Powers of a Fraction: Variable in the exponent of the power

Let's determine the corresponding expression for $\left(\frac{2}{3}\right)^a$:

We apply the property of exponentiation for fractions, which states:
$\left(\frac{x}{y}\right)^n = \frac{x^n}{y^n}$.

Substituting $x = 2$, $y = 3$, and $n = a$, we have:

$\left(\frac{2}{3}\right)^a = \frac{2^a}{3^a}$.

Therefore, the correct expression is $\frac{2^a}{3^a}$.

Assessing the possible choices:

• Choice 1: $\frac{2}{3^a}$ - This is incorrect as it does not raise the numerator to $a$.
• Choice 2: $\frac{2a}{3a}$ - This is incorrect as it misuses the exponent rule.
• Choice 3: $\frac{2^a}{3^a}$ - This is correct, as it follows the exponentiation property.
• Choice 4: $\frac{2^a}{3}$ - This is incorrect as it does not raise the denominator to $a$.

Thus, the correct choice is Choice 3: $\frac{2^a}{3^a}$.

To solve this problem, we will apply the rules of exponents:

• Step 1: Identify the Given Expression
• Step 2: Apply the Exponent Rule

Now, let's work through these steps:

Step 1: We are given the expression $\left(\frac{3}{4}\right)^x$.

Step 2: According to the rule of exponents, when a fraction is raised to a power, this is equivalent to raising both the numerator and the denominator to that power. Therefore, we have:

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

Therefore, the expression $\left(\frac{3}{4}\right)^x$ is equivalent to $\frac{3^x}{4^x}$.

Thus, the correct answer is option 1, which is $\frac{3^x}{4^x}$.

The solution to the problem is $\frac{3^x}{4^x}$.

To solve this problem, we need to rewrite the expression $\left(\frac{11}{19}\right)^{a+3b}$ using the rules for powers of fractions. Specifically, we apply the exponent to both the numerator and the denominator separately.

According to the rule $\left(\frac{x}{y}\right)^n = \frac{x^n}{y^n}$, we apply the exponent $a+3b$ to both 11 and 19:

$\left(\frac{11}{19}\right)^{a+3b} = \frac{11^{a+3b}}{19^{a+3b}}$

Therefore, the expression can be rewritten as $\frac{11^{a+3b}}{19^{a+3b}}$, matching choice 3 in the provided options.

Hence, the solution to the problem is $\frac{11^{a+3b}}{19^{a+3b}}$.

To solve this problem, we'll apply the power of a fraction rule, which states that $\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$.

Step 1: Recognize the given expression: $\left(\frac{5}{9}\right)^{2x+1}$.

Step 2: Apply the exponent rule to rewrite the expression. According to the rule, this becomes:

$\left(\frac{5}{9}\right)^{2x+1} = \frac{5^{2x+1}}{9^{2x+1}}$.

Therefore, the expression can be rewritten as $\frac{5^{2x+1}}{9^{2x+1}}$.

Among the given choices, the correct option is choice 1: $\frac{5^{2x+1}}{9^{2x+1}}$.

Thus, the solution to the problem is $\frac{5^{2x+1}}{9^{2x+1}}$.

To solve the problem, follow these steps:

• Identify the given expression: $\left(\frac{5}{7}\right)^{ax}$.
• Apply the rule for exponentiation of a fraction: $\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$.
• Exponentiate both the numerator and the denominator separately by $ax$:
• Calculate $\frac{5^{ax}}{7^{ax}}$ which follows directly from the application of the property.

Therefore, the rewritten expression is $\frac{5^{ax}}{7^{ax}}$.

Among the given choices, the correct one is:

• Choice 4: $\frac{5^{ax}}{7^{ax}}$

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:

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

To solve this problem, we'll follow these steps:

• Step 1: Understand the expression and the exponent rules.
• Step 2: Apply the exponent to both the numerator and denominator.
• Step 3: Simplify each part of the expression by distributing the exponent correctly.

Now, let's work through each step:
Step 1: The original expression is $\left(\frac{2\times7\times21}{5\times6}\right)^x$. It needs to be rewritten by applying the exponent to each part of the fraction according to the rules of exponents.
Step 2: Using $\left(\frac{a}{b}\right)^x = \frac{a^x}{b^x}$, we apply the exponent $x$ to both the numerator and denominator:
$\left(\frac{2\times7\times21}{5\times6}\right)^x = \frac{(2\times7\times21)^x}{(5\times6)^x}$
Step 3: Distribute the exponent $x$ across each multiplication in the numerator and the denominator according to the rule $(a \times b)^x = a^x \times b^x$:
In the numerator: $(2 \times 7 \times 21)^x = 2^x \times 7^x \times 21^x$.
In the denominator: $(5 \times 6)^x = 5^x \times 6^x$.
Thus, the expression becomes:
$\frac{2^x \times 7^x \times 21^x}{5^x \times 6^x}$

Therefore, the solution to the problem is $\frac{2^x\times7^x\times21^x}{5^x\times6^x}$.

To solve this problem, we'll follow these steps:

• Step 1: Identify the given expression.
• Step 2: Apply the exponent rule $\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$ to distribute $x$ to numerator and denominator.
• Step 3: Simplify the denominator expression by distributing exponent $x$ to each factor.

Now, let's work through each step:
Step 1: We have the original expression $\left(\frac{3}{2 \times 5}\right)^x$.
Step 2: Apply the rule to get $\frac{3^x}{(2 \times 5)^x}$.
Step 3: Expand the denominator: $(2 \times 5)^x = 2^x \times 5^x$. This leads us to $\frac{3^x}{2^x \times 5^x}$.

Therefore, the solution to the problem is $\frac{3^x}{2^x \times 5^x}$.

To simplify the given expression, we start with the original expression:

$\left(\frac{3\times4}{5\times11\times9}\right)^y$.

Using the property for powers of a fraction, we distribute the exponent $y$ to the numerator and the denominator:

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

First, apply the formula:

$\left(\frac{3\times4}{5\times11\times9}\right)^y = \frac{(3\times4)^y}{(5\times11\times9)^y}$.

Next, apply the power of a product property, $(ab)^c = a^c \times b^c$, to both the numerator and the denominator:

The numerator becomes $(3\times4)^y = 3^y \times 4^y$.

The denominator becomes $(5\times11\times9)^y = 5^y \times 11^y \times 9^y$.

Thus, the fully simplified expression is:

$\frac{3^y\times4^y}{5^y\times11^y\times9^y}$.

After comparing with the given options, this matches choice 1 and 2, so option 4 is the right one: A+B are correct

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:

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

To solve this problem, we need to express $\left(\frac{2}{3 \times 5 \times 7}\right)^x$ by applying the rule for powers of a fraction.

Using the exponent rule $\left(\frac{a}{b}\right)^x = \frac{a^x}{b^x}$, we proceed as follows:

• Step 1: Express the numerator and denominator with the exponent $x$.
  The expression $\left(\frac{2}{3 \times 5 \times 7}\right)^x$ becomes $\frac{2^x}{(3 \times 5 \times 7)^x}$.
• Step 2: Apply the power of a product rule to the denominator.
  This results in $(3 \times 5 \times 7)^x = 3^x \times 5^x \times 7^x$.
• Step 3: Substitute back into the fraction from Step 1.
  We get $\frac{2^x}{3^x \times 5^x \times 7^x}$.

Therefore, the original expression $\left(\frac{2}{3 \times 5 \times 7}\right)^x$ simplifies to $\frac{2^x}{3^x \times 5^x \times 7^x}$.

The correct answer is: $\frac{2^x}{3^x \times 5^x \times 7^x}$.

To solve this problem, we'll apply the exponent rule for fractions and products.

• The given expression is $\left(\frac{2 \times 4 \times 5}{7}\right)^a$. We need to simplify this using exponent rules.
• First, apply the exponent to both the numerator $2 \times 4 \times 5$ and the denominator $7$:
  $\left(\frac{2 \times 4 \times 5}{7}\right)^a = \frac{(2 \times 4 \times 5)^a}{7^a}$.
• Now, apply the rule $(ab)^n = a^n \times b^n$ to distribute the exponent on the numerator:
• $(2 \times 4 \times 5)^a = 2^a \times 4^a \times 5^a$.
• Therefore, the expression simplifies to:
  $\frac{2^a \times 4^a \times 5^a}{7^a}$.

Therefore, the expression simplifies to $\frac{2^a \times 4^a \times 5^a}{7^a}$.

To solve the expression $\left(\frac{3\times7}{4\times8}\right)^{b+1}$, we follow these steps:

• Step 1: Apply the power $(b+1)$ to the entire fraction, $\left(\frac{3 \times 7}{4 \times 8}\right)^{b+1}$.
• Step 2: Use the exponentiation rule $(\frac{a}{b})^n = \frac{a^n}{b^n}$ to apply the power to both numerator and denominator separately.
• Step 3: Expand the numerator and the denominator: $(3 \times 7)^{b+1} = 3^{b+1} \times 7^{b+1}$ and $(4 \times 8)^{b+1} = 4^{b+1} \times 8^{b+1}$.
• Step 4: Combine these results into one fraction: $\frac{3^{b+1} \times 7^{b+1}}{4^{b+1} \times 8^{b+1}}$.

Therefore, the simplified expression is $\frac{3^{b+1}\times7^{b+1}}{4^{b+1}\times8^{b+1}}$.

Upon examining the choices, the correct option is choice 3: $\frac{3^{b+1}\times7^{b+1}}{4^{b+1}\times8^{b+1}}$.

To solve this problem, follow these steps:

• Identify the given expression: $\left(\frac{5 \times 11}{3 \times 7}\right)^a$.

• Apply the exponent rule for powers of a fraction: $\left(\frac{x}{y}\right)^a = \frac{x^a}{y^a}$.

• Apply this rule separately to the numerator and the denominator:

The numerator is $5 \times 11$ and the denominator is $3 \times 7$. When the fraction is raised to a power $a$, we apply the power to both the numerator and denominator:

$\left(\frac{5 \times 11}{3 \times 7}\right)^a = \frac{(5 \times 11)^a}{(3 \times 7)^a}$

Which corresponds to option 1.

Each product is raised to the power $a$. By exponent rules $(xy)^a = x^a \times y^a$, this expression becomes:

$\frac{5^a \times 11^a}{3^a \times 7^a}$

Thus, the expression can be rewritten as: $\frac{5^a \times 11^a}{3^a \times 7^a}$.

Referring to the provided choices, this matches choice 3.

Therefore, the correct choice is 4, A+C are correct.

To solve this problem, we'll use exponentiation properties to simplify the given expression:

• Step 1: Apply the exponent to each term in the numerator: $(5 \times 6 \times 7)^b = 5^b \times 6^b \times 7^b$.
• Step 2: Apply the exponent to each term in the denominator: $(9 \times 11 \times 13)^b = 9^b \times 11^b \times 13^b$.
• Step 3: Use the property of exponents for fractions: $\left(\frac{5 \times 6 \times 7}{9 \times 11 \times 13}\right)^b = \frac{5^b \times 6^b \times 7^b}{9^b \times 11^b \times 13^b}$.

As we have followed the rules of exponents and simplified accordingly, the corresponding expression is:

$\frac{5^b \times 6^b \times 7^b}{9^b \times 11^b \times 13^b}$.

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

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 solve the problem, we will leverage the rules of exponents:

• Apply the rule $\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$ to the expression $\left(\frac{2}{3\times7\times5}\right)^{a+2}$.
• Distribute the exponent $(a+2)$ to both the numerator and the denominator.

First, distribute the exponent to the numerator:

$2^{a+2}$

Now, distribute the exponent to the entire denominator:

$(3\times7\times5)^{a+2}$

Thus, the expression becomes:

$\frac{2^{a+2}}{(3\times7\times5)^{a+2}}$

This matches choice 3: $\frac{2^{a+2}}{\left(3\times7\times5\right)^{a+2}}$.

Let's solve the problem step-by-step:

We begin with the expression: $\left(\frac{5 \times 6 \times 7}{9}\right)^{2x+1}$.

Step 1: Apply the exponent to both the numerator and the denominator using the rule $\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$.

This gives: $\frac{(5 \times 6 \times 7)^{2x+1}}{9^{2x+1}}$, as in choice b.

Step 2: Distribute the exponent across each factor in the numerator: $(a \times b \times c)^n = a^n \times b^n \times c^n$.

This results in: $\frac{5^{2x+1} \times 6^{2x+1} \times 7^{2x+1}}{9^{2x+1}}$.

Therefore, the expression $\left(\frac{5\times6\times7}{9}\right)^{2x+1}$ evaluates to:

$\frac{5^{2x+1} \times 6^{2x+1} \times 7^{2x+1}}{9^{2x+1}}$.

This corresponds to choice 1. Hence, choices a' and b' are equivalent.

The correct answer is: a'+b' are correct.

First, let's apply the exponent to the entire fraction:

$\left(\frac{6}{11 \times 13 \times 15}\right)^{xy} = \frac{6^{xy}}{(11 \times 13 \times 15)^{xy}}$

Now, distribute the $xy$ exponent in the denominator to each factor:

$\frac{6^{xy}}{11^{xy} \times 13^{xy} \times 15^{xy}}$

Thus, the rewritten expression is $\frac{6^{xy}}{11^{xy} \times 13^{xy} \times 15^{xy}}$.

Comparing our expression with the options given and based on our simplification, option 3: $\frac{6^{xy}}{\left(11 \times 13 \times 15\right)^{xy}}$ makes sense, as well as option 2: $\frac{6^{xy}}{11^{xy}\times13^{xy}\times15^{xy}}$ after distributing the exponent within the denominator.

Therefore, both options B and C are correct, making the right choice option 4.

Therefore, the correct answer is: B+C are correct.