-4}\times15^{-4}}$

This matches precisely the provided solution.

The solution to the question is:$\frac{7^{-4}\times11^{-4}\times19^{-4}}{3^{-4}\times12^{-4}\times15^{-4}}$

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:

$8^3\times1=8^3$

We are given the expression: $\left(\frac{1}{2\times3\times4}\right)^{-2}$. We need to simplify it using the rules of exponents.

• Step 1: Identify the base of the exponent.
  The base is $\frac{1}{2\times3\times4}$.
  

• Step 2: Apply the rule for negative exponents.
  For a fraction $\frac{1}{a}$ with a negative exponent, $\left( \frac{1}{a} \right)^{-n} = a^n$. Therefore, $\left(\frac{1}{2\times3\times4}\right)^{-2} = (2\times3\times4)^2$.
  

• Step 3: Expand the expression.
  $(2\times3\times4)^2 = 2^2 \times 3^2 \times 4^2$.

Thus, the simplified expression is: $2^2\times3^2\times4^2$

We begin with the expression: $\frac{1}{a^{-x}}$.
Our goal is to simplify this expression while converting any negative exponents into positive ones.

• Recall the rule for negative exponents: $a^{-n} = \frac{1}{a^n}$.
• Correspondingly, $\frac{1}{a^{-n}} = a^n$.
• Thus, in our expression $\frac{1}{a^{-x}}$, the negative exponent can be converted and flipped to the numerator by the rule: $\frac{1}{a^{-x}} = a^x$.

To determine which expression has the greatest value, we apply the exponent rules to simplify each choice:

• For $x^3 \times x^4$, using the product rule: $x^3 \times x^4 = x^{3+4} = x^7$.
• For $(x^3)^5$, using the power of a power rule: $(x^3)^5 = x^{3 \times 5} = x^{15}$.
• $x^{10}$ is already in its simplest form.
• For $\frac{x^9}{x^2}$, using the quotient rule: $\frac{x^9}{x^2} = x^{9-2} = x^7$.

To identify the greater value, we compare the exponents:

• $x^7$ from choices 1 and 4.
• $x^{15}$ from choice 2.
• $x^{10}$ from choice 3.

The expression with the largest exponent is $(x^3)^5$ or $x^{15}$.

Therefore, the expression with the greatest value is $(x^3)^5$.

We'll use the law of exponents for negative exponents, but in the opposite direction:

$\frac{1}{a^n} =a^{-n}$Let's apply this law to the problem:

$4^5-4^6\cdot\frac{1}{4}= 4^5-4^6\cdot4^{-1}$When we apply the above law to the second term from the left in the sum, and convert the fraction to a term with a negative exponent,

Next, we'll use the law of exponents for multiplying terms with identical bases:

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

$4^5-4^6\cdot4^{-1} =4^5-4^{6+(-1)}=4^5-4^{6-1}=4^5-4^{5}=0$When we apply the above law of exponents to the second term from the left in the expression we got in the last step, then we'll simplify the resulting expression,

Let's summarize the solution steps:

$4^5-4^6\cdot\frac{1}{4}= 4^5-4^6\cdot4^{-1} =4^5-4^{5}=0$

We got that the answer is 0,

Therefore the correct answer is answer A.

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