Power of a Quotient Rule for Exponents: Identify the greater value

To solve this problem, we need to simplify and compare the given expressions.

Let's simplify each:

• $y^7 \times y^2$:
  Using the product of powers rule, $y^7 \times y^2 = y^{7+2} = y^9$.
• $(y^4)^3$:
  Using the power of a power rule, $(y^4)^3 = y^{4 \times 3} = y^{12}$.
• $y^9$:
  This is already in its simplest form, $y^9$.
• $\frac{y^{11}}{y^4}$:
  Using the power of a quotient rule, $\frac{y^{11}}{y^4} = y^{11-4} = y^7$.

Now that all the expressions are in the form $y^n$, we can compare the exponents to see which is greatest: $y^9$, $y^{12}$, $y^9$, and $y^7$.

The expression with the highest power is $y^{12}$, which corresponds to the choice $(y^4)^3$.

Thus, the greater value among the choices is $(y^4)^3$.

To determine which value is greater, let's simplify each choice:

Choice 1: $(a^2)^4$
By using the power of a power rule: $(x^m)^n = x^{m \times n}$, it simplifies to:
$(a^2)^4 = a^{2 \times 4} = a^8$.

Choice 2: $a^2 + a^0$
Evaluate using the zero exponent rule, $a^0 = 1$:
This expression becomes $a^2 + 1$.

Choice 3: $a^2 \times a^1$
Apply the product of powers rule: $x^m \times x^n = x^{m+n}$:
This simplifies to $a^{2+1} = a^3$.

Choice 4: $\frac{a^{14}}{a^9}$
Apply the quotient of powers rule: $\frac{x^m}{x^n} = x^{m-n}$:
This simplifies to $a^{14-9} = a^5$.

Now, let's compare these simplified forms:
We have $a^8$, $a^2 + 1$, $a^3$, and $a^5$.

For $a > 1$, exponential functions grow rapidly, thus:
- $a^8$ is greater than $a^5$.
- $a^8$ is greater than $a^3$.
- $a^8$ is greater than $a^2 + 1$ for sufficiently large $a$.

Thus, the expression with the highest power, and therefore the greatest value, is $(a^2)^4$.

To determine which of the given expressions is the greatest, we will use the relevant exponent rules to simplify each one:

• Simplify $y^7 \times y^2$:
  Using the Product of Powers rule, we have $y^7 \times y^2 = y^{7+2} = y^9$.
• Simplify $(y^4)^3$:
  Using the Power of a Power rule, we have $(y^4)^3 = y^{4 \times 3} = y^{12}$.
• Simplify $y^9$:
  This expression is already simplified and is $y^9$.
• Simplify $\frac{y^{11}}{y^4}$:
  Using the Division of Powers rule, we have $\frac{y^{11}}{y^4} = y^{11-4} = y^7$.

After simplifying, we compare the powers of $y$ from each expression:

• $y^9$ from $y^7 \times y^2$
• $y^{12}$ from $(y^4)^3$
• $y^9$ from $y^9$
• $y^7$ from $\frac{y^{11}}{y^4}$

Clearly, $y^{12}$ is the largest power among the expressions, meaning that $(y^4)^3$ is the greatest value.

Therefore, the correct choice is $(y^4)^3$.

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 $(x^3)^5$.

Note that in all options there are fractions where both numerator and denominator have terms with identical bases, therefore we will use the division law between terms with identical bases to solve the exercise:

$\frac{c^m}{c^n}=c^{m-n}$Let's apply it to the problem, first we'll simplify each of the suggested options using the above law (options in order):

$\frac{a^{17}}{a^{20}}=a^{17-20}=a^{-3}$$\frac{a^{10}}{a^1}=a^{10-1}=a^9$$\frac{a^3}{a^{-2}}=a^{3-(-2)}=a^{3+2}=a^5$$\frac{a^2}{a^4}=a^{2-4}=a^{-2}$Let's return to the problem, given that:

a>1 therefore the option with the largest value will be the one where $a$has the largest exponent (for emphasis - a positive exponent is greater than a negative exponent),

meaning the option:

$a^9$above is correct, it came from option B in the answers,

therefore answer B is correct.

Note that in almost all options there are fractions where both the numerator and denominator have identical bases, therefore we will use the division law between terms with identical bases to solve the exercise:

$\frac{c^m}{c^n}=c^{m-n}$Let's apply this to the problem, first we'll simplify each of the given options using the above law (options in order):

$\frac{a^{12}}{a^{10}}=a^{12-10}=a^2$$\frac{a^4}{a^2}=a^{4-2}=a^2$$\frac{a^{-7}}{a^5}=a^{-7-5}=a^{-12}$$a^3$Back to the problem, given that:

a>1 therefore the option with the largest value will be the one where $a$has the largest exponent (for emphasis - a positive exponent is greater than a negative exponent),

meaning the option:

$a^3$above is correct,

therefore answer D is correct.

Notice that in almost all options there are fractions where both numerator and denominator have identical bases, therefore we will use the division law between terms with identical bases to solve the exercise:

$\frac{c^m}{c^n}=c^{m-n}$Let's apply this to the problem. First, let's simplify each of the given options using the above law (options in order):

$\frac{a^9}{a^8}=a^{9-8}=a^1$$\frac{a^{-2}}{a^{10}}=a^{-2-10}=a^{-12}$$a^2$$\frac{a^{20}}{a^{30}}=a^{20-30}=a^{-10}$Let's return to the problem, given that:

a>1 Therefore, the option with the largest value will be the one where $a$has the largest exponent (for emphasis - a positive exponent is greater than a negative exponent),

Which means the option:

$a^2$above is correct, it is option C,

Therefore answer C is correct.

Note that in all options there are fractions where both numerator and denominator have terms with identical bases, therefore we will use the division law between terms with identical bases to solve the exercise:

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

Let's apply it to the problem, first we'll simplify each of the suggested options using the above law (options in order):

$\frac{a^4}{a^{-4}}=a^{4-(-4)}=a^{4+4}=a^8$$\frac{a^{10}}{a^9}=a^{10-9}=a^1=a$$\frac{a^4}{a^{-1}}=a^{4-(-1)}=a^{4+1}=a^5$$\frac{a^6}{a^7}=a^{6-7}=a^{-1}$

where we also used the fact that any number to the power of 1 equals the number itself, meaning that:

$b^1=b$

Back to the problem, given that:

a>1

therefore the option with the largest value will be the one where $a$has the largest exponent (for emphasis - a positive exponent is greater than a negative exponent),

meaning the option:$a^8$above is correct, it came from option A in the answers,

therefore answer A is correct.