Power of a Quotient Rule for Exponents: Variables in the exponent of the power

To solve this problem, we apply the Power of a Quotient Rule for Exponents, which states that for any non-zero base $a$ and integers $m$ and $n$, the expression $\frac{a^m}{a^n} = a^{m-n}$. In this case, our base $a$ is 11.

Given the expression $\frac{11^{2a}}{11^5}$, let's simplify it using the rule:

• The numerator is $11^{2a}$.
• The denominator is $11^5$.

Applying the rule:

$\frac{11^{2a}}{11^5} = 11^{2a-5}$

Thus, the expression simplifies to $11^{2a-5}$.

So, the solution to the question is: $11^{2a-5}$

To solve the expression $\frac{2^a}{2^2}$, we will apply the Power of a Quotient Rule for Exponents, which states that when you divide two powers with the same base, you subtract the exponents.

Here are the steps:

• Identify the base: Both the numerator and the denominator have the same base, which is 2.

• Apply the quotient rule of exponents: $\frac{b^m}{b^n} = b^{m-n}$. By applying this rule:

  $\frac{2^a}{2^2} = 2^{a-2}$.

Thus, by utilizing the rule, we find that:

$\frac{2^a}{2^2} = 2^{a-2}$.

The solution to the question is: $2^{a-2}$

We need to simplify the expression $\frac{4^5}{4^x}$.

According to the rules of exponents, specifically the power of a quotient rule, when you divide like bases you subtract the exponents. The rule is written as:

• $\frac{a^m}{a^n} = a^{m-n}$

This means we take the exponent in the numerator and subtract the exponent in the denominator. Let's apply this rule to our expression:

$\frac{4^5}{4^x} = 4^{5-x}$

Hence, the simplified form of the expression is $4^{5-x}$.

The solution to the question is: $4^{5-x}$

To solve the expression $\frac{7^{5b}}{7^{2b}}$, we will use the Power of a Quotient Rule for Exponents, which states that $\frac{a^m}{a^n} = a^{m-n}$ when $a$ is a nonzero number. This rule allows us to simplify expressions where the bases are the same.

1. Identify the base and the exponents in the expression. Here, the base is 7, and the exponents are $5b$ and $2b$.

2. Apply the Power of a Quotient Rule:
$\frac{7^{5b}}{7^{2b}} = 7^{5b - 2b}$

3. Simplify the expression in the exponent:
Calculate $5b - 2b = 3b$.

4. Therefore, the expression simplifies to $7^{3b}$.

However, according to the given correct answer, we are asked to provide the intermediate expression as well – that is, before calculating the difference:
So, the solution as an intermediate step is:
$7^{5b - 2b}$

The explicit step-by-step answer provided in the question's solution matches our intermediate form.

The solution to the question is:

$7^{5b - 2b}$

We start with the expression: $\frac{9^x}{9^y}$.
We need to simplify this expression using the Power of a Quotient Rule for exponents, which states that $\frac{a^m}{a^n} = a^{m-n}$. Here, the base $a$ must be the same in both the numerator and the denominator, and we subtract the exponent of the denominator from the exponent of the numerator.

Applying this rule to our expression, we identify $a = 9$, $m = x$, and $n = y$. So we have:

• $a = 9$
• $m = x$
• $n = y$

Using the Power of a Quotient Rule, we therefore rewrite the expression as:

$a^{m-n} = 9^{x-y}$

Hence, the simplified expression of $\frac{9^x}{9^y}$ is $9^{x-y}$.

The solution to the question is: $9^{x-y}$

To solve the question, let's apply the Power of a Quotient Rule for Exponents. This rule states that when dividing like bases with exponents, you can subtract the exponent of the denominator from the exponent of the numerator. Mathematically, this is expressed as:

$\frac{b^m}{b^n} = b^{m-n}$

In our case, the base $b$ is 3, the exponent $m$ for the numerator is $2a$, and the exponent $n$ for the denominator is $a$. Thus, we can substitute these into the formula:

$\frac{3^{2a}}{3^a} = 3^{2a-a}$

Now, simplify the exponent:

$3^{2a-a} = 3^{a}$

Therefore, the expression simplifies to:

$3^a$

The solution to the question is: $3^a$

To solve the given expression $\frac{6^{4x}}{6^{x+1}}$, we must apply the Power of a Quotient Rule for Exponents. This rule states that $\frac{a^m}{a^n} = a^{m-n}$.

Using this rule, the given expression can be rewritten as follows:

• The numerator is $6^{4x}$.
• The denominator is $6^{x+1}$.

Apply the Power of a Quotient Rule:

$\frac{6^{4x}}{6^{x+1}} = 6^{4x - (x + 1)}$

We need to simplify the exponent by performing the subtraction $4x - (x + 1)$:

Step 1: Distribute the subtraction sign to the terms inside the parenthesis:

• $4x - x - 1$

Step 2: Combine like terms:

• $3x - 1$

The expression simplifies to:

$6^{3x-1}$

Therefore, the solution to the question is: $6^{3x-1}$.

To solve the problem $\frac{11^{5a}}{11^{a-4}}$, we need to use the Power of a Quotient Rule for exponents, which states that $\frac{b^m}{b^n} = b^{m-n}$.

Let's apply this rule to the given expression:

• The base is $11$, which is the same for both the numerator and the denominator.
• The exponent in the numerator is $5a$.
• The exponent in the denominator is $a - 4$.

According to the formula $\frac{b^m}{b^n} = b^{m-n}$, we can subtract the exponent in the denominator from the exponent in the numerator:

$5a - (a - 4) = 5a - a + 4$.

This simplifies to $4a + 4$.

Therefore, $\frac{11^{5a}}{11^{a-4}} = 11^{4a + 4}$.

The correct answer provided was $11^{4a-4}$.

Therefore, the final expression we arrived at using the Power of a Quotient Rule is: $11^{4a + 4}$.

I couldn't get to the shown answer.

To solve this problem, we need to simplify the given expression using the rules of exponents. The expression we have is:

$\frac{16^{x+4}}{16^3}$

According to the quotient rule for exponents, when dividing like bases, you subtract the exponent of the denominator from the exponent of the numerator. This is expressed as:

• $\frac{a^m}{a^n} = a^{m-n}$

In our problem, the base is 16, so we apply the quotient rule. We subtract the exponent 3 in the denominator from the exponent $x+4$ in the numerator:

$\frac{16^{x+4}}{16^3} = 16^{(x+4)-3}$

Simplify the exponent by performing the subtraction within the exponent:

$16^{(x+4)-3} = 16^{x+4-3} = 16^{x+1}$

Thus, the expression simplifies to:

$16^{x+1}$

The solution to the question is: $16^{x+1}$

Let's start by analyzing the given expression $\frac{20^{x+y}}{20^{a+y}}$.

We have a fraction with the same base number in both the numerator and the denominator.

According to the Power of a Quotient Rule for Exponents, for any non-zero number $b$ and integers $m$ and $n$, $\frac{b^m}{b^n} = b^{m-n}$.

This means we can subtract the exponents of the denominator from the exponents of the numerator. First, write down the exponents explicitly:

• Numerator: $x + y$
• Denominator: $a + y$

Next, apply the rule:

$\frac{20^{x+y}}{20^{a+y}} = 20^{(x+y)-(a+y)}$

Distribute the subtraction in the exponent:

$20^{x+y-a-y}$

Simplify the terms:

As the variable $y$ is present in both terms, they cancel each other out, resulting in:

$20^{x-a}$

This is the simplest form for the expression, and it matches the provided correct answer.

The solution to the question is: $20^{x-a}$

We start with the expression: $\frac{\left(4\times7\right)^{2x}}{\left(4\times7\right)^4}$.

According to the Power of a Quotient Rule for Exponents, which states that $\frac{a^m}{a^n} = a^{m-n}$, we can simplify the expression by subtracting the exponents.

The base here is $4 \times 7$, and it is common in both the numerator and the denominator.

Thus, using the exponent rule, we have:

• Exponent in the numerator: $2x$
• Exponent in the denominator: $4$

Now, apply the rule:

$\frac{\left(4\times7\right)^{2x}}{\left(4\times7\right)^4} = \left(4\times7\right)^{2x-4}$

The solution to the question is: $\left(4\times7\right)^{2x-4}$.

To solve the given expression $\frac{\left(12\times2\right)^5}{\left(2\times12\right)^{3y}}$, we need to apply the rule for the power of a quotient for exponents: $\frac{a^m}{a^n} = a^{m-n}$.

The expressions in both the numerator and the denominator have the same base $\left(12 \times 2\right)$. Therefore, the expression can be rewritten as:

• Base: $\left(12 \times 2\right)$
• Exponent in the numerator: $5$
• Exponent in the denominator: $3y$

Now, applying the quotient rule:

$\frac{\left(12\times2\right)^5}{\left(2\times12\right)^{3y}} = \left(12\times2\right)^{5-3y}$

The solution to the question is:

$\left(12\times2\right)^{5-3y}$

We're given the expression:

$\frac{\left(3\times14\right)^{a+1}}{\left(3\times14\right)^2}$

The problem requires us to simplify this expression using the power of a quotient rule for exponents. This rule states that:

• $\frac{x^m}{x^n} = x^{m-n}$

In our case, we identify:

• $x = 3\times14$
• $m = a+1$
• $n = 2$

Applying the power of a quotient rule, we get:

$\frac{\left(3\times14\right)^{a+1}}{\left(3\times14\right)^2} = \left(3\times14\right)^{a+1-2}$

Therefore, the solution to the question is:

$\left(3\times14\right)^{a-1}$

To solve the equation, you're required to simplify the expression $\frac{\left(9\times7\right)^{2x}}{\left(7\times9\right)^{2y}}$. This expression contains powers of quotients, and you can apply the properties of exponents to simplify it.

Let's go through the solution step by step:

• Both the numerator and the denominator are raised to some powers. Notice that the base of both the numerator and the denominator is $(9 \times 7)$, since $9 \times 7 = 63$ and $7 \times 9 = 63$.
• Therefore, you can rewrite the expression as $\frac{(63)^{2x}}{(63)^{2y}}$.
• According to the property of exponents, $\frac{a^m}{a^n} = a^{m-n}$, where $a$ is a non-zero number. Apply this rule:
• $\frac{(63)^{2x}}{(63)^{2y}} = (63)^{2x-2y}$

With the simplification completed, you get $\left(63\right)^{2x-2y}$.

Finally, substitute back $63$ for $9 \times 7$, and you can express the result as $\left(9 \times 7\right)^{2x-2y}$.

The solution to the question is: $\left(9\times7\right)^{2x-2y}$.

We are given the expression $\frac{(15 \times 6)^{ax}}{(15 \times 6)^{y+1}}$ and need to simplify it using the rules of exponents.

First, let's recall the rule: the "Power of a Quotient Rule for Exponents" which states that for any real number $a$ and any integers $m$ and $n$, $\frac{a^m}{a^n} = a^{m-n}$.

Applying this rule to our expression, we have the same base $(15 \times 6)$ in both the numerator and the denominator. Thus, we can subtract the exponent in the denominator from the exponent in the numerator.

The exponents in the numerator and the denominator are $ax$ and $y+1$ respectively. Therefore, we subtract the exponent $y+1$ from $ax$:

• Numerator's exponent: $ax$
• Denominator's exponent: $y+1$
• Exponent of the result: $ax - (y+1)$

Simplifying the exponent, we have:

$ax - (y+1) = ax - y - 1$

Therefore, the expression simplifies to:

$\left(15 \times 6\right)^{ax-y-1}$

The solution to the question is: $\left(15 \times 6\right)^{ax-y-1}$

To solve the problem $\frac{\left(3\times7\right)^{2x+5}}{\left(3\times7\right)^{x+3}}=$ , we need to apply the Power of a Quotient Rule for Exponents.

The Power of a Quotient Rule states that $\frac{a^m}{a^n} = a^{m-n}$ where $a$ is a nonzero number and $m$ and $n$ are integers. In this expression, $a$ will be equal to $(3 \times 7)$.

Start by writing the expression in a simplified form using the rule:

• The numerator is $\left(3 \times 7\right)^{2x + 5}$
• The denominator is $\left(3 \times 7\right)^{x + 3}$

Applying the quotient rule:

$\frac{\left(3 \times 7\right)^{2x+5}}{\left(3 \times 7\right)^{x+3}} = \left(3 \times 7\right)^{(2x+5) - (x+3)}$

Now we simplify the exponent:

• $(2x + 5) - (x + 3) = 2x + 5 - x - 3$
• Combine like terms: $2x - x + 5 - 3 = x + 2$

Thus, $\frac{\left(3 \times 7\right)^{2x+5}}{\left(3 \times 7\right)^{x+3}} = \left(3 \times 7\right)^{x+2}$.

The solution to the question is: $\left(3 \times 7\right)^{x+2}$

Let's start solving the expression:

$\frac{\left(2\times9\right)^{x+7}}{\left(9\times2\right)^{y+4}}=$

First, observe the base of the numerators and denominators. Both are essentially equal since $2\times9 = 9\times2 = 18$.

Thus, the expression can be written as:

$\frac{18^{x+7}}{18^{y+4}}$

According to the rule of exponents \/, where $\frac{a^m}{a^n} = a^{m-n}$, we can subtract the exponents in such a situation:

Therefore, our expression becomes:

$18^{(x+7)-(y+4)}$

Now simplify the exponent:

$x + 7 - y - 4 = x - y + 3$

Thus, the final simplified expression is:

$18^{x-y+3}$

Observe that $18 = 9 \times 2$, hence the expression can also be rewritten as:

$\left(9\times2\right)^{x-y+3}$

The solution to the question is: $\left(9\times2\right)^{x-y+3}$

Let's begin by examining the given expression: $\frac{\left(5\times8\right)^{3+5y}}{\left(8\times5\right)^{3y+1}}=$

Both the numerator and the denominator share the same base, $5 \times 8$, which can be expressed as $(40)$.

Next, we apply the quotient rule for exponents, which states that $\frac{a^m}{a^n} = a^{m-n}$, provided that $a \neq 0$.

We have:

• Numerator exponent: $(3 + 5y)$
• Denominator exponent: $(3y + 1)$

By applying the quotient rule, we can subtract the exponent in the denominator from the exponent in the numerator:

$(3 + 5y) - (3y + 1) = 3 + 5y - 3y - 1$

Simplifying the expression, we get:

• $3 - 1 = 2$
• $5y - 3y = 2y$

Combining these, we have:

$(40)^{2y + 2}$

Thus, the simplified form of the expression is:

$(5 \times 8)^{2y + 2}$

The solution to the question is: $(5 \times 8)^{2y + 2}$

To solve this problem, we need to simplify the given expression by applying the Power of a Quotient Rule for Exponents, which states that: $\frac{a^m}{a^n} = a^{m-n}$.

We are given the expression:

$\frac{(10 \times 4)^{6+ax}}{(4 \times 10)^{4+ax}}$

Notice that $10 \times 4$ and $4 \times 10$ are equivalent, therefore:

• $a = 10 \times 4$
• $m = 6 + ax$
• $n = 4 + ax$

Applying the quotient rule, we can write:

$\frac{a^m}{a^n} = a^{m-n}$

Inserting the values:

$a = (10 \times 4)$, $m = 6 + ax$, and $n = 4 + ax$, we obtain:

$a^{6+ax-(4+ax)}$

By simplifying the exponents:

• $(6 + ax) - (4 + ax) = 6 + ax - 4 - ax = 2$

Therefore, the expression becomes:

$(10 \times 4)^2$

The solution to the question is: $(10 \times 4)^2$

We are given the expression: $\frac{\left(8\times12\right)^{7+x+y}}{\left(8\times12\right)^{a+2}}$.
Our goal is to simplify this expression using the properties of exponents, particularly the quotient rule.

The quotient rule for exponents states that $\frac{a^m}{a^n} = a^{m-n}$, where both expressions have the same base.

In the expression $\frac{\left(8\times12\right)^{7+x+y}}{\left(8\times12\right)^{a+2}}$, the base is $8\times12$ and is common for both the numerator and the denominator.

Using the quotient rule, we subtract the exponents:

• Numerator exponent: $7+x+y$
• Denominator exponent: $a+2$